HC Verma ii Solutions for Class 12 Science Physics Chapter 34 Magnetic Field are provided here with simple step-by-step explanations. These solutions for Magnetic Field are extremely popular among class 12 Science students for Physics Magnetic Field Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the HC Verma ii Book of class 12 Science Physics Chapter 34 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s HC Verma ii Solutions. All HC Verma ii Solutions for class 12 Science Physics are prepared by experts and are 100% accurate.

#### Page No 228:

#### Question 1:

#### Answer:

Magnetic force becomes zero as the particle is at rest in this frame of reference and we know that force on a particle,

*F *= *qvB*sin($\theta $), where *v* is the velocity of the particle.

So, when *v *= 0,* F* = 0.

Magnetic field on the particles still exists because the current is independent of the frame of reference. For any reason, if the electrons of the wire seem to be at rest in a frame of reference, the protons are still flowing opposite to the frame of reference. Due to this, the current and the magnetic fields still exist.

#### Page No 229:

#### Question 2:

Magnetic force becomes zero as the particle is at rest in this frame of reference and we know that force on a particle,

*F *= *qvB*sin($\theta $), where *v* is the velocity of the particle.

So, when *v *= 0,* F* = 0.

Magnetic field on the particles still exists because the current is independent of the frame of reference. For any reason, if the electrons of the wire seem to be at rest in a frame of reference, the protons are still flowing opposite to the frame of reference. Due to this, the current and the magnetic fields still exist.

#### Answer:

Yes, a charged particle can be accelerated by a magnetic field. A magnetic field exerts force on the charged particle, which is perpendicular to both the magnetic field and velocity. If initially the charged particle is moving at right angle to the magnetic field, then the resultant trajectory of the particle is circular motion. In circular motion, the magnitude of the velocity remains constant but direction changes continuously. So, the motion is accelerated but speed remains constant.

#### Page No 229:

#### Question 3:

Yes, a charged particle can be accelerated by a magnetic field. A magnetic field exerts force on the charged particle, which is perpendicular to both the magnetic field and velocity. If initially the charged particle is moving at right angle to the magnetic field, then the resultant trajectory of the particle is circular motion. In circular motion, the magnitude of the velocity remains constant but direction changes continuously. So, the motion is accelerated but speed remains constant.

#### Answer:

No, it depend on the magnetic field, i.e. whether the field is a uniform or a non-uniform magnetic field and also on the orientation of the current loop. In case of a uniform magnetic field, the force on the circular loop is zero if the magnetic field is parallel to the plane of the loop and in case of a non-uniform magnetic field, the force may or may not be zero.

#### Page No 229:

#### Question 4:

No, it depend on the magnetic field, i.e. whether the field is a uniform or a non-uniform magnetic field and also on the orientation of the current loop. In case of a uniform magnetic field, the force on the circular loop is zero if the magnetic field is parallel to the plane of the loop and in case of a non-uniform magnetic field, the force may or may not be zero.

#### Answer:

As the electrons are in motion, there is a magnetic force acting on them individually. But the current through the wire represents the collective motion of all the electrons that are moving or vibrating very randomly; so, overall effect is negligible. Hence there is no net magnetic force on the wire.

Also, $F=ILB\mathrm{sin}\left(\theta \right)$

So, if the current in the wire is zero, then the force experienced by the wire will also be zero.

#### Page No 229:

#### Question 5:

As the electrons are in motion, there is a magnetic force acting on them individually. But the current through the wire represents the collective motion of all the electrons that are moving or vibrating very randomly; so, overall effect is negligible. Hence there is no net magnetic force on the wire.

Also, $F=ILB\mathrm{sin}\left(\theta \right)$

So, if the current in the wire is zero, then the force experienced by the wire will also be zero.

#### Answer:

Let us assume that the magnetic field is uniform and is acting along positive x axis in the cubical region.

Now if we project a charged particle inside this cube along positive y axis then as the direction of velocity and magnetic field is perpendicular to each other so the resultant trajectory of the particle will be a circle.

#### Page No 229:

#### Question 6:

Let us assume that the magnetic field is uniform and is acting along positive x axis in the cubical region.

Now if we project a charged particle inside this cube along positive y axis then as the direction of velocity and magnetic field is perpendicular to each other so the resultant trajectory of the particle will be a circle.

#### Answer:

As the particle gets deflected towards the positive y-axis, we can conclude that force is acting on the particle along the positive y-axis. Now, as the electron is moving along the positive x-axis, the current can be assumed to be flowing along the negative x-axis. Applying Fleming's left-hand rule, we find that the thumb points in the direction of force, i.e. the positive y-axis and the middle finger points in the direction of current, i.e. negative x-axis. Consequently, the forefinger gives us the direction of magnetic field, i.e. out of the plane of the paper or in the positive z-direction. So, we can conclude that the magnetic field is pointing along the positive z-axis.

#### Page No 229:

#### Question 7:

As the particle gets deflected towards the positive y-axis, we can conclude that force is acting on the particle along the positive y-axis. Now, as the electron is moving along the positive x-axis, the current can be assumed to be flowing along the negative x-axis. Applying Fleming's left-hand rule, we find that the thumb points in the direction of force, i.e. the positive y-axis and the middle finger points in the direction of current, i.e. negative x-axis. Consequently, the forefinger gives us the direction of magnetic field, i.e. out of the plane of the paper or in the positive z-direction. So, we can conclude that the magnetic field is pointing along the positive z-axis.

#### Answer:

Yes, if the direction of the area vector coincides with the direction of the magnetic field, the torque acting on the loop due to the magnetic field will become zero. Hence, no rotation will be produced in the coil.

It follows from the fact that torque acting on the loop is directly proportional to sin$\theta $, where $\theta $ is the angle made by the area vector with the direction of the magnetic field. So, we can see from this correlation that torque is zero if

$\theta $ = 0 or $\theta $ = 180^{0}.

$\tau =\stackrel{\rightharpoonup}{m}\times \stackrel{\rightharpoonup}{B}\phantom{\rule{0ex}{0ex}}=mB\mathrm{sin}\left(\theta \right)\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{For}\theta =0\mathrm{or}\mathrm{integral}\mathrm{multiple}\mathrm{of}\mathrm{\pi},\phantom{\rule{0ex}{0ex}}\tau =0\phantom{\rule{0ex}{0ex}}$

Which implies that the coil will not rotate.

#### Page No 229:

#### Question 8:

Yes, if the direction of the area vector coincides with the direction of the magnetic field, the torque acting on the loop due to the magnetic field will become zero. Hence, no rotation will be produced in the coil.

It follows from the fact that torque acting on the loop is directly proportional to sin$\theta $, where $\theta $ is the angle made by the area vector with the direction of the magnetic field. So, we can see from this correlation that torque is zero if

$\theta $ = 0 or $\theta $ = 180^{0}.

$\tau =\stackrel{\rightharpoonup}{m}\times \stackrel{\rightharpoonup}{B}\phantom{\rule{0ex}{0ex}}=mB\mathrm{sin}\left(\theta \right)\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{For}\theta =0\mathrm{or}\mathrm{integral}\mathrm{multiple}\mathrm{of}\mathrm{\pi},\phantom{\rule{0ex}{0ex}}\tau =0\phantom{\rule{0ex}{0ex}}$

Which implies that the coil will not rotate.

#### Answer:

The net charge in a current- carrying wire is zero. Yet, negative charge, i.e. electrons are moving in the wire towards the positive terminal. It is this motion of electrons in the conductor which produces the current in the wire and is also responsible for the magnetic force acting on the wire.*F = **qVB*sin($\theta $), where *F* is the force, *q* is the charge of electrons, *V *is the velocity of electrons and *B* is the magnetic field.

Moreover, the positive charges on the wire are due to nucleus containing proton. As they are not moving so there is no force on them, so the force is only due to the moving electrons in the wire.

#### Page No 229:

#### Question 9:

The net charge in a current- carrying wire is zero. Yet, negative charge, i.e. electrons are moving in the wire towards the positive terminal. It is this motion of electrons in the conductor which produces the current in the wire and is also responsible for the magnetic force acting on the wire.*F = **qVB*sin($\theta $), where *F* is the force, *q* is the charge of electrons, *V *is the velocity of electrons and *B* is the magnetic field.

Moreover, the positive charges on the wire are due to nucleus containing proton. As they are not moving so there is no force on them, so the force is only due to the moving electrons in the wire.

#### Answer:

If the angle between the positive normal and the magnetic field is 0, then the equilibrium is stable. It follows directly from the fact that *U* = - *mBc*os$\theta $, where *m* is the magnetic moment. So, when $\theta $ is 0, Potential energy, i.e. *U *of the system is negative, the system is more stable. But if $\theta $ is ${180}^{\xb0}$, *U* is positive or the system is unstable.

Stability of a system depends on its energy and every system tries to minimise its energy. The configuration of the system with least energy is most stable and the configuration with the most energy is least stable or unstable.

#### Page No 229:

#### Question 10:

If the angle between the positive normal and the magnetic field is 0, then the equilibrium is stable. It follows directly from the fact that *U* = - *mBc*os$\theta $, where *m* is the magnetic moment. So, when $\theta $ is 0, Potential energy, i.e. *U *of the system is negative, the system is more stable. But if $\theta $ is ${180}^{\xb0}$, *U* is positive or the system is unstable.

Stability of a system depends on its energy and every system tries to minimise its energy. The configuration of the system with least energy is most stable and the configuration with the most energy is least stable or unstable.

#### Answer:

Force experienced by the charge *q* moving with velocity v in a magnetic field *B* is given by*F = qVB*

Hence, *B* = $\frac{F}{qV}$

Also, weber/*m ^{â€‹2} *is the unit for magnetic field

*B.*

Now, equating both the units of the magnetic field

*B,*we get:

$\frac{Weber}{{m}^{2}}=\frac{F}{q\times V}\phantom{\rule{0ex}{0ex}}\Rightarrow Weber=\frac{F\times {m}^{2}}{q\times m\times {s}^{-1}}=\frac{F\times m}{q\times {s}^{-1}}=\frac{W}{q}\times s=Volt\times s$

Thus, the units weber and volt second are same.

#### Page No 229:

#### Question 1:

Force experienced by the charge *q* moving with velocity v in a magnetic field *B* is given by*F = qVB*

Hence, *B* = $\frac{F}{qV}$

Also, weber/*m ^{â€‹2} *is the unit for magnetic field

*B.*

Now, equating both the units of the magnetic field

*B,*we get:

$\frac{Weber}{{m}^{2}}=\frac{F}{q\times V}\phantom{\rule{0ex}{0ex}}\Rightarrow Weber=\frac{F\times {m}^{2}}{q\times m\times {s}^{-1}}=\frac{F\times m}{q\times {s}^{-1}}=\frac{W}{q}\times s=Volt\times s$

Thus, the units weber and volt second are same.

#### Answer:

(d) downwards

A positively-charged particle projected towards east can be considered as current in the eastern direction. Here, the positive charge is deflected towards the north by a magnetic field, i.e. the positively-charged particle experiences a force in the northern direction.

Hence, in order to determine the direction of the magnetic field, we apply Fleming's left-hand rule. According to this rule, when we stretch the thumb, the fore-finger and the middle finger mutually perpendicular to each other, then the thumb gives the direction of the force experienced by the charged particle, the fore-finger gives the direction of the magnetic field and the middle finger gives the direction of the current. Thus, if we direct the middle finger in the eastern direction, the thumb in the northern direction, we see that the fore-finger points in the downward direction.

Thus, the direction of the magnetic field is found to be in the downward direction.

#### Page No 229:

#### Question 2:

(d) downwards

A positively-charged particle projected towards east can be considered as current in the eastern direction. Here, the positive charge is deflected towards the north by a magnetic field, i.e. the positively-charged particle experiences a force in the northern direction.

Hence, in order to determine the direction of the magnetic field, we apply Fleming's left-hand rule. According to this rule, when we stretch the thumb, the fore-finger and the middle finger mutually perpendicular to each other, then the thumb gives the direction of the force experienced by the charged particle, the fore-finger gives the direction of the magnetic field and the middle finger gives the direction of the current. Thus, if we direct the middle finger in the eastern direction, the thumb in the northern direction, we see that the fore-finger points in the downward direction.

Thus, the direction of the magnetic field is found to be in the downward direction.

#### Answer:

When the charged particle is whirled in a horizontal circle, at any moment, the current direction can be taken along the tangent of the circle. Also, the magnetic field is in the vertical direction. So, using Fleming's left-hand rule, the force can be radially outward or inward, depending on the direction of the magnetic field, i.e. either upward or downward. Also, the direction of force depends on the direction of the whirl, i.e. clockwise or anticlockwise and obviously on the charge of the particle, i.e. whether it is positive or negative. So, the correct answer is that the tension may increase or decrease.

#### Page No 229:

#### Question 3:

When the charged particle is whirled in a horizontal circle, at any moment, the current direction can be taken along the tangent of the circle. Also, the magnetic field is in the vertical direction. So, using Fleming's left-hand rule, the force can be radially outward or inward, depending on the direction of the magnetic field, i.e. either upward or downward. Also, the direction of force depends on the direction of the whirl, i.e. clockwise or anticlockwise and obviously on the charge of the particle, i.e. whether it is positive or negative. So, the correct answer is that the tension may increase or decrease.

#### Answer:

(d) Li^{++}

Force on a moving charged particle,*F = **qVB*sin$\theta $

â€‹As velocity *V* and magnetic field *B *are constant, and angle between the magnetic field and charged particle is $90\xb0$, the only thing on which force* F* depends is charge *q*.

Now, the charge on $L{i}^{++}$ > charge on electron, proton or $H{e}^{+}$. So the force is maximum for $L{i}^{++}$.

#### Page No 229:

#### Question 4:

(d) Li^{++}

Force on a moving charged particle,*F = **qVB*sin$\theta $

â€‹As velocity *V* and magnetic field *B *are constant, and angle between the magnetic field and charged particle is $90\xb0$, the only thing on which force* F* depends is charge *q*.

Now, the charge on $L{i}^{++}$ > charge on electron, proton or $H{e}^{+}$. So the force is maximum for $L{i}^{++}$.

#### Answer:

(a) Electron

When a particle moves in a magnetic field, the necessary centripetal force, for the particle to move in a circle, is provided by the magnetic force acting on the particle.

So, equating the Lorentz force with the cetripetal force for a charged particle of charge *q *describing a circle of radius *r,* we get:

$qvB=\hspace{0.17em}\frac{m{v}^{2}}{r}\phantom{\rule{0ex}{0ex}}\Rightarrow r=\frac{mv}{qB}$

Here, *m* is the mass of the particle and *v* is its velocity.

The amount of charge of all the given particles is same; hence, for a given charge, $r\propto m$. And since the electron is the lightest of all the particles, it describes the smallest circle when projected with the same velocity perpendicular to the magnetic field.

#### Page No 229:

#### Question 5:

(a) Electron

When a particle moves in a magnetic field, the necessary centripetal force, for the particle to move in a circle, is provided by the magnetic force acting on the particle.

So, equating the Lorentz force with the cetripetal force for a charged particle of charge *q *describing a circle of radius *r,* we get:

$qvB=\hspace{0.17em}\frac{m{v}^{2}}{r}\phantom{\rule{0ex}{0ex}}\Rightarrow r=\frac{mv}{qB}$

Here, *m* is the mass of the particle and *v* is its velocity.

The amount of charge of all the given particles is same; hence, for a given charge, $r\propto m$. And since the electron is the lightest of all the particles, it describes the smallest circle when projected with the same velocity perpendicular to the magnetic field.

#### Answer:

(d) Li^{+}

Time period of the revolution of the particle, $T=\frac{2\mathrm{\pi}m}{qB}$

As frequency is the reciprocal of time period, so $f=\frac{qB}{2\mathrm{\pi}m}$

The charge on all the four particles is same. But the mass is maximum for Li^{+}. So, it will have the smallest frequency of revolution.

#### Page No 229:

#### Question 6:

(d) Li^{+}

Time period of the revolution of the particle, $T=\frac{2\mathrm{\pi}m}{qB}$

As frequency is the reciprocal of time period, so $f=\frac{qB}{2\mathrm{\pi}m}$

The charge on all the four particles is same. But the mass is maximum for Li^{+}. So, it will have the smallest frequency of revolution.

#### Answer:

(a) zero

When a circular loop is placed in a uniform magnetic field, it always experiences zero toque. We all know that a current-carrying wire experiences a force when placed in an external magnetic field. But in the case of a circular loop, forces are present in pairs, i.e. they are equal and opposite in magnitude. So, for every point on the loop, there exists another point on the diametrically opposite edge for which the force is equal and opposite to the force acting on first point. So, these two forces cancel in pair. In this way, the net torque on the loop is always zero when placed in a uniform magnetic field.

#### Page No 229:

#### Question 7:

(a) zero

When a circular loop is placed in a uniform magnetic field, it always experiences zero toque. We all know that a current-carrying wire experiences a force when placed in an external magnetic field. But in the case of a circular loop, forces are present in pairs, i.e. they are equal and opposite in magnitude. So, for every point on the loop, there exists another point on the diametrically opposite edge for which the force is equal and opposite to the force acting on first point. So, these two forces cancel in pair. In this way, the net torque on the loop is always zero when placed in a uniform magnetic field.

#### Answer:

(c) will be deviated by different angles and hence will separate

Force on a charged particle, *F = qVB*

For an electron, this force is *F = $-$eVB*, whereas on a proton this force is *F* =* eVB.*

Here, '*e*' is the charge of the electron. So, from the above formulas, we can see that electrons and protons will experience equal force but in opposite directions; so they separate out. In other words, we can say that they are deviated by different angles and causes them to separate.

#### Page No 229:

#### Question 8:

(c) will be deviated by different angles and hence will separate

Force on a charged particle, *F = qVB*

For an electron, this force is *F = $-$eVB*, whereas on a proton this force is *F* =* eVB.*

Here, '*e*' is the charge of the electron. So, from the above formulas, we can see that electrons and protons will experience equal force but in opposite directions; so they separate out. In other words, we can say that they are deviated by different angles and causes them to separate.

#### Answer:

d) helix with uniform pitch

From the figure, the velocity of the particle can be resolved in two components,Vcos$\theta $ parallel to the magnetic field and Vsin $\theta $ perpendicular to the magnetic field. We know that magnetic field does not change the speed of a particle; rather, it changes the direction of its velocity. So, a magnetic force acts on the particle due to the vertical component of velocity, which tries to move the particle in a circle.This force tries to rotate the particle in a circle. But as there is a horizontal component of velocity also, the particle will move helically with a constant pitch because no force acts on the particle along the direction of the horizontal component of velocity.

#### Page No 229:

#### Question 9:

d) helix with uniform pitch

From the figure, the velocity of the particle can be resolved in two components,Vcos$\theta $ parallel to the magnetic field and Vsin $\theta $ perpendicular to the magnetic field. We know that magnetic field does not change the speed of a particle; rather, it changes the direction of its velocity. So, a magnetic force acts on the particle due to the vertical component of velocity, which tries to move the particle in a circle.This force tries to rotate the particle in a circle. But as there is a horizontal component of velocity also, the particle will move helically with a constant pitch because no force acts on the particle along the direction of the horizontal component of velocity.

#### Answer:

(d) a helix with non-uniform pitch

Here, the total Lorentz force on the particle,*F = qE + qVB *

We all know that magnetic field B does not change the speed of the particle but changes its direction. But as an electric field is also present that accelerate the particle in the direction of the field, the resultant path is a helix with a non-uniform pitch.

#### Page No 229:

#### Question 10:

(d) a helix with non-uniform pitch

Here, the total Lorentz force on the particle,*F = qE + qVB *

We all know that magnetic field B does not change the speed of the particle but changes its direction. But as an electric field is also present that accelerate the particle in the direction of the field, the resultant path is a helix with a non-uniform pitch.

#### Answer:

(d) zero

We can use the right-hand thumb rule to get the direction of magnetic field due to the current-carrying wire. Based on this, it can be determined that the direction of magnetic field is along the axis of the wire. Also, the charged particle is moving along the axis. So, no magnetic force will act on it, as the angle between the magnetic field and the velocity of the charged particle may be ${0}^{\xb0}$^{}or ${180}^{\xb0}$. So, sin $\theta $ of the angles between velocity and magnetic field is zero.

Also, the force, *F=qVB* sin $\theta $.

So, the force on the charged particle is zero.

#### Page No 229:

#### Question 1:

(d) zero

We can use the right-hand thumb rule to get the direction of magnetic field due to the current-carrying wire. Based on this, it can be determined that the direction of magnetic field is along the axis of the wire. Also, the charged particle is moving along the axis. So, no magnetic force will act on it, as the angle between the magnetic field and the velocity of the charged particle may be ${0}^{\xb0}$^{}or ${180}^{\xb0}$. So, sin $\theta $ of the angles between velocity and magnetic field is zero.

Also, the force, *F=qVB* sin $\theta $.

So, the force on the charged particle is zero.

#### Answer:

(a) the electric field must be zero

(d) the magnetic field may or may not be zero

As the charged particle is at rest, its velocity, V = 0 and magnetic force, F = qVB = 0. Hence, we cannot determine whether a magnetic field is present or not.

But as the particle at rest experiences no electromagnetic force, the electric field must be zero. This is because electric force acts on a particle whether it is at rest or in motion.

#### Page No 230:

#### Question 2:

(a) the electric field must be zero

(d) the magnetic field may or may not be zero

As the charged particle is at rest, its velocity, V = 0 and magnetic force, F = qVB = 0. Hence, we cannot determine whether a magnetic field is present or not.

But as the particle at rest experiences no electromagnetic force, the electric field must be zero. This is because electric force acts on a particle whether it is at rest or in motion.

#### Answer:

(a) the electric field must not be zero

(d) the magnetic field may or may not be zero

As the charged particle is at rest, its velocity, V = 0 and magnetic force, F = qVB = 0. Hence, we cannot determine whether a magnetic field is present or not. But as the particle at rest experiences an electromagnetic force, the electric field must be non-zero. As electric force acts on a particle, whether it is at rest or in motion, an electric force must be present.

#### Page No 230:

#### Question 3:

(a) the electric field must not be zero

(d) the magnetic field may or may not be zero

As the charged particle is at rest, its velocity, V = 0 and magnetic force, F = qVB = 0. Hence, we cannot determine whether a magnetic field is present or not. But as the particle at rest experiences an electromagnetic force, the electric field must be non-zero. As electric force acts on a particle, whether it is at rest or in motion, an electric force must be present.

#### Answer:

(c) both fields cannot be zero

(d) both fields can be nonzero

As the particle gets deflected, a force acts on the particle. So, either it has got deflected due to the magnetic force or electric force; so, both the fields cannot be zero. Also, the particle can be deflected under the combined effect of magnetic and electric forces; so, both fields can be non-zero.

#### Page No 230:

#### Question 4:

(c) both fields cannot be zero

(d) both fields can be nonzero

As the particle gets deflected, a force acts on the particle. So, either it has got deflected due to the magnetic force or electric force; so, both the fields cannot be zero. Also, the particle can be deflected under the combined effect of magnetic and electric forces; so, both fields can be non-zero.

#### Answer:

(a) *E* = 0, *B* = 0

(b) *E* = 0, *B* ≠ 0

(d) *E* ≠ 0, *B* ≠ 0

A charged particle can move in a gravity-free space without any change in velocity in the following three ways:

(1) *E = *0, *B = *0, i.e. no force is acting on the particle and hence, it moves with a constant velocity.

(2) *E *= 0, *B *≠â€‹ 0. If magnetic field is along the direction of the velocity *v,* then the force acting on the charged particle will be zero, as *F = q v $\times $ B = *0. Hence, the particle will not accelerate.

(3) If the force due to magnetic field and the force due to electric field counterbalance each other, then the net force acting on the particle will be zero and hence, the particle will move with a constant velocity.

#### Page No 230:

#### Question 5:

(a) *E* = 0, *B* = 0

(b) *E* = 0, *B* ≠ 0

(d) *E* ≠ 0, *B* ≠ 0

A charged particle can move in a gravity-free space without any change in velocity in the following three ways:

(1) *E = *0, *B = *0, i.e. no force is acting on the particle and hence, it moves with a constant velocity.

(2) *E *= 0, *B *≠â€‹ 0. If magnetic field is along the direction of the velocity *v,* then the force acting on the charged particle will be zero, as *F = q v $\times $ B = *0. Hence, the particle will not accelerate.

(3) If the force due to magnetic field and the force due to electric field counterbalance each other, then the net force acting on the particle will be zero and hence, the particle will move with a constant velocity.

#### Answer:

(b) *E* = 0, *B* ≠ 0

The electric field exerts a force *q E* on the charged particle, which always accelerates (increases the speed) the particle. The particle can never be rotated in a circle by the electric field because then the radius of the orbit will keep on increasing due to the acceleration, which is not possible. So, options (c) and (d) are incorrect. On the other hand, a magnetic field does not change the magnitude of the velocity but changes only the direction of the velocity. Since the particle is moving in a circle, where its speed remains constant and only the direction of velocity changes, so it can only be achieved if E = 0 and *B* ≠ 0.

#### Page No 230:

#### Question 6:

(b) *E* = 0, *B* ≠ 0

The electric field exerts a force *q E* on the charged particle, which always accelerates (increases the speed) the particle. The particle can never be rotated in a circle by the electric field because then the radius of the orbit will keep on increasing due to the acceleration, which is not possible. So, options (c) and (d) are incorrect. On the other hand, a magnetic field does not change the magnitude of the velocity but changes only the direction of the velocity. Since the particle is moving in a circle, where its speed remains constant and only the direction of velocity changes, so it can only be achieved if E = 0 and *B* ≠ 0.

#### Answer:

(a) $\overrightarrow{E}\left|\right|\overrightarrow{B}$, $\overrightarrow{v}\left|\right|\overrightarrow{E}$

(b) $\overrightarrow{E}$ is not parallel to $\overrightarrow{B}$

In option (a) velocity, electric field and magnetic field are parallel to each other. So, the particle may accelerate but always continue to travel in the same straight path or go undeflected.

Another possibility of the particle to go undeviated is that magnetic force acting on it is counterbalanced by electric force. This is possible if all the three, i.e. velocity, magnetic field and electric field are perpendicular to each other, so that magnetic force is balanced by electric force. So option (b) can also be one possibility. But (c) and (d) are wrong statements.

#### Page No 230:

#### Question 7:

(a) $\overrightarrow{E}\left|\right|\overrightarrow{B}$, $\overrightarrow{v}\left|\right|\overrightarrow{E}$

(b) $\overrightarrow{E}$ is not parallel to $\overrightarrow{B}$

In option (a) velocity, electric field and magnetic field are parallel to each other. So, the particle may accelerate but always continue to travel in the same straight path or go undeflected.

Another possibility of the particle to go undeviated is that magnetic force acting on it is counterbalanced by electric force. This is possible if all the three, i.e. velocity, magnetic field and electric field are perpendicular to each other, so that magnetic force is balanced by electric force. So option (b) can also be one possibility. But (c) and (d) are wrong statements.

#### Answer:

(a) $\underset{E}{\to}$ must be perpendicular to $\underset{B}{\to}$

(b) $\underset{v}{\to}$ must be perpendicular to $\underset{E}{\to}$

As the charged particle is not accelerated, the field $\underset{E}{\to}$ cannot be parallel to velocity $\underset{v}{\to}$. Hence, the velocity $\underset{v}{\to}$ is perpendicular to the electric field $\underset{E}{\to}$.

The magnetic force, i.e. force due to the magnetic field, acts in a direction perpendicular to the plane containing $\underset{v}{\to}$ and $\underset{B}{\to}$. Hence, in order to counterbalance the magnetic force, an equal and opposite electric force must be applied along the same axis in which the magnetic force is acting. Hence $\underset{E}{\to}$ must be perpendicular to $\underset{v}{\to}$ and $\underset{B}{\to}$.

#### Page No 230:

#### Question 8:

(a) $\underset{E}{\to}$ must be perpendicular to $\underset{B}{\to}$

(b) $\underset{v}{\to}$ must be perpendicular to $\underset{E}{\to}$

As the charged particle is not accelerated, the field $\underset{E}{\to}$ cannot be parallel to velocity $\underset{v}{\to}$. Hence, the velocity $\underset{v}{\to}$ is perpendicular to the electric field $\underset{E}{\to}$.

The magnetic force, i.e. force due to the magnetic field, acts in a direction perpendicular to the plane containing $\underset{v}{\to}$ and $\underset{B}{\to}$. Hence, in order to counterbalance the magnetic force, an equal and opposite electric force must be applied along the same axis in which the magnetic force is acting. Hence $\underset{E}{\to}$ must be perpendicular to $\underset{v}{\to}$ and $\underset{B}{\to}$.

#### Answer:

(b) The circle described by the singly-ionised charge will have a radius that is double that of the other circle.

(d) The two circles touch each other.

The radius of the orbit of a charged particle in an external magnetic field,

$r=\frac{mV}{qB}$,

where *r* is the radius of the circle, *m* is the mass of the ion, *V* is the velocity with which the ion is projected, *q* is the charge on the ion and *B* is the uniform magnetic field.

Since the mass *m*, the velocity *V* and the magnetic field *B* are same for both the ions, *r* is inversely proportional to the charge on the ion.

Hence, the radius of the circle described by the singly-charged ion will be twice the radius of the circle described by doubly-ionised ion.

Moreover, as both the charges are projected from the same place, the two circles described by them will touch each other at the point of projection.

#### Page No 230:

#### Question 9:

(b) The circle described by the singly-ionised charge will have a radius that is double that of the other circle.

(d) The two circles touch each other.

The radius of the orbit of a charged particle in an external magnetic field,

$r=\frac{mV}{qB}$,

where *r* is the radius of the circle, *m* is the mass of the ion, *V* is the velocity with which the ion is projected, *q* is the charge on the ion and *B* is the uniform magnetic field.

Since the mass *m*, the velocity *V* and the magnetic field *B* are same for both the ions, *r* is inversely proportional to the charge on the ion.

Hence, the radius of the circle described by the singly-charged ion will be twice the radius of the circle described by doubly-ionised ion.

Moreover, as both the charges are projected from the same place, the two circles described by them will touch each other at the point of projection.

#### Answer:

(a) *y*-axis

(b) *z*-axis

Any magnetic field, except one parallel to the direction of velocity can change the direction of the particle. Therefore, either the magnetic field along y-axis or along z-axis can reverse the direction of the particle, as the velocity is along the *x* direction

#### Page No 230:

#### Question 10:

(a) *y*-axis

(b) *z*-axis

Any magnetic field, except one parallel to the direction of velocity can change the direction of the particle. Therefore, either the magnetic field along y-axis or along z-axis can reverse the direction of the particle, as the velocity is along the *x* direction

#### Answer:

(b) *E _{y}^{' }= E_{y} - $\frac{v{B}_{z}}{{c}^{2}}$*

(c) ${B}_{\mathrm{y}}^{\text{'}}={B}_{\mathrm{y}}+v{E}_{\mathrm{z}}$

Electric force due to a charged particle is

*q E*and magnetic force is

*q V B.*

We can sort out the two wrong equations using dimensional analysis. Now, equating the above two forces. we get:

*E = V B*

Hence, analysing the answers using dimensional analysis, we see that the second term on the RHS of the equations (b) and (c) are not dimensionally correct. Thus, the options (b) and (c) are wrong.

#### Page No 230:

#### Question 1:

(b) *E _{y}^{' }= E_{y} - $\frac{v{B}_{z}}{{c}^{2}}$*

(c) ${B}_{\mathrm{y}}^{\text{'}}={B}_{\mathrm{y}}+v{E}_{\mathrm{z}}$

Electric force due to a charged particle is

*q E*and magnetic force is

*q V B.*

We can sort out the two wrong equations using dimensional analysis. Now, equating the above two forces. we get:

*E = V B*

Hence, analysing the answers using dimensional analysis, we see that the second term on the RHS of the equations (b) and (c) are not dimensionally correct. Thus, the options (b) and (c) are wrong.

#### Answer:

Given:

Upward speed of the alpha particle, *v* = 3 × 10^{4} km/s = 3 × 10^{7} m/s

Magnetic field, *B =* 1.0 T

The direction of the magnetic field is from south to north.

Charge of the alpha particle, *q* = 2*e**,*

where *e* is the charge of an electron.*q *= 2 × 1.6 × 10^{−19} C,

Magnetic force acting on the α-particle,

$\overrightarrow{F}=q\overrightarrow{v}\times \overrightarrow{B}\mathrm{sin}90\xb0$

= 2 × 1.6 × 10^{−19} × 3 × 10^{7} × 1

= 9.6 × 10^{−12} N, towards west

The direction of magnetic force can be found using Fleming's left-hand rule.

#### Page No 230:

#### Question 2:

Given:

Upward speed of the alpha particle, *v* = 3 × 10^{4} km/s = 3 × 10^{7} m/s

Magnetic field, *B =* 1.0 T

The direction of the magnetic field is from south to north.

Charge of the alpha particle, *q* = 2*e**,*

where *e* is the charge of an electron.*q *= 2 × 1.6 × 10^{−19} C,

Magnetic force acting on the α-particle,

$\overrightarrow{F}=q\overrightarrow{v}\times \overrightarrow{B}\mathrm{sin}90\xb0$

= 2 × 1.6 × 10^{−19} × 3 × 10^{7} × 1

= 9.6 × 10^{−12} N, towards west

The direction of magnetic force can be found using Fleming's left-hand rule.

#### Answer:

Given:

The kinetic energy of the electron projected in the horizontal direction, *K.E* = 10 keV = 1.6 × 10^{−15} J

Magnetic field, *B* = 1 × 10^{−7} T

The direction of magnetic field is vertically upward.

(a) The direction can be found by the right-hand screw rule. So, the electron will be deflected towards left.

(b) Kinetic energy,

$\mathrm{K}.\mathrm{E}.=\frac{1}{2}m{v}^{2}\phantom{\rule{0ex}{0ex}}v=\sqrt{\frac{K.E.\times 2}{m}}$

Magnetic force,

$\overrightarrow{F}=q\overrightarrow{v}\times \overrightarrow{B}$ ....(i)

and *F* = *ma*

Therefore, equation (i) will be

$a=\frac{q\overrightarrow{v}\times \overrightarrow{B}}{m}$

Applying equation of motion*s* = *ut* + $\frac{1}{2}$*at*^{2},

t= Time taken to cross the magnetic field

As there is no force acting on the electron in the horizontal direction, the velocity of the electron remain constant in this direction.

So, the time taken to cross a distance of 1m in the horizontal direction in the magnetic field,

$t=\frac{d}{{v}_{\mathrm{horizontal}}}\phantom{\rule{0ex}{0ex}}=\frac{1}{\sqrt{{\displaystyle \frac{2\times K.E}{m}}}}$

Putting the respective value of the variables, we get:

$y=\frac{1}{2}\frac{qvB}{m}{t}^{2}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\frac{qvB}{m}\frac{1}{{\displaystyle \frac{2\times K.E}{m}}}\phantom{\rule{0ex}{0ex}}=\frac{qvB}{4\times K.E}\phantom{\rule{0ex}{0ex}}=\frac{1}{4}\frac{1.6\times {10}^{-19}\times 1\times {10}^{-7}\times \sqrt{{\displaystyle \frac{2\times \mathrm{K}.\mathrm{E}}{\mathrm{m}}}}}{10\times {10}^{-16}}\phantom{\rule{0ex}{0ex}}=0.148\times {10}^{-3}m\phantom{\rule{0ex}{0ex}}=0.01482\mathrm{cm}$*s* = 0.0148 = 1.5 × 10^{−2} cm

#### Page No 230:

#### Question 3:

Given:

The kinetic energy of the electron projected in the horizontal direction, *K.E* = 10 keV = 1.6 × 10^{−15} J

Magnetic field, *B* = 1 × 10^{−7} T

The direction of magnetic field is vertically upward.

(a) The direction can be found by the right-hand screw rule. So, the electron will be deflected towards left.

(b) Kinetic energy,

$\mathrm{K}.\mathrm{E}.=\frac{1}{2}m{v}^{2}\phantom{\rule{0ex}{0ex}}v=\sqrt{\frac{K.E.\times 2}{m}}$

Magnetic force,

$\overrightarrow{F}=q\overrightarrow{v}\times \overrightarrow{B}$ ....(i)

and *F* = *ma*

Therefore, equation (i) will be

$a=\frac{q\overrightarrow{v}\times \overrightarrow{B}}{m}$

Applying equation of motion*s* = *ut* + $\frac{1}{2}$*at*^{2},

t= Time taken to cross the magnetic field

As there is no force acting on the electron in the horizontal direction, the velocity of the electron remain constant in this direction.

So, the time taken to cross a distance of 1m in the horizontal direction in the magnetic field,

$t=\frac{d}{{v}_{\mathrm{horizontal}}}\phantom{\rule{0ex}{0ex}}=\frac{1}{\sqrt{{\displaystyle \frac{2\times K.E}{m}}}}$

Putting the respective value of the variables, we get:

$y=\frac{1}{2}\frac{qvB}{m}{t}^{2}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\frac{qvB}{m}\frac{1}{{\displaystyle \frac{2\times K.E}{m}}}\phantom{\rule{0ex}{0ex}}=\frac{qvB}{4\times K.E}\phantom{\rule{0ex}{0ex}}=\frac{1}{4}\frac{1.6\times {10}^{-19}\times 1\times {10}^{-7}\times \sqrt{{\displaystyle \frac{2\times \mathrm{K}.\mathrm{E}}{\mathrm{m}}}}}{10\times {10}^{-16}}\phantom{\rule{0ex}{0ex}}=0.148\times {10}^{-3}m\phantom{\rule{0ex}{0ex}}=0.01482\mathrm{cm}$*s* = 0.0148 = 1.5 × 10^{−2} cm

#### Answer:

Given:

Magnetic field, *B* = (4 × 10^{−3}$\hat{k}$)T

Force exerted by the magnetic field on the charged particle, *F* = (4 $\hat{i}$ + 3 $\hat{j}$) × 10^{−10} N

Charge of the particle, *q* = 1 × 10^{−9} C

As per the question, the charge is going in the X-Y plane.

So, the x-component of force, *F*_{x} = 4 × 10^{−10} N

and the y-component of force, *F*_{y} = 3 × 10^{−10} N

Considering the motion along x-axis:*F*_{x }= *qv*_{y}×*B*

On putting the respective values, we get:

*v*_{y} = 100 m/s

Motion along y-axis:*F*_{y}_{}= *qv*_{x}×*B*

⇒ *v*_{x} = 75 m/s

Thus, total velocity = (−75 $\overrightarrow{i}$ + 100 $\overrightarrow{j}$) m/s

#### Page No 230:

#### Question 4:

Given:

Magnetic field, *B* = (4 × 10^{−3}$\hat{k}$)T

Force exerted by the magnetic field on the charged particle, *F* = (4 $\hat{i}$ + 3 $\hat{j}$) × 10^{−10} N

Charge of the particle, *q* = 1 × 10^{−9} C

As per the question, the charge is going in the X-Y plane.

So, the x-component of force, *F*_{x} = 4 × 10^{−10} N

and the y-component of force, *F*_{y} = 3 × 10^{−10} N

Considering the motion along x-axis:*F*_{x }= *qv*_{y}×*B*

On putting the respective values, we get:

*v*_{y} = 100 m/s

Motion along y-axis:*F*_{y}_{}= *qv*_{x}×*B*

⇒ *v*_{x} = 75 m/s

Thus, total velocity = (−75 $\overrightarrow{i}$ + 100 $\overrightarrow{j}$) m/s

#### Answer:

Given,

Magnetic field, *B* = (7.0*i* − 3.0*j*) × 10^{−3} T

Acceleration of the particle, *a* = (*xi* + 7*j*) × 10^{−6} m/s^{2}

We have denoted the unidentified number as *x*.*B* and *a* are perpendicular to each other. (Because magnetic force always acts perpendicular to the motion of the particle)

So, the dot product of the two quantities should be zero.

That is, B.*a* = 0

⇒ 7*x* × 10^{−3}^{}× 10^{−6} − 3 × 10^{−3} × 7 × 10^{−6} = 0

⇒ 7*x* − 21 = 0

$x=\frac{21}{7}=3$

Acceleration of the particle is (3*i* + 7*j*) × 10^{−6} m/s^{2}.

#### Page No 230:

#### Question 5:

Given,

Magnetic field, *B* = (7.0*i* − 3.0*j*) × 10^{−3} T

Acceleration of the particle, *a* = (*xi* + 7*j*) × 10^{−6} m/s^{2}

We have denoted the unidentified number as *x*.*B* and *a* are perpendicular to each other. (Because magnetic force always acts perpendicular to the motion of the particle)

So, the dot product of the two quantities should be zero.

That is, B.*a* = 0

⇒ 7*x* × 10^{−3}^{}× 10^{−6} − 3 × 10^{−3} × 7 × 10^{−6} = 0

⇒ 7*x* − 21 = 0

$x=\frac{21}{7}=3$

Acceleration of the particle is (3*i* + 7*j*) × 10^{−6} m/s^{2}.

#### Answer:

Given:

Mass of the bullet, *m* = 10g

Charge of the bullet, *q* = 4.00 μC

Speed of the bullet in horizontal direction, *v *= 270 m/s

Vertical magnetic field, *B* = 500 μT

Distance travelled by the bullet, *d* = 100 m

Magnetic force,

$\overrightarrow{F}=q\overrightarrow{v}\times \overrightarrow{B}$ ....(i)

Also,*F* = *ma*

Using equation (i) we can write:

$ma=q\overrightarrow{v}\times \overrightarrow{B}$

$a=\frac{qvB}{m}$

Time taken by the bullet to travel 100 m horizontally,

$t=\frac{d}{v}=\frac{100}{270}\mathrm{s}$

Deflection due to the magnetic field in this time interval,

$y=\frac{1}{2}a{t}^{2}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\times \frac{4.00\times {10}^{-6}\times 270\times 500\times {10}^{-6}}{10\times {10}^{-3}}\times {\left(\frac{100}{270}\right)}^{2}\phantom{\rule{0ex}{0ex}}=3.7\times {10}^{-6}\mathrm{m}.$

#### Page No 230:

#### Question 6:

Given:

Mass of the bullet, *m* = 10g

Charge of the bullet, *q* = 4.00 μC

Speed of the bullet in horizontal direction, *v *= 270 m/s

Vertical magnetic field, *B* = 500 μT

Distance travelled by the bullet, *d* = 100 m

Magnetic force,

$\overrightarrow{F}=q\overrightarrow{v}\times \overrightarrow{B}$ ....(i)

Also,*F* = *ma*

Using equation (i) we can write:

$ma=q\overrightarrow{v}\times \overrightarrow{B}$

$a=\frac{qvB}{m}$

Time taken by the bullet to travel 100 m horizontally,

$t=\frac{d}{v}=\frac{100}{270}\mathrm{s}$

Deflection due to the magnetic field in this time interval,

$y=\frac{1}{2}a{t}^{2}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\times \frac{4.00\times {10}^{-6}\times 270\times 500\times {10}^{-6}}{10\times {10}^{-3}}\times {\left(\frac{100}{270}\right)}^{2}\phantom{\rule{0ex}{0ex}}=3.7\times {10}^{-6}\mathrm{m}.$

#### Answer:

Given:

The initial acceleration of a proton, when it is released from rest, is *a*_{0} towards west.

*F* =* qE* ....(i)

*F* = *ma*_{0} ....(ii)

Here, *q *is the charge, *E *is the electric field and *m* is the mass.

On equating both the forces *F* of equations (i) and (ii), we get:*qE* = *ma*_{0}

$\Rightarrow E=\frac{m{a}_{0}}{q}$ towards west.

When the proton is projected towards north with a speed *v*_{0}, it moves with an initial acceleration 3*a*_{0}_{}towards west.

$\overrightarrow{F}=q{\overrightarrow{v}}_{0}\times \overrightarrow{B}\phantom{\rule{0ex}{0ex}}$,

where *B* is the magnetic field.

$\Rightarrow B=\frac{\left|\overrightarrow{F}\right|}{q{v}_{0}}$

Again, an electric force will act on the proton in the west direction, due to which, an acceleration *a*_{0} will act on the proton.Now, as the proton was initially moving with a velocity v, a magnetic force is also acting on the proton.So, the change in acceleration of the proton will be solely due to the magnetic force acting on it.

Change in acceleration towards west due to the magnetic force

= 3*a*_{0} − *a*_{0} = 2*a*_{0}

So, *F* = *m2a*_{0}

Hence, required magnetic field,

$B=\frac{2m{a}_{0}}{q{v}_{0}}$

#### Page No 230:

#### Question 7:

Given:

The initial acceleration of a proton, when it is released from rest, is *a*_{0} towards west.

*F* =* qE* ....(i)

*F* = *ma*_{0} ....(ii)

Here, *q *is the charge, *E *is the electric field and *m* is the mass.

On equating both the forces *F* of equations (i) and (ii), we get:*qE* = *ma*_{0}

$\Rightarrow E=\frac{m{a}_{0}}{q}$ towards west.

When the proton is projected towards north with a speed *v*_{0}, it moves with an initial acceleration 3*a*_{0}_{}towards west.

$\overrightarrow{F}=q{\overrightarrow{v}}_{0}\times \overrightarrow{B}\phantom{\rule{0ex}{0ex}}$,

where *B* is the magnetic field.

$\Rightarrow B=\frac{\left|\overrightarrow{F}\right|}{q{v}_{0}}$

Again, an electric force will act on the proton in the west direction, due to which, an acceleration *a*_{0} will act on the proton.Now, as the proton was initially moving with a velocity v, a magnetic force is also acting on the proton.So, the change in acceleration of the proton will be solely due to the magnetic force acting on it.

Change in acceleration towards west due to the magnetic force

= 3*a*_{0} − *a*_{0} = 2*a*_{0}

So, *F* = *m2a*_{0}

Hence, required magnetic field,

$B=\frac{2m{a}_{0}}{q{v}_{0}}$

#### Answer:

Given:

A straight wire of length, *l *= 10 cm

Electric current flowing through the wire, *I* = 10 A

Magnetic field, *B* = 0.1 T

Angle between the wire and magnetic field, *θ *= 53Ëš

Magnetic force,

$\overrightarrow{F}=I\overrightarrow{l}\times \overrightarrow{B}$

∴$\left|\overrightarrow{F}\right|$ = *IlB*sin*θ*

∴$\left|\overrightarrow{\mathit{F}}\right|$ = *IlB*sin53Ëš*$\Rightarrow $F* = 10 × 10 × 10^{−2} × 0.1 × 0.798

$\Rightarrow $$\overrightarrow{F}$= 0.0798$\approx $0.08 N

The direction of force can be found using Fleming's left-hand rule.

Therefore, the direction of magnetic force is perpendicular to the wire as well as the magnetic field.

#### Page No 230:

#### Question 8:

Given:

A straight wire of length, *l *= 10 cm

Electric current flowing through the wire, *I* = 10 A

Magnetic field, *B* = 0.1 T

Angle between the wire and magnetic field, *θ *= 53Ëš

Magnetic force,

$\overrightarrow{F}=I\overrightarrow{l}\times \overrightarrow{B}$

∴$\left|\overrightarrow{F}\right|$ = *IlB*sin*θ*

∴$\left|\overrightarrow{\mathit{F}}\right|$ = *IlB*sin53Ëš*$\Rightarrow $F* = 10 × 10 × 10^{−2} × 0.1 × 0.798

$\Rightarrow $$\overrightarrow{F}$= 0.0798$\approx $0.08 N

The direction of force can be found using Fleming's left-hand rule.

Therefore, the direction of magnetic force is perpendicular to the wire as well as the magnetic field.

#### Answer:

Given:

A square frame abcd of side, *l *= 20 cm

Electric current through the wire, *I* = 2 A

Magnetic field, *B* = 0.1 T

The direction of magnetic field is perpendicular to the plane of the frame, coming out of the plane.

As per the question, current enters at the corner *d* of the square frame and leaves at the opposite corner *b.*

Angle between the frame and magnetic field, *θ *= 90Ëš

Magnetic force,

$\overrightarrow{F}=i\overrightarrow{l}\times \overrightarrow{B}$

For wire *da *and *cb*,

$\left|\overrightarrow{F}\right|$ = *ilB*sin*θ*

∴$\left|\overrightarrow{\mathit{F}}\right|$ = *ilB*sin90Ëš

= $\frac{2}{2}$ × 20 × 10^{−2} × 0.1

= 0.02 N

The direction of force can be found using Fleming's left-hand rule.

Thus, the direction of magnetic force is towards the left.

For wires *dc *and *ab*,

$\left|\overrightarrow{F}\right|$ = *ilB*sin*θ*

∴$\left|\overrightarrow{\mathit{F}}\right|$ = *ilB*sin90Ëš

= $\frac{2}{2}$ × 20 × 10^{−2} × 0.1

= 0.02 N

The direction of force can be found using Fleming's left-hand rule.

Thus, the direction of magnetic force is downwards.

#### Page No 231:

#### Question 9:

Given:

A square frame abcd of side, *l *= 20 cm

Electric current through the wire, *I* = 2 A

Magnetic field, *B* = 0.1 T

The direction of magnetic field is perpendicular to the plane of the frame, coming out of the plane.

As per the question, current enters at the corner *d* of the square frame and leaves at the opposite corner *b.*

Angle between the frame and magnetic field, *θ *= 90Ëš

Magnetic force,

$\overrightarrow{F}=i\overrightarrow{l}\times \overrightarrow{B}$

For wire *da *and *cb*,

$\left|\overrightarrow{F}\right|$ = *ilB*sin*θ*

∴$\left|\overrightarrow{\mathit{F}}\right|$ = *ilB*sin90Ëš

= $\frac{2}{2}$ × 20 × 10^{−2} × 0.1

= 0.02 N

The direction of force can be found using Fleming's left-hand rule.

Thus, the direction of magnetic force is towards the left.

For wires *dc *and *ab*,

$\left|\overrightarrow{F}\right|$ = *ilB*sin*θ*

∴$\left|\overrightarrow{\mathit{F}}\right|$ = *ilB*sin90Ëš

= $\frac{2}{2}$ × 20 × 10^{−2} × 0.1

= 0.02 N

The direction of force can be found using Fleming's left-hand rule.

Thus, the direction of magnetic force is downwards.

#### Answer:

Given:

Magnetic field, (*B*) = 1 T

Radius of the cylindrical region, *r* = 4.0 cm

Electric current through the wire, *I* = 2 A

The direction of magnetic field is perpendicular to the plane of the wire.

So, angle between wire and magnetic field, *θ *= 90Ëš

Magnetic force,

$\overrightarrow{F}=i\overrightarrow{l}\times \overrightarrow{B}$

∴$\left|\overrightarrow{\mathit{F}}\right|$ = *ilB*sin*θ*

$\left|\overrightarrow{F}\right|$ = *ilB*sin90Ëš

Here, *l* = 2*r*

∴$\left|\overrightarrow{F}\right|$ = *i*2*rB*sin90Ëš

= 2 × 8 × 10^{−2} × 1.0 × 1

= 0.16N

#### Page No 231:

#### Question 10:

Given:

Magnetic field, (*B*) = 1 T

Radius of the cylindrical region, *r* = 4.0 cm

Electric current through the wire, *I* = 2 A

The direction of magnetic field is perpendicular to the plane of the wire.

So, angle between wire and magnetic field, *θ *= 90Ëš

Magnetic force,

$\overrightarrow{F}=i\overrightarrow{l}\times \overrightarrow{B}$

∴$\left|\overrightarrow{\mathit{F}}\right|$ = *ilB*sin*θ*

$\left|\overrightarrow{F}\right|$ = *ilB*sin90Ëš

Here, *l* = 2*r*

∴$\left|\overrightarrow{F}\right|$ = *i*2*rB*sin90Ëš

= 2 × 8 × 10^{−2} × 1.0 × 1

= 0.16N

#### Answer:

Given:

A wire of length *l* cm

Electric current through the wire = *i*

Magnetic field, $\overrightarrow{B}={B}_{0}(\overrightarrow{i}+\overrightarrow{j}+\overrightarrow{k})$ T.

As per the question, the current is passing along the *X*-axis.

Magnetic force,

$\overrightarrow{F}=i\overrightarrow{l}\times \overrightarrow{\mathrm{B}}$

Putting the repective values in the above equation, we get:

= *i*{(*l* $\overrightarrow{i}$) × (*B*_{0}$\overrightarrow{i}$ + *B*_{0}$\overrightarrow{j}$ + *B*_{0}$\overrightarrow{k}$)}

= *i*{*lB*_{0}$\overrightarrow{k}$ − *lB*_{0}$\overrightarrow{j}$}

= *ilB*_{0}{−$\overrightarrow{j}$ + $\overrightarrow{k}$}

The magnitude of the magnetic force,

$\left|\overrightarrow{\mathit{F}}\right|$ = $\sqrt{2{i}^{2}{l}^{2}{B}_{0}^{2}}$

∴ $\left|\overrightarrow{\mathit{F}}\right|$ = $\sqrt{2}$*ilB*_{0}

#### Page No 231:

#### Question 11:

Given:

A wire of length *l* cm

Electric current through the wire = *i*

Magnetic field, $\overrightarrow{B}={B}_{0}(\overrightarrow{i}+\overrightarrow{j}+\overrightarrow{k})$ T.

As per the question, the current is passing along the *X*-axis.

Magnetic force,

$\overrightarrow{F}=i\overrightarrow{l}\times \overrightarrow{\mathrm{B}}$

Putting the repective values in the above equation, we get:

= *i*{(*l* $\overrightarrow{i}$) × (*B*_{0}$\overrightarrow{i}$ + *B*_{0}$\overrightarrow{j}$ + *B*_{0}$\overrightarrow{k}$)}

= *i*{*lB*_{0}$\overrightarrow{k}$ − *lB*_{0}$\overrightarrow{j}$}

= *ilB*_{0}{−$\overrightarrow{j}$ + $\overrightarrow{k}$}

The magnitude of the magnetic force,

$\left|\overrightarrow{\mathit{F}}\right|$ = $\sqrt{2{i}^{2}{l}^{2}{B}_{0}^{2}}$

∴ $\left|\overrightarrow{\mathit{F}}\right|$ = $\sqrt{2}$*ilB*_{0}

#### Answer:

Given:

Length of the wire PQ inside the magnetic field, *l *= 50 cm

Electric current through the wire, *I* = 5 A

Magnetic field, *B* = 0.2 T

The direction of magnetic field is perpendicular to the plane of the frame and it is going into the plane of the circuit.

As per the question,

Angle between the plane of the wire and the magnetic field, *θ *= 90Ëš

Magnetic force,

$\overrightarrow{F}=i\overrightarrow{l}\times \overrightarrow{\mathrm{B}}$

∴$\left|\overrightarrow{\mathit{F}}\right|$ = *il*Bsin*θ*

∴$\left|\overrightarrow{F}\right|$ = *il*Bsin90Ëš

= 5 × 50 × 10^{−2}^{}× 0.2 × 1

= 0.50 N

The direction of force can be found using Fleming's left-hand rule.

Thus, the direction of magnetic force is upwards in the plane of the paper.

#### Page No 231:

#### Question 12:

Given:

Length of the wire PQ inside the magnetic field, *l *= 50 cm

Electric current through the wire, *I* = 5 A

Magnetic field, *B* = 0.2 T

The direction of magnetic field is perpendicular to the plane of the frame and it is going into the plane of the circuit.

As per the question,

Angle between the plane of the wire and the magnetic field, *θ *= 90Ëš

Magnetic force,

$\overrightarrow{F}=i\overrightarrow{l}\times \overrightarrow{\mathrm{B}}$

∴$\left|\overrightarrow{\mathit{F}}\right|$ = *il*Bsin*θ*

∴$\left|\overrightarrow{F}\right|$ = *il*Bsin90Ëš

= 5 × 50 × 10^{−2}^{}× 0.2 × 1

= 0.50 N

The direction of force can be found using Fleming's left-hand rule.

Thus, the direction of magnetic force is upwards in the plane of the paper.

#### Answer:

Given:

A circular loop of radius* = a*

So, the length of the loop, *l *= 2π*a*

Electric current through the loop = *i*

As per the question,

The loop is placed in a two-dimensional magnetic field. The centre of the loop coincides with the centre of the field. The strength of the magnetic field at the periphery of the loop is *B*

Therefore, the magnetic field points radially outwards.

Here, the angle between the length of the loop and the magnetic field, *θ *= 90Ëš

Magnetic force is given by

$\overrightarrow{F}=i\overrightarrow{l}\times \overrightarrow{B}\phantom{\rule{0ex}{0ex}}\overrightarrow{F}\mathit{=}\mathit{}i\left(2\mathrm{\pi}a\mathit{\times}\overrightarrow{B}\right)\phantom{\rule{0ex}{0ex}}\overrightarrow{F}\mathit{=}\mathit{}i2\mathrm{\pi}aB\mathrm{sin}90\xb0\phantom{\rule{0ex}{0ex}}=i2\mathrm{\pi}aB$

Direction of the force can be found using Fleming's left-hand rule.

Thus, the direction of magnetic force is perpendicular to the plane of the figure and pointing inside.

#### Page No 231:

#### Question 13:

Given:

A circular loop of radius* = a*

So, the length of the loop, *l *= 2π*a*

Electric current through the loop = *i*

As per the question,

The loop is placed in a two-dimensional magnetic field. The centre of the loop coincides with the centre of the field. The strength of the magnetic field at the periphery of the loop is *B*

Therefore, the magnetic field points radially outwards.

Here, the angle between the length of the loop and the magnetic field, *θ *= 90Ëš

Magnetic force is given by

$\overrightarrow{F}=i\overrightarrow{l}\times \overrightarrow{B}\phantom{\rule{0ex}{0ex}}\overrightarrow{F}\mathit{=}\mathit{}i\left(2\mathrm{\pi}a\mathit{\times}\overrightarrow{B}\right)\phantom{\rule{0ex}{0ex}}\overrightarrow{F}\mathit{=}\mathit{}i2\mathrm{\pi}aB\mathrm{sin}90\xb0\phantom{\rule{0ex}{0ex}}=i2\mathrm{\pi}aB$

Direction of the force can be found using Fleming's left-hand rule.

Thus, the direction of magnetic force is perpendicular to the plane of the figure and pointing inside.

#### Answer:

Given:

A hypothetical magnetic field existing in a region, $\overrightarrow{B}={B}_{0}{\overrightarrow{e}}_{r}$

where ${\overrightarrow{e}}_{r}$ denotes the unit vector along the radial direction.

A circular loop of radius* a*

So, the length of the loop, *l *= 2π*a*

Electric current through loop = *i*

As per the question, the loop is placed with its plane parallel to the *X*−*Y* plane and its centre is at (0, 0, *d*).

Here, angle between the length of the loop and the magnetic field is *θ. *

Magnetic force is given by

$\overrightarrow{F}=i\overrightarrow{l}\times \overrightarrow{B}\phantom{\rule{0ex}{0ex}}\overrightarrow{F}\mathit{=}\mathit{}i\left(2\mathrm{\pi}a\mathit{\times}\overrightarrow{B}\right)\phantom{\rule{0ex}{0ex}}\overrightarrow{F}\mathit{=}\mathit{}i2\mathrm{\pi}aB\mathrm{sin\theta}\phantom{\rule{0ex}{0ex}}=\frac{i2\mathrm{\pi}a{B}_{0}a}{\sqrt{{a}^{2}+{d}^{2}}}\phantom{\rule{0ex}{0ex}}=\frac{i2\mathrm{\pi}{B}_{0}{a}^{2}}{\sqrt{{a}^{2}+{d}^{2}}}$

#### Page No 231:

#### Question 14:

Given:

A hypothetical magnetic field existing in a region, $\overrightarrow{B}={B}_{0}{\overrightarrow{e}}_{r}$

where ${\overrightarrow{e}}_{r}$ denotes the unit vector along the radial direction.

A circular loop of radius* a*

So, the length of the loop, *l *= 2π*a*

Electric current through loop = *i*

As per the question, the loop is placed with its plane parallel to the *X*−*Y* plane and its centre is at (0, 0, *d*).

Here, angle between the length of the loop and the magnetic field is *θ. *

Magnetic force is given by

$\overrightarrow{F}=i\overrightarrow{l}\times \overrightarrow{B}\phantom{\rule{0ex}{0ex}}\overrightarrow{F}\mathit{=}\mathit{}i\left(2\mathrm{\pi}a\mathit{\times}\overrightarrow{B}\right)\phantom{\rule{0ex}{0ex}}\overrightarrow{F}\mathit{=}\mathit{}i2\mathrm{\pi}aB\mathrm{sin\theta}\phantom{\rule{0ex}{0ex}}=\frac{i2\mathrm{\pi}a{B}_{0}a}{\sqrt{{a}^{2}+{d}^{2}}}\phantom{\rule{0ex}{0ex}}=\frac{i2\mathrm{\pi}{B}_{0}{a}^{2}}{\sqrt{{a}^{2}+{d}^{2}}}$

#### Answer:

Given,

A rectangular wire loop of width* a*

Electric current through the loop = *i*

Direction of the current is anti-clockwise.

Strength of the magnetic field in the lower region = *B*

Direction of the magnetic field is into the plane of the loop.

Here, angle between the length of the loop and magnetic field, *θ *= 90Ëš

Magnetic force is given by

$\overrightarrow{F}=i\overrightarrow{a}\times \overrightarrow{B}\phantom{\rule{0ex}{0ex}}$

The magnetic force will act only on side AD and BC.

As side AD is outside the magnetic field, so *F* = 0

Magnetic force on side BC is

$\overrightarrow{F}=i\overrightarrow{a}\times \overrightarrow{B}\phantom{\rule{0ex}{0ex}}=iaB\mathrm{sin}\theta \phantom{\rule{0ex}{0ex}}=iaB\phantom{\rule{0ex}{0ex}}$

Direction of force can be found using Fleming's left-hand rule.

Thus, the direction of the magnetic force is upward.

Similarly if we change the direction of current to clockwise,

the force along BC,

$\overrightarrow{F}=i\overrightarrow{a}\times \overrightarrow{B}\phantom{\rule{0ex}{0ex}}=-iaB\phantom{\rule{0ex}{0ex}}$

Thus, the change in force is equal to the change in tension

= *iaB* − (− *iaB*) = 2*ia*B.

#### Page No 231:

#### Question 15:

Given,

A rectangular wire loop of width* a*

Electric current through the loop = *i*

Direction of the current is anti-clockwise.

Strength of the magnetic field in the lower region = *B*

Direction of the magnetic field is into the plane of the loop.

Here, angle between the length of the loop and magnetic field, *θ *= 90Ëš

Magnetic force is given by

$\overrightarrow{F}=i\overrightarrow{a}\times \overrightarrow{B}\phantom{\rule{0ex}{0ex}}$

The magnetic force will act only on side AD and BC.

As side AD is outside the magnetic field, so *F* = 0

Magnetic force on side BC is

$\overrightarrow{F}=i\overrightarrow{a}\times \overrightarrow{B}\phantom{\rule{0ex}{0ex}}=iaB\mathrm{sin}\theta \phantom{\rule{0ex}{0ex}}=iaB\phantom{\rule{0ex}{0ex}}$

Direction of force can be found using Fleming's left-hand rule.

Thus, the direction of the magnetic force is upward.

Similarly if we change the direction of current to clockwise,

the force along BC,

$\overrightarrow{F}=i\overrightarrow{a}\times \overrightarrow{B}\phantom{\rule{0ex}{0ex}}=-iaB\phantom{\rule{0ex}{0ex}}$

Thus, the change in force is equal to the change in tension

= *iaB* − (− *iaB*) = 2*ia*B.

#### Answer:

Given:

Uniform magnetic field existing in the region of the arbitrary loop = *B*

Let the electric current flowing through the loop be *i*.

Length of each side of the loop is *l*.

Assume that the direction of the current is clockwise.

Direction of the magnetic field is going into the plane of the loop.

Magnetic force is given by

$\overrightarrow{F}=i\overrightarrow{l}\times \overrightarrow{B}\phantom{\rule{0ex}{0ex}}\overrightarrow{F}\mathit{=}\mathit{}ilB\mathrm{sin}\theta \phantom{\rule{0ex}{0ex}}$

Here, *θ* = 90Ëš

Direction of force can be found using Fleming's lef- hand rule.

Force *F*_{1} acting on AB = *il*B upwards

Force *F*_{2} acting on DC = *il*B downwards

So*, **F*_{1}_{}and *F*_{2} cancel each other.

Force *F*_{3} acting on AD = *il*B outwards (Pointing towards the left from AB)

Force *F*_{4} acting on BC = *il*B outwards (Pointing towards the right from BC)

So*, **F*_{3}_{}and *F*_{4} cancel each other.

Therefore, the net force acting on the arbitrary loop is 0.

#### Page No 231:

#### Question 16:

Given:

Uniform magnetic field existing in the region of the arbitrary loop = *B*

Let the electric current flowing through the loop be *i*.

Length of each side of the loop is *l*.

Assume that the direction of the current is clockwise.

Direction of the magnetic field is going into the plane of the loop.

Magnetic force is given by

$\overrightarrow{F}=i\overrightarrow{l}\times \overrightarrow{B}\phantom{\rule{0ex}{0ex}}\overrightarrow{F}\mathit{=}\mathit{}ilB\mathrm{sin}\theta \phantom{\rule{0ex}{0ex}}$

Here, *θ* = 90Ëš

Direction of force can be found using Fleming's lef- hand rule.

Force *F*_{1} acting on AB = *il*B upwards

Force *F*_{2} acting on DC = *il*B downwards

So*, **F*_{1}_{}and *F*_{2} cancel each other.

Force *F*_{3} acting on AD = *il*B outwards (Pointing towards the left from AB)

Force *F*_{4} acting on BC = *il*B outwards (Pointing towards the right from BC)

So*, **F*_{3}_{}and *F*_{4} cancel each other.

Therefore, the net force acting on the arbitrary loop is 0.

#### Answer:

Given:

Uniform magnetic field existing in the region of the wire = *B*

Let the electric current flowing through the wire be *i*.

Length of the wire between two points *a* and *b = l *

Magnetic force is given by

$\overrightarrow{F}=i\overrightarrow{l}\times \overrightarrow{B}\phantom{\rule{0ex}{0ex}}\overrightarrow{F}\mathit{=}ilB\mathrm{sin}\theta \phantom{\rule{0ex}{0ex}}$

Let us consider two wires of length l, one straight and the other circular.

The circular wire is of radius a such that $2\mathrm{\pi}a=l$

Suppose the magnetic field is pointing along the z direction and both the wires are lying in the xy plane, so that the angle between the area vector and the magnetic field is $90\xb0$.

For the straight wire of length *l* lying in a uniform magnetic field of strength B:

Force, *F* = $ilB\mathrm{sin}(90\xb0)$ = ilB

For the circular wire:

Length, *l *= $2\mathrm{\pi}a$

Angel between the area vector and magnetic field will again be $90\xb0$.

Force acting on the circular wire*,
F =* $i\left(2\mathrm{\pi}a\right)B\mathrm{sin}(90\xb0)$

= $i2\mathrm{\pi}aB\mathit{}\mathit{=}\mathit{}ilB$

Both the forces are equal in magnitude. This implies that the magnetic force is independent of the shape of the wire and depends on the length and orientation of the wire.

Therefore, the magnetic force is independent of the shape of the wire.

#### Page No 231:

#### Question 17:

Given:

Uniform magnetic field existing in the region of the wire = *B*

Let the electric current flowing through the wire be *i*.

Length of the wire between two points *a* and *b = l *

Magnetic force is given by

$\overrightarrow{F}=i\overrightarrow{l}\times \overrightarrow{B}\phantom{\rule{0ex}{0ex}}\overrightarrow{F}\mathit{=}ilB\mathrm{sin}\theta \phantom{\rule{0ex}{0ex}}$

Let us consider two wires of length l, one straight and the other circular.

The circular wire is of radius a such that $2\mathrm{\pi}a=l$

Suppose the magnetic field is pointing along the z direction and both the wires are lying in the xy plane, so that the angle between the area vector and the magnetic field is $90\xb0$.

For the straight wire of length *l* lying in a uniform magnetic field of strength B:

Force, *F* = $ilB\mathrm{sin}(90\xb0)$ = ilB

For the circular wire:

Length, *l *= $2\mathrm{\pi}a$

Angel between the area vector and magnetic field will again be $90\xb0$.

Force acting on the circular wire*,
F =* $i\left(2\mathrm{\pi}a\right)B\mathrm{sin}(90\xb0)$

= $i2\mathrm{\pi}aB\mathit{}\mathit{=}\mathit{}ilB$

Both the forces are equal in magnitude. This implies that the magnetic force is independent of the shape of the wire and depends on the length and orientation of the wire.

Therefore, the magnetic force is independent of the shape of the wire.

#### Answer:

Given,

Radius of semicircular wire, *r* = 5.0 cm

Thus, the length of the wire = 2*r*

Electric current flowing through wire = 5.0 A

Magnetic field, *B = *0.50 T

Direction of magnetic field is perpendicular to the plane of the wire which implies that angle between length of the wire and magnetic field, *θ *= 90Ëš

As we know the magnetic force is given by

$\overrightarrow{F}=I\overrightarrow{l}\times \overrightarrow{B}\phantom{\rule{0ex}{0ex}}\overrightarrow{F}\mathit{=}\mathit{}I2rB\mathrm{sin}90\xb0$

=5× 2 × 0.05 × 0.5 = 0.25 N

#### Page No 231:

#### Question 18:

Given,

Radius of semicircular wire, *r* = 5.0 cm

Thus, the length of the wire = 2*r*

Electric current flowing through wire = 5.0 A

Magnetic field, *B = *0.50 T

Direction of magnetic field is perpendicular to the plane of the wire which implies that angle between length of the wire and magnetic field, *θ *= 90Ëš

As we know the magnetic force is given by

$\overrightarrow{F}=I\overrightarrow{l}\times \overrightarrow{B}\phantom{\rule{0ex}{0ex}}\overrightarrow{F}\mathit{=}\mathit{}I2rB\mathrm{sin}90\xb0$

=5× 2 × 0.05 × 0.5 = 0.25 N

#### Answer:

Given:

Electric current flowing through the wire = *i*

The wire is kept in the *x*−*y* plane along the curve, $y=A\mathrm{sin}\left(\frac{2\mathrm{\pi}}{\lambda}x\right)$

Magnetic field (*B*) exists in the *z* direction.

We have to find the magnetic force on the portion of the wire between *x* = 0 and *x* = *λ*.

Magnetic force is given by

$\overrightarrow{F}=i\overrightarrow{l}\times \overrightarrow{B}$

For a small element *dl*,

$d\overrightarrow{F}\mathit{=}i\left(\mathrm{d}l\mathit{\times}\overrightarrow{B}\right)$

The effective force on the whole wire is equivalent to the force on a starlight wire of length λ placed along the x axis.

So,

$F\mathit{=}iB{\int}_{0}^{\lambda}dl$

$\Rightarrow F=i\lambda B$

#### Page No 231:

#### Question 19:

Given:

Electric current flowing through the wire = *i*

The wire is kept in the *x*−*y* plane along the curve, $y=A\mathrm{sin}\left(\frac{2\mathrm{\pi}}{\lambda}x\right)$

Magnetic field (*B*) exists in the *z* direction.

We have to find the magnetic force on the portion of the wire between *x* = 0 and *x* = *λ*.

Magnetic force is given by

$\overrightarrow{F}=i\overrightarrow{l}\times \overrightarrow{B}$

For a small element *dl*,

$d\overrightarrow{F}\mathit{=}i\left(\mathrm{d}l\mathit{\times}\overrightarrow{B}\right)$

The effective force on the whole wire is equivalent to the force on a starlight wire of length λ placed along the x axis.

So,

$F\mathit{=}iB{\int}_{0}^{\lambda}dl$

$\Rightarrow F=i\lambda B$

#### Answer:

Given:

Radius of the semi-circular portion of the rigid wire = *R*

Magnetic field = *B*

Electric current flowing through the wire = *i*

As per the question, the wire is partially immersed in a perpendicular magnetic field.

As PQ and RS are straight wires of length *l* each and strength of the magnetic field is also same on both the wires, the force acting on these wires will be equal in magnitude but their directions will be opposite to each other.(Direction of force can be found out using Fleming's left hand rule.)

So, the magnetic force on the wire PQ and the force on the wire RS are equal and opposite to each other. Both the forces cancel each other.

Therefore, only the semicircular loop PR will experience a net magnetic force.

Here, angle between the length of the wire and magnetic field, *θ *= 90Ëš

Magnetic force in the loop PR,

$\overrightarrow{F}=i\overrightarrow{l}\times \overrightarrow{B}$

Here, *l* = 2*R*

$\overrightarrow{F}\mathit{=}\mathit{}i2R\mathit{\times}\overrightarrow{B}\phantom{\rule{0ex}{0ex}}\overrightarrow{F}\mathit{=}\mathit{}i2RB\mathrm{sin}90\xb0=i2RB$

#### Page No 231:

#### Question 20:

Given:

Radius of the semi-circular portion of the rigid wire = *R*

Magnetic field = *B*

Electric current flowing through the wire = *i*

As per the question, the wire is partially immersed in a perpendicular magnetic field.

As PQ and RS are straight wires of length *l* each and strength of the magnetic field is also same on both the wires, the force acting on these wires will be equal in magnitude but their directions will be opposite to each other.(Direction of force can be found out using Fleming's left hand rule.)

So, the magnetic force on the wire PQ and the force on the wire RS are equal and opposite to each other. Both the forces cancel each other.

Therefore, only the semicircular loop PR will experience a net magnetic force.

Here, angle between the length of the wire and magnetic field, *θ *= 90Ëš

Magnetic force in the loop PR,

$\overrightarrow{F}=i\overrightarrow{l}\times \overrightarrow{B}$

Here, *l* = 2*R*

$\overrightarrow{F}\mathit{=}\mathit{}i2R\mathit{\times}\overrightarrow{B}\phantom{\rule{0ex}{0ex}}\overrightarrow{F}\mathit{=}\mathit{}i2RB\mathrm{sin}90\xb0=i2RB$

#### Answer:

Given:

Mass of the wire, *M* = 10 mg = 10^{−5} Kg

Length of the wire, *l =* 1.0 m

Electric current flowing through wire, *I* = 2.0 A

As per the question, the weight of the wire should be balanced by the magnetic force acting on the wire.Also angle between the length of the wire and magnetic field is $90\xb0$.

Thus, *Mg* = *IlB*, where*g* is the acceleration due to gravity = 9.8 m/s^{2}

*B* is the applied magnetic field

So,

$B=\frac{M\mathrm{g}}{Il}\phantom{\rule{0ex}{0ex}}=\frac{{10}^{-5}\times 9.8}{2\times 1}\phantom{\rule{0ex}{0ex}}=4.9\times {10}^{-5}\mathrm{T}$

#### Page No 231:

#### Question 21:

Given:

Mass of the wire, *M* = 10 mg = 10^{−5} Kg

Length of the wire, *l =* 1.0 m

Electric current flowing through wire, *I* = 2.0 A

As per the question, the weight of the wire should be balanced by the magnetic force acting on the wire.Also angle between the length of the wire and magnetic field is $90\xb0$.

Thus, *Mg* = *IlB*, where*g* is the acceleration due to gravity = 9.8 m/s^{2}

*B* is the applied magnetic field

So,

$B=\frac{M\mathrm{g}}{Il}\phantom{\rule{0ex}{0ex}}=\frac{{10}^{-5}\times 9.8}{2\times 1}\phantom{\rule{0ex}{0ex}}=4.9\times {10}^{-5}\mathrm{T}$

#### Answer:

Given:

Length of the rod PQ = 20.0 cm

Mass of the rod, *M* = 200 g

Length of the two threads, *l** *= 20.0 cm

Applied magnetic field, *B* = 0.500 T

As per the question,

(a) When the circuit is open:

The weight of the rod is balanced by the tension in the rod. So,

2*T*cos30° = *M*g

$T=\frac{M\mathrm{g}}{2\mathrm{cos}30\xb0}\phantom{\rule{0ex}{0ex}}=\frac{0.2\times 9.8}{2{\displaystyle \frac{\sqrt{3}}{2}}}\phantom{\rule{0ex}{0ex}}=1.13\mathrm{N}$

(b) When the circuit is closed and current flowing through the circuit = 2 A:

Then,

2*T*cos 30°= *M*g + *ilB*

= 0.200× 9.8 + 2 × 0.2 × 0.5

= 1.95 + 0.2 = 2.16

⇒ 2*T* = $2.16\times \frac{2}{\sqrt{3}}$

⇒ *T* = 1.245 = 1.25 N

#### Page No 232:

#### Question 22:

Given:

Length of the rod PQ = 20.0 cm

Mass of the rod, *M* = 200 g

Length of the two threads, *l** *= 20.0 cm

Applied magnetic field, *B* = 0.500 T

As per the question,

(a) When the circuit is open:

The weight of the rod is balanced by the tension in the rod. So,

2*T*cos30° = *M*g

$T=\frac{M\mathrm{g}}{2\mathrm{cos}30\xb0}\phantom{\rule{0ex}{0ex}}=\frac{0.2\times 9.8}{2{\displaystyle \frac{\sqrt{3}}{2}}}\phantom{\rule{0ex}{0ex}}=1.13\mathrm{N}$

(b) When the circuit is closed and current flowing through the circuit = 2 A:

Then,

2*T*cos 30°= *M*g + *ilB*

= 0.200× 9.8 + 2 × 0.2 × 0.5

= 1.95 + 0.2 = 2.16

⇒ 2*T* = $2.16\times \frac{2}{\sqrt{3}}$

⇒ *T* = 1.245 = 1.25 N

#### Answer:

Given:

Length of the two metal strips =* l*

Separation between the strips = *b*

Mass of the wire = *m*

Strength magnetic field = B

Coefficient of friction between the wire and the floor = µ

Let the distance covered by the wire be *x. *

Due to the presence of the magnetic field, a net magnetic force will act on the wire towards the right.

Let the wire has moved through a distance *x* after travelling the length of the strips.

As the contact between the wire and strip is smooth so coefficient of friction between them is zero. Under the influence of magnetic force,firstly the wire will travel a distance equal to the length of the strips. After this, it travels a distance x and also now,a frictional force will act on the wire in a direction opposite to its direction of motion.

So we can equate the work done by the magnetic force and the frictional force.

Thus, *F* × *l* = µ*mg* × *x*,

where *g* is the acceleration due to gravity

⇒ *i**b*B*l* = µ*mg**x**$\Rightarrow x=\frac{ibl\mathrm{B}}{\mathrm{\mu}m\mathrm{g}}$*

#### Page No 232:

#### Question 23:

Given:

Length of the two metal strips =* l*

Separation between the strips = *b*

Mass of the wire = *m*

Strength magnetic field = B

Coefficient of friction between the wire and the floor = µ

Let the distance covered by the wire be *x. *

Due to the presence of the magnetic field, a net magnetic force will act on the wire towards the right.

Let the wire has moved through a distance *x* after travelling the length of the strips.

As the contact between the wire and strip is smooth so coefficient of friction between them is zero. Under the influence of magnetic force,firstly the wire will travel a distance equal to the length of the strips. After this, it travels a distance x and also now,a frictional force will act on the wire in a direction opposite to its direction of motion.

So we can equate the work done by the magnetic force and the frictional force.

Thus, *F* × *l* = µ*mg* × *x*,

where *g* is the acceleration due to gravity

⇒ *i**b*B*l* = µ*mg**x**$\Rightarrow x=\frac{ibl\mathrm{B}}{\mathrm{\mu}m\mathrm{g}}$*

#### Answer:

Given:

Mass of the metal wire, *M* = 10 g

Distance between the two horizontal metal rails, *l* = 4.90 cm

Vertically-downward magnetic field, *B* = 0.800 T

As per the question, when the resistance of the circuit is slowly decreased below 20.0 â„¦, the wire *PQ* starts sliding on the rails. At that moment,

current in the wire, *i =$\frac{V}{R}=\frac{6}{20}\mathrm{A}$*

Using Fleming's left-hand rule, the magnetic force will act towards the right. So, due to this magnetic force, the wire will try to slide on the rails.The frictional force will try to oppose this motion of the wire.When the wire just starts sliding on the rails, the frictional force acting on the wire will just balance the magnetic force acting on the wire.This implies

µ*R* = *F*, where

µ is the coffiecent of friction*R *is the normal reaction force and*F *is the magnetic force

⇒ µ × *M* × *g* = *ilB**µ* × 10 × 10^{−3} × 9.8 = $\frac{6}{20}$ × 4.9 × 10^{−2} × 0.8

$\mu =\frac{6\times 4.9\times {10}^{-2}\times 0.8}{20\times 10\times {10}^{-3}\times 9.8}$*µ* = 0.12

#### Page No 232:

#### Question 24:

Given:

Mass of the metal wire, *M* = 10 g

Distance between the two horizontal metal rails, *l* = 4.90 cm

Vertically-downward magnetic field, *B* = 0.800 T

As per the question, when the resistance of the circuit is slowly decreased below 20.0 â„¦, the wire *PQ* starts sliding on the rails. At that moment,

current in the wire, *i =$\frac{V}{R}=\frac{6}{20}\mathrm{A}$*

Using Fleming's left-hand rule, the magnetic force will act towards the right. So, due to this magnetic force, the wire will try to slide on the rails.The frictional force will try to oppose this motion of the wire.When the wire just starts sliding on the rails, the frictional force acting on the wire will just balance the magnetic force acting on the wire.This implies

µ*R* = *F*, where

µ is the coffiecent of friction*R *is the normal reaction force and*F *is the magnetic force

⇒ µ × *M* × *g* = *ilB**µ* × 10 × 10^{−3} × 9.8 = $\frac{6}{20}$ × 4.9 × 10^{−2} × 0.8

$\mu =\frac{6\times 4.9\times {10}^{-2}\times 0.8}{20\times 10\times {10}^{-3}\times 9.8}$*µ* = 0.12

#### Answer:

Given:

Length of the wire = *l*

Distance between the plastic rails = *d*

The coefficient of friction between the wire and the rails = µ

Electric current flowing through the wire = *i*

The minimum magnetic field that should exist in the space, in order to slide the wire on the rails, should be such that the net magnetic force acting on the wire is equal to the frictional force on the wire.*µR* = *F*,

where

µ is the coefficient of friction between the wire and the rail*R *is the normal reaction force*F *is the magnetic force*µM*g = *iBl*

$B=\frac{\mu M\mathrm{g}}{il}$

#### Page No 232:

#### Question 25:

Given:

Length of the wire = *l*

Distance between the plastic rails = *d*

The coefficient of friction between the wire and the rails = µ

Electric current flowing through the wire = *i*

The minimum magnetic field that should exist in the space, in order to slide the wire on the rails, should be such that the net magnetic force acting on the wire is equal to the frictional force on the wire.*µR* = *F*,

where

µ is the coefficient of friction between the wire and the rail*R *is the normal reaction force*F *is the magnetic force*µM*g = *iBl*

$B=\frac{\mu M\mathrm{g}}{il}$

#### Answer:

Given:

Radius of the circular wire loop = *a*

Electric current flowing through the loop = *i*

Perpendicular magnetic field = *B*

(a) The force exerted by the magnetic field on a small element d*l *of the wire*,**F*_{dl} = *i*. (d*l* × *B*) = *idlB*

Using Fleming's left-hand rule, we can say that the direction of magnetic force is towards the centre for any small element of length *dl* of the wire (d*l* and B are perpendicular to each other).

(b) Suppose some part of wire loop subtends a small angle 2*θ** *at the centre of a circular loop

So,

2*T*sin*θ* = *i*.(*dl* × *B*)

= *i*.2*aθ B* (Using length of any arc, *l =rθ)*

âˆµ *θ* is very small, sin*θ* = *θ*

2*Tθ* = *i*.2*aθB*

∴ *T* = *iaB*

#### Page No 232:

#### Question 26:

Given:

Radius of the circular wire loop = *a*

Electric current flowing through the loop = *i*

Perpendicular magnetic field = *B*

(a) The force exerted by the magnetic field on a small element d*l *of the wire*,**F*_{dl} = *i*. (d*l* × *B*) = *idlB*

Using Fleming's left-hand rule, we can say that the direction of magnetic force is towards the centre for any small element of length *dl* of the wire (d*l* and B are perpendicular to each other).

(b) Suppose some part of wire loop subtends a small angle 2*θ** *at the centre of a circular loop

So,

2*T*sin*θ* = *i*.(*dl* × *B*)

= *i*.2*aθ B* (Using length of any arc, *l =rθ)*

âˆµ *θ* is very small, sin*θ* = *θ*

2*Tθ* = *i*.2*aθB*

∴ *T* = *iaB*

#### Answer:

Given:

Radius of cross-section of the wire = *r*

Young's modulus of the material of the wire is *Y*.

As per the question, when the applied magnetic field is switched off, the tension in the wire increased and so did its length.

Young's modulus,

$Y=\frac{\mathrm{Stress}}{\mathrm{Strain}}\phantom{\rule{0ex}{0ex}}Y=\frac{T/A}{\u2206l/l}$

Here,*T *is the tension*A *is the area of cross-section

Δ*l** *is the increse in length of the wire

$\u2206l=\frac{Tl}{YA}=\frac{iaB\mathit{\times}l}{Y\times \mathrm{\pi}{r}^{2}}\phantom{\rule{0ex}{0ex}}2\mathrm{\pi}\times \u2206r=\frac{iaB\times 2\mathrm{\pi}a}{Y\times \mathrm{\pi}{r}^{2}}\phantom{\rule{0ex}{0ex}}\u2206r=\frac{i{a}^{2}B}{\mathrm{\pi}{r}^{2}Y}$

#### Page No 232:

#### Question 27:

Given:

Radius of cross-section of the wire = *r*

Young's modulus of the material of the wire is *Y*.

As per the question, when the applied magnetic field is switched off, the tension in the wire increased and so did its length.

Young's modulus,

$Y=\frac{\mathrm{Stress}}{\mathrm{Strain}}\phantom{\rule{0ex}{0ex}}Y=\frac{T/A}{\u2206l/l}$

Here,*T *is the tension*A *is the area of cross-section

Δ*l** *is the increse in length of the wire

$\u2206l=\frac{Tl}{YA}=\frac{iaB\mathit{\times}l}{Y\times \mathrm{\pi}{r}^{2}}\phantom{\rule{0ex}{0ex}}2\mathrm{\pi}\times \u2206r=\frac{iaB\times 2\mathrm{\pi}a}{Y\times \mathrm{\pi}{r}^{2}}\phantom{\rule{0ex}{0ex}}\u2206r=\frac{i{a}^{2}B}{\mathrm{\pi}{r}^{2}Y}$

#### Answer:

Given:

Magnetic field, $\overrightarrow{B}={B}_{0}\left(1+\frac{x}{l}\right)\overrightarrow{k}$

Length of the edge of a square loop = *l*

Electric current flowing through it = *i*

As per the question, the loop is placed with its edges parallel to the *X*−*Y* axes.

In the figure, arrow denotes the direction of force on different sides of the square.

The net magnetic force experienced by the loop,

$\overrightarrow{F}=i\overrightarrow{l}\times \overrightarrow{B}$

Force on AB:

Consider a small element of length dx at a distance x from the origin on line AB.

Force on this small element,

$dF=i{B}_{0}(1+\frac{x}{l})dx$

Force on the full length of AB,*${F}_{\mathrm{AB}}={\int}_{x=0}^{x=l}i{B}_{0}(1+\frac{x}{l})dx\phantom{\rule{0ex}{0ex}}=i{B}_{0}{\int}_{x=0}^{x=l}(dx+\frac{1}{l}xdx)\phantom{\rule{0ex}{0ex}}=i{B}_{0}(l+\frac{l}{2})\phantom{\rule{0ex}{0ex}}=\frac{3i{B}_{0}l}{2}$*

Force on AB will be acting downwards.

Similarly, force on CD,

${F}_{2}=i{B}_{0}\left(l+\frac{l}{2}\right)\phantom{\rule{0ex}{0ex}}=\frac{3i{B}_{0}l}{2}$

So, the net vertical force =* **F*_{1}_{}−* **F*_{2} = 0

Force on AD,

${F}_{3}=i{B}_{0}l\left(1+\frac{0}{l}\right)\phantom{\rule{0ex}{0ex}}=i{B}_{0}l$

Force on BC

${F}_{4}=i{B}_{0}l\left(1+\frac{1}{l}\right)\phantom{\rule{0ex}{0ex}}=2i{B}_{0}l$

So, the net horizontal force =* **F*_{4}−*F*_{3} = *iB*_{0}*l*

#### Page No 232:

#### Question 28:

Given:

Magnetic field, $\overrightarrow{B}={B}_{0}\left(1+\frac{x}{l}\right)\overrightarrow{k}$

Length of the edge of a square loop = *l*

Electric current flowing through it = *i*

As per the question, the loop is placed with its edges parallel to the *X*−*Y* axes.

In the figure, arrow denotes the direction of force on different sides of the square.

The net magnetic force experienced by the loop,

$\overrightarrow{F}=i\overrightarrow{l}\times \overrightarrow{B}$

Force on AB:

Consider a small element of length dx at a distance x from the origin on line AB.

Force on this small element,

$dF=i{B}_{0}(1+\frac{x}{l})dx$

Force on the full length of AB,*${F}_{\mathrm{AB}}={\int}_{x=0}^{x=l}i{B}_{0}(1+\frac{x}{l})dx\phantom{\rule{0ex}{0ex}}=i{B}_{0}{\int}_{x=0}^{x=l}(dx+\frac{1}{l}xdx)\phantom{\rule{0ex}{0ex}}=i{B}_{0}(l+\frac{l}{2})\phantom{\rule{0ex}{0ex}}=\frac{3i{B}_{0}l}{2}$*

Force on AB will be acting downwards.

Similarly, force on CD,

${F}_{2}=i{B}_{0}\left(l+\frac{l}{2}\right)\phantom{\rule{0ex}{0ex}}=\frac{3i{B}_{0}l}{2}$

So, the net vertical force =* **F*_{1}_{}−* **F*_{2} = 0

Force on AD,

${F}_{3}=i{B}_{0}l\left(1+\frac{0}{l}\right)\phantom{\rule{0ex}{0ex}}=i{B}_{0}l$

Force on BC

${F}_{4}=i{B}_{0}l\left(1+\frac{1}{l}\right)\phantom{\rule{0ex}{0ex}}=2i{B}_{0}l$

So, the net horizontal force =* **F*_{4}−*F*_{3} = *iB*_{0}*l*

#### Answer:

Given:

Length of the conducting wire = *l*

Inward magnetic field = *B*

Velocity of the conducting wire = *v*

As the wire is moving with velocity *v,* we can consider this as the net motion of electrons inside the wire with velocity *v.*

(a) The average magnetic force on a free electron of the wire

= *e*(*v* × *B*) = evB, where *e* is the charge of an electron.

(b) The redistribution stops when the electric force is just balanced by the magnetic force.

Electric force, *F* = *eE *and also magnetic force, *F* = evB

On equatinging the two forces, we get:*eE* = *evB*

$\Rightarrow $*E* = *vB*

(c) The potential difference is developed between the ends of the wire:*V* = *lE* = *lvB, w*here *V* is the potential difference across the ends of the wire.

#### Page No 232:

#### Question 29:

Given:

Length of the conducting wire = *l*

Inward magnetic field = *B*

Velocity of the conducting wire = *v*

As the wire is moving with velocity *v,* we can consider this as the net motion of electrons inside the wire with velocity *v.*

(a) The average magnetic force on a free electron of the wire

= *e*(*v* × *B*) = evB, where *e* is the charge of an electron.

(b) The redistribution stops when the electric force is just balanced by the magnetic force.

Electric force, *F* = *eE *and also magnetic force, *F* = evB

On equatinging the two forces, we get:*eE* = *evB*

$\Rightarrow $*E* = *vB*

(c) The potential difference is developed between the ends of the wire:*V* = *lE* = *lvB, w*here *V* is the potential difference across the ends of the wire.

#### Answer:

Given:

Width of the silver strip = *d*

Area of cross-section *= A*

Electric current flowing through the strip = *i*

The number of free electrons per unit volume = *n*

(a) The relation between the drift velocity and current through any wire,*i* = *vnAe*, where *e *= charge of an electron and v is the drift velocity.

$v=\frac{i}{nAe}$

(b)The magnetic field existing in the region is *B.*

The average magnetic force on a current-carrying conductor,*F* = i*lB*

So, the force on a free electron

= $\frac{ilB}{nAl}=\frac{iB}{nA}$ upwards (Using Fleming's left-hand rule)

(c) Let us take the electric field as *E*.

So, further accumulation of electrons will stop when the electric force is just balanced by the magnetic force.

$e\mathrm{E}=\mathrm{F}=\frac{i\mathrm{B}}{\mathrm{A}n}\phantom{\rule{0ex}{0ex}}\Rightarrow E=\frac{i\mathrm{B}}{e\mathrm{A}n}$

(d) The potential difference developed across the width of the conductor due to the electron-accumulation will be*V* = *E**r*

$\Rightarrow E\times d=\frac{i\mathrm{B}d}{e\mathrm{A}n}$

#### Page No 232:

#### Question 30:

Given:

Width of the silver strip = *d*

Area of cross-section *= A*

Electric current flowing through the strip = *i*

The number of free electrons per unit volume = *n*

(a) The relation between the drift velocity and current through any wire,*i* = *vnAe*, where *e *= charge of an electron and v is the drift velocity.

$v=\frac{i}{nAe}$

(b)The magnetic field existing in the region is *B.*

The average magnetic force on a current-carrying conductor,*F* = i*lB*

So, the force on a free electron

= $\frac{ilB}{nAl}=\frac{iB}{nA}$ upwards (Using Fleming's left-hand rule)

(c) Let us take the electric field as *E*.

So, further accumulation of electrons will stop when the electric force is just balanced by the magnetic force.

$e\mathrm{E}=\mathrm{F}=\frac{i\mathrm{B}}{\mathrm{A}n}\phantom{\rule{0ex}{0ex}}\Rightarrow E=\frac{i\mathrm{B}}{e\mathrm{A}n}$

(d) The potential difference developed across the width of the conductor due to the electron-accumulation will be*V* = *E**r*

$\Rightarrow E\times d=\frac{i\mathrm{B}d}{e\mathrm{A}n}$

#### Answer:

Given:

Charge of the particle, *q* = 2.0 × 10^{−8} C

Mass of the particle, *m* = 2.0 × 10^{−10} g

Projected speed of the particle*, v* = 2.0 × 10^{3} m s^{−1}

Uniform magnetic field, *B* = 0.10 T.

As per the question, the velocity is perpendicular to the field.

So, for the particle to move in a circle,the centrifugal force to the particle will be provided by the magnetic force acting on it.

Using *qvB* = $\frac{m{v}^{2}}{r}$, where *r* is the radius of the circle formed,

$r=\frac{mv}{qB}\phantom{\rule{0ex}{0ex}}=\frac{2\times {10}^{-13}\times 2\times {10}^{3}}{2\times {10}^{-8}\times 0.10}\phantom{\rule{0ex}{0ex}}=20\mathrm{cm}$

Time period,

$T=\frac{2\mathrm{\pi}m}{qB}\phantom{\rule{0ex}{0ex}}=\frac{2\times 3.14\times 2\times {10}^{-13}}{2\times {10}^{-8}\times 0.10}\phantom{\rule{0ex}{0ex}}=6.28\times {10}^{-4}\mathrm{s}$

#### Page No 232:

#### Question 31:

Given:

Charge of the particle, *q* = 2.0 × 10^{−8} C

Mass of the particle, *m* = 2.0 × 10^{−10} g

Projected speed of the particle*, v* = 2.0 × 10^{3} m s^{−1}

Uniform magnetic field, *B* = 0.10 T.

As per the question, the velocity is perpendicular to the field.

So, for the particle to move in a circle,the centrifugal force to the particle will be provided by the magnetic force acting on it.

Using *qvB* = $\frac{m{v}^{2}}{r}$, where *r* is the radius of the circle formed,

$r=\frac{mv}{qB}\phantom{\rule{0ex}{0ex}}=\frac{2\times {10}^{-13}\times 2\times {10}^{3}}{2\times {10}^{-8}\times 0.10}\phantom{\rule{0ex}{0ex}}=20\mathrm{cm}$

Time period,

$T=\frac{2\mathrm{\pi}m}{qB}\phantom{\rule{0ex}{0ex}}=\frac{2\times 3.14\times 2\times {10}^{-13}}{2\times {10}^{-8}\times 0.10}\phantom{\rule{0ex}{0ex}}=6.28\times {10}^{-4}\mathrm{s}$

#### Answer:

Given:

Radius of the circle, *r *= 1 cm

Magnetic field = 0.10 T

We know that the charge of a proton is *e* and that of an alpha particle is 2*e*. Also, the mass of a proton is *m *and that of an aplha particle is 4*m**. *

Suppose, both the particles are moving with speed *v.*

According to the question,

${r}_{\mathrm{p}}=\frac{mv}{eB},\text{where}{r}_{\mathrm{p}}\text{istheradiusofthecircledescribedbytheproton.}\phantom{\rule{0ex}{0ex}}0.01=\frac{mv}{e\times 0.1}...\left(\mathrm{i}\right)$

${r}_{\mathrm{\alpha}}=\frac{4mv}{2eB}\phantom{\rule{0ex}{0ex}}{r}_{\mathrm{\alpha}}=\frac{4mv}{2e\times 0.1}...\left(\mathrm{ii}\right)$

On dividing equation (ii) by (i), we get:

$\frac{{r}_{\mathrm{\alpha}}}{0.01}=\frac{4mv\times e\times 0.1}{2e\times 0.1\times mv}\phantom{\rule{0ex}{0ex}}{r}_{\mathrm{\alpha}}=0.02\mathrm{m}=2\mathrm{cm}$

#### Page No 232:

#### Question 32:

Given:

Radius of the circle, *r *= 1 cm

Magnetic field = 0.10 T

We know that the charge of a proton is *e* and that of an alpha particle is 2*e*. Also, the mass of a proton is *m *and that of an aplha particle is 4*m**. *

Suppose, both the particles are moving with speed *v.*

According to the question,

${r}_{\mathrm{p}}=\frac{mv}{eB},\text{where}{r}_{\mathrm{p}}\text{istheradiusofthecircledescribedbytheproton.}\phantom{\rule{0ex}{0ex}}0.01=\frac{mv}{e\times 0.1}...\left(\mathrm{i}\right)$

${r}_{\mathrm{\alpha}}=\frac{4mv}{2eB}\phantom{\rule{0ex}{0ex}}{r}_{\mathrm{\alpha}}=\frac{4mv}{2e\times 0.1}...\left(\mathrm{ii}\right)$

On dividing equation (ii) by (i), we get:

$\frac{{r}_{\mathrm{\alpha}}}{0.01}=\frac{4mv\times e\times 0.1}{2e\times 0.1\times mv}\phantom{\rule{0ex}{0ex}}{r}_{\mathrm{\alpha}}=0.02\mathrm{m}=2\mathrm{cm}$

#### Answer:

Given:

Kinetic energy of an electron = 100 eV

Radius of the circle = 10 cm

$\frac{1}{2}m{v}^{2}$ = 100 eV = 1.6 × 10^{−17} J (1 eV = 1.6 × 10^{−19} J)

Here,*m* is the mass of an electron and *v* is the speed of an electron. Thus,

$\frac{1}{2}$ × 9.1 × 10^{−31} × *v*^{2} = 1.6 × 10^{−17} J

⇒ *v*^{2} = 0.35 × 10^{14}*v* = 0.591 × 10^{7} m/s

Now,

$r=\frac{mv}{eB}\phantom{\rule{0ex}{0ex}}\Rightarrow B=\frac{mv}{er}$

$=\frac{9.1\times {10}^{-31}\times 0.591\times {10}^{7}}{1.6\times {10}^{-19}\times 0.1}$*B* = 3.3613 × 10^{−4} T

Therefore, the applied magnetic field = 3.4 × 10^{−4} T

Number of revolutions per second of the electron,

$f=\frac{1}{T}$

$T=\frac{2\pi r}{v}=\frac{2\pi m}{eB}$

$T=\frac{2\mathrm{\pi}m}{Be}\phantom{\rule{0ex}{0ex}}$

$f=\frac{Be}{2\mathrm{\pi}m}\phantom{\rule{0ex}{0ex}}=\frac{3.4\times {10}^{-4}\times 1.6\times {10}^{-19}}{2\times 3.14\times 9.1\times {10}^{-31}}\phantom{\rule{0ex}{0ex}}$

= 0.094 × 10^{8}

= 9.4 × 10^{6}* f* = 9.4 × 10^{6}

#### Page No 232:

#### Question 33:

Given:

Kinetic energy of an electron = 100 eV

Radius of the circle = 10 cm

$\frac{1}{2}m{v}^{2}$ = 100 eV = 1.6 × 10^{−17} J (1 eV = 1.6 × 10^{−19} J)

Here,*m* is the mass of an electron and *v* is the speed of an electron. Thus,

$\frac{1}{2}$ × 9.1 × 10^{−31} × *v*^{2} = 1.6 × 10^{−17} J

⇒ *v*^{2} = 0.35 × 10^{14}*v* = 0.591 × 10^{7} m/s

Now,

$r=\frac{mv}{eB}\phantom{\rule{0ex}{0ex}}\Rightarrow B=\frac{mv}{er}$

$=\frac{9.1\times {10}^{-31}\times 0.591\times {10}^{7}}{1.6\times {10}^{-19}\times 0.1}$*B* = 3.3613 × 10^{−4} T

Therefore, the applied magnetic field = 3.4 × 10^{−4} T

Number of revolutions per second of the electron,

$f=\frac{1}{T}$

$T=\frac{2\pi r}{v}=\frac{2\pi m}{eB}$

$T=\frac{2\mathrm{\pi}m}{Be}\phantom{\rule{0ex}{0ex}}$

$f=\frac{Be}{2\mathrm{\pi}m}\phantom{\rule{0ex}{0ex}}=\frac{3.4\times {10}^{-4}\times 1.6\times {10}^{-19}}{2\times 3.14\times 9.1\times {10}^{-31}}\phantom{\rule{0ex}{0ex}}$

= 0.094 × 10^{8}

= 9.4 × 10^{6}* f* = 9.4 × 10^{6}

#### Answer:

Given:

Kinetic energy of proton = *K*

Distance of the target from the accelerator = *l*

Therefore, radius of the circular orbit $\le $*l*

As per the question, the beam is bent by a perpendicular magnetic field.

We know:

$r=\frac{mv}{eB}$

For a proton, the above equation can be written as:

$l=\frac{{m}_{\mathrm{p}}v}{eB}$ (As *r=l)*....(i)

Here,*m*_{p}_{}is the mass of a proton

*v *is the velocity

*e *is the charge

*B *is the magnetic field

$\frac{1}{2}{m}_{\mathrm{p}}{v}^{2}=K\phantom{\rule{0ex}{0ex}}\Rightarrow v=\sqrt{\frac{2K}{{m}_{\mathrm{p}}}}$

Putting the value of *v* in the equation (i), we get:

$l=\frac{\sqrt{2K{m}_{\mathrm{p}}}}{eB}\phantom{\rule{0ex}{0ex}}B=\frac{\sqrt{2K{m}_{\mathrm{p}}}}{el}$

#### Page No 233:

#### Question 34:

Given:

Kinetic energy of proton = *K*

Distance of the target from the accelerator = *l*

Therefore, radius of the circular orbit $\le $*l*

As per the question, the beam is bent by a perpendicular magnetic field.

We know:

$r=\frac{mv}{eB}$

For a proton, the above equation can be written as:

$l=\frac{{m}_{\mathrm{p}}v}{eB}$ (As *r=l)*....(i)

Here,*m*_{p}_{}is the mass of a proton

*v *is the velocity

*e *is the charge

*B *is the magnetic field

$\frac{1}{2}{m}_{\mathrm{p}}{v}^{2}=K\phantom{\rule{0ex}{0ex}}\Rightarrow v=\sqrt{\frac{2K}{{m}_{\mathrm{p}}}}$

Putting the value of *v* in the equation (i), we get:

$l=\frac{\sqrt{2K{m}_{\mathrm{p}}}}{eB}\phantom{\rule{0ex}{0ex}}B=\frac{\sqrt{2K{m}_{\mathrm{p}}}}{el}$

#### Answer:

Given:

Applied potential difference, *V *= 12 kV = 12 × 10^{3} V

Speed of a charged particle, *v *=1.0 × 10^{6} m s^{−1}

Magnetic field strength, *B* = 0.2 T

As per the question, a charged particle is injected perpendicularly into the magnetic field.

We know:

$qV=\frac{1}{2}m{v}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{m}{q}=\frac{2V}{{v}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{2\times 12\times {10}^{3}}{{\left(1\times {10}^{6}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=24\times {10}^{-9}$

and $r=\frac{mv}{qB}$

$\Rightarrow r=\frac{24\times {10}^{-9}\times {10}^{6}}{0.2}$*$\Rightarrow $r* = 12 × 10^{−2} m = 12 cm

#### Page No 233:

#### Question 35:

Given:

Applied potential difference, *V *= 12 kV = 12 × 10^{3} V

Speed of a charged particle, *v *=1.0 × 10^{6} m s^{−1}

Magnetic field strength, *B* = 0.2 T

As per the question, a charged particle is injected perpendicularly into the magnetic field.

We know:

$qV=\frac{1}{2}m{v}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{m}{q}=\frac{2V}{{v}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{2\times 12\times {10}^{3}}{{\left(1\times {10}^{6}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=24\times {10}^{-9}$

and $r=\frac{mv}{qB}$

$\Rightarrow r=\frac{24\times {10}^{-9}\times {10}^{6}}{0.2}$*$\Rightarrow $r* = 12 × 10^{−2} m = 12 cm

#### Answer:

Given:

Speed of the helium ions, *v* = 10 km s^{−1} = 10^{4} m/s

Uniform magnetic field, *B* = 1.0 T

Charge on the helium ions = 2*e*

Mass of the helium ions, m = $4\times 1.6\times {10}^{-27}\mathrm{Kg}$

(a) The force acting on an ion,*F *= *qvB*sin*θ*

= 2 × 1.6 × 10^{−19}^{}× 10^{4} × 1.0

= 3.2 × 10^{−15} N

(b) The radius of the circle in which it circulates,

$r=\frac{mv}{qB}\phantom{\rule{0ex}{0ex}}=\frac{4\times 1.6\times {10}^{-27}\times {10}^{4}}{2\times 1.6\times {10}^{-19}\times 1}\phantom{\rule{0ex}{0ex}}=\frac{2\times {10}^{-23}}{{10}^{-19}}\phantom{\rule{0ex}{0ex}}=2\times {10}^{-4}\mathrm{m}$

(c) The time taken by an ion to complete the circle,

$T=\frac{2\mathrm{\pi}r}{v}\phantom{\rule{0ex}{0ex}}=\frac{6.28\times 2.1\times {10}^{-4}}{{10}^{4}}$

= 1.31 × 10^{−7} s

#### Page No 233:

#### Question 36:

Given:

Speed of the helium ions, *v* = 10 km s^{−1} = 10^{4} m/s

Uniform magnetic field, *B* = 1.0 T

Charge on the helium ions = 2*e*

Mass of the helium ions, m = $4\times 1.6\times {10}^{-27}\mathrm{Kg}$

(a) The force acting on an ion,*F *= *qvB*sin*θ*

= 2 × 1.6 × 10^{−19}^{}× 10^{4} × 1.0

= 3.2 × 10^{−15} N

(b) The radius of the circle in which it circulates,

$r=\frac{mv}{qB}\phantom{\rule{0ex}{0ex}}=\frac{4\times 1.6\times {10}^{-27}\times {10}^{4}}{2\times 1.6\times {10}^{-19}\times 1}\phantom{\rule{0ex}{0ex}}=\frac{2\times {10}^{-23}}{{10}^{-19}}\phantom{\rule{0ex}{0ex}}=2\times {10}^{-4}\mathrm{m}$

(c) The time taken by an ion to complete the circle,

$T=\frac{2\mathrm{\pi}r}{v}\phantom{\rule{0ex}{0ex}}=\frac{6.28\times 2.1\times {10}^{-4}}{{10}^{4}}$

= 1.31 × 10^{−7} s

#### Answer:

Given:

Velocity of the proton, *v* = 3 × 10^{6} m s^{−1}

Uniform magnetic field, B = 0.6 T

As per the question, the proton is projected perpendicular to a uniform magnetic field.

We know,*F* = *m*_{p}*a** *....(i)

and*F* = *evB*sin*θ** *....(ii)

On equating (i) and (ii), we get:*ma* = *evB*sin*θ* (As *θ* = 90Ëš)

$a=\frac{evB}{m}\phantom{\rule{0ex}{0ex}}=\frac{1.6\times {10}^{-19}\times 3\times {10}^{6}\times 0.6}{1.67\times {10}^{-27}}$

= 1.72 × 10^{14} m/s^{2}

#### Page No 233:

#### Question 37:

Given:

Velocity of the proton, *v* = 3 × 10^{6} m s^{−1}

Uniform magnetic field, B = 0.6 T

As per the question, the proton is projected perpendicular to a uniform magnetic field.

We know,*F* = *m*_{p}*a** *....(i)

and*F* = *evB*sin*θ** *....(ii)

On equating (i) and (ii), we get:*ma* = *evB*sin*θ* (As *θ* = 90Ëš)

$a=\frac{evB}{m}\phantom{\rule{0ex}{0ex}}=\frac{1.6\times {10}^{-19}\times 3\times {10}^{6}\times 0.6}{1.67\times {10}^{-27}}$

= 1.72 × 10^{14} m/s^{2}

#### Answer:

Given:

(a) Radius of the circle = 1 m

Magnetic field strength = 0.50 T

We know:

$r=\frac{{m}_{\mathrm{e}}{v}_{\mathrm{e}}}{Be}\phantom{\rule{0ex}{0ex}}{v}_{\mathrm{e}}=\frac{rBe}{{m}_{\mathrm{e}}},\text{where}{m}_{\mathrm{e}}\text{isthemassoftheelectron}\text{and}{v}_{\mathrm{e}}\text{isthespeedoftheelectron}.$

$=\frac{1\times 0.50\times 1.6\times {10}^{-19}}{9.1\times {10}^{-31}}\phantom{\rule{0ex}{0ex}}\approx 8.8\times {10}^{10}\mathrm{m}/\mathrm{s}$

Since, the speed of the electron moving along the circle is greater than the speed of light, it is not reasonable.

(b) For a proton,

$r=\frac{{m}_{\mathrm{p}}{v}_{\mathrm{p}}}{Be}\phantom{\rule{0ex}{0ex}}{v}_{\mathrm{p}}=\frac{rBe}{{m}_{\mathrm{p}}},$

where *m*_{p} is the mass of the proton and *v*_{p} is its speed.

$=\frac{1\times 0.50\times 1.6\times {10}^{-19}}{1.6\times {10}^{-27}}\phantom{\rule{0ex}{0ex}}=5\times {10}^{7}\mathrm{m}/\mathrm{s}$

#### Page No 233:

#### Question 38:

Given:

(a) Radius of the circle = 1 m

Magnetic field strength = 0.50 T

We know:

$r=\frac{{m}_{\mathrm{e}}{v}_{\mathrm{e}}}{Be}\phantom{\rule{0ex}{0ex}}{v}_{\mathrm{e}}=\frac{rBe}{{m}_{\mathrm{e}}},\text{where}{m}_{\mathrm{e}}\text{isthemassoftheelectron}\text{and}{v}_{\mathrm{e}}\text{isthespeedoftheelectron}.$

$=\frac{1\times 0.50\times 1.6\times {10}^{-19}}{9.1\times {10}^{-31}}\phantom{\rule{0ex}{0ex}}\approx 8.8\times {10}^{10}\mathrm{m}/\mathrm{s}$

Since, the speed of the electron moving along the circle is greater than the speed of light, it is not reasonable.

(b) For a proton,

$r=\frac{{m}_{\mathrm{p}}{v}_{\mathrm{p}}}{Be}\phantom{\rule{0ex}{0ex}}{v}_{\mathrm{p}}=\frac{rBe}{{m}_{\mathrm{p}}},$

where *m*_{p} is the mass of the proton and *v*_{p} is its speed.

$=\frac{1\times 0.50\times 1.6\times {10}^{-19}}{1.6\times {10}^{-27}}\phantom{\rule{0ex}{0ex}}=5\times {10}^{7}\mathrm{m}/\mathrm{s}$

#### Answer:

Given:

Mass of the particle = *m*

Positive charge on the particle = *q*

Uniform velocity of the particle = *v*

Magnetic field = *B*

(a) The radius of the circular arc described by the particle in the magnetic field:

We know,

$r=\frac{mv}{qB}$

(b)

The angle subtended by the arc at the centre:

Line MAB is tangent to arc ABC, so the angle described by the charged particle,

∠MAO = 90°

Now, ∠NAC = 90°

OAC = OCA = *θ*

[by geometry]

Then, AOC = ${180}^{\xb0}$ − (*θ* + *θ*) (By angle-sum property of a triangle)

= π − 2*θ*

(c) The time for which the particle stay inside the magnetic field:

Distance covered by the particle inside the magnetic field,*l* = *r*θ

= *r*(π − 2θ)

$t=\frac{l}{v}=\frac{m}{qB}(\mathrm{\pi}-2\mathrm{\theta})\left(\text{U}\mathrm{sing}\mathrm{r}=\frac{mv}{qB}\right)$

(d) If the charge *q* on the particle is negative, then

(i) Radius of circular arc, $r=\frac{mv}{qB}$

(ii) The centre of the arc will lie within the magnetic field. Therefore, the angle subtended by the arc = π + 2*θ*

(iii) Similarly, the time taken by the particle to cover the path inside the magnetic field = $\frac{m}{qB}(\mathrm{\pi}+2\theta )$

#### Page No 233:

#### Question 39:

Given:

Mass of the particle = *m*

Positive charge on the particle = *q*

Uniform velocity of the particle = *v*

Magnetic field = *B*

(a) The radius of the circular arc described by the particle in the magnetic field:

We know,

$r=\frac{mv}{qB}$

(b)

The angle subtended by the arc at the centre:

Line MAB is tangent to arc ABC, so the angle described by the charged particle,

∠MAO = 90°

Now, ∠NAC = 90°

OAC = OCA = *θ*

[by geometry]

Then, AOC = ${180}^{\xb0}$ − (*θ* + *θ*) (By angle-sum property of a triangle)

= π − 2*θ*

(c) The time for which the particle stay inside the magnetic field:

Distance covered by the particle inside the magnetic field,*l* = *r*θ

= *r*(π − 2θ)

$t=\frac{l}{v}=\frac{m}{qB}(\mathrm{\pi}-2\mathrm{\theta})\left(\text{U}\mathrm{sing}\mathrm{r}=\frac{mv}{qB}\right)$

(d) If the charge *q* on the particle is negative, then

(i) Radius of circular arc, $r=\frac{mv}{qB}$

(ii) The centre of the arc will lie within the magnetic field. Therefore, the angle subtended by the arc = π + 2*θ*

(iii) Similarly, the time taken by the particle to cover the path inside the magnetic field = $\frac{m}{qB}(\mathrm{\pi}+2\theta )$

#### Answer:

Given:

Mass of the particle = *m*

Charge of the particle = *q*

Magnetic field = *B*

As per the question, the particle is projected into a perpendicular magnetic field.

(a) When the width, *d* = $\frac{mv}{qB}$:*d* is equal to the radius and *θ* is the angle between the radius and tangent, which is equal to $\frac{\mathrm{\pi}}{2}$.

(b) When the width, *d = *$\frac{mv}{2qB}$

Width of the region in which a magnetic field is applied is half of the radius of the circular path described by the particle.

As the magnetic force is acting only along the y direction, the velocity of the particle will remain constant along the x direction. So, if *d* is the distance travelled along the x axis, then*d* = *v*_{x}*t*

$\mathrm{t}=\frac{d}{{\mathrm{v}}_{\mathrm{x}}}...\left(\mathrm{i}\right)$

(i) (ii)

The acceleration along the x direction is zero. The force will act only along the y direction.

Using the equation of motion for motion along the y axis:*v _{y}* =

*u*+

_{y}*a*

_{y}t${v}_{\mathrm{y}}=0+\frac{q{u}_{\mathrm{x}}Bt}{m}\phantom{\rule{0ex}{0ex}}=\frac{q{u}_{\mathrm{x}}Bt}{m}$

Putting the value of

*t*from equation (i), we get:

$\frac{q{u}_{\mathrm{x}}Bd}{m{v}_{\mathrm{x}}}$

We know:

$\mathrm{tan\theta}=\frac{{v}_{\mathrm{y}}}{{v}_{\mathrm{x}}}\phantom{\rule{0ex}{0ex}}\frac{qBd}{m{v}_{\mathrm{x}}}=\frac{qBm{v}_{\mathrm{x}}}{2qBm{v}_{\mathrm{x}}}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \theta ={\mathrm{tan}}^{-1}\left(\frac{1}{2}\right)\phantom{\rule{0ex}{0ex}}=26.4=30\xb0=\frac{\mathrm{\pi}}{6}$

(c) When the width,

*d =*$\frac{2mv}{qB}$

From the figure, it can be seen that the angle between the initial direction and final direction of velocity is π.

#### Page No 233:

#### Question 40:

Given:

Mass of the particle = *m*

Charge of the particle = *q*

Magnetic field = *B*

As per the question, the particle is projected into a perpendicular magnetic field.

(a) When the width, *d* = $\frac{mv}{qB}$:*d* is equal to the radius and *θ* is the angle between the radius and tangent, which is equal to $\frac{\mathrm{\pi}}{2}$.

(b) When the width, *d = *$\frac{mv}{2qB}$

Width of the region in which a magnetic field is applied is half of the radius of the circular path described by the particle.

As the magnetic force is acting only along the y direction, the velocity of the particle will remain constant along the x direction. So, if *d* is the distance travelled along the x axis, then*d* = *v*_{x}*t*

$\mathrm{t}=\frac{d}{{\mathrm{v}}_{\mathrm{x}}}...\left(\mathrm{i}\right)$

(i) (ii)

The acceleration along the x direction is zero. The force will act only along the y direction.

Using the equation of motion for motion along the y axis:*v _{y}* =

*u*+

_{y}*a*

_{y}t${v}_{\mathrm{y}}=0+\frac{q{u}_{\mathrm{x}}Bt}{m}\phantom{\rule{0ex}{0ex}}=\frac{q{u}_{\mathrm{x}}Bt}{m}$

Putting the value of

*t*from equation (i), we get:

$\frac{q{u}_{\mathrm{x}}Bd}{m{v}_{\mathrm{x}}}$

We know:

$\mathrm{tan\theta}=\frac{{v}_{\mathrm{y}}}{{v}_{\mathrm{x}}}\phantom{\rule{0ex}{0ex}}\frac{qBd}{m{v}_{\mathrm{x}}}=\frac{qBm{v}_{\mathrm{x}}}{2qBm{v}_{\mathrm{x}}}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \theta ={\mathrm{tan}}^{-1}\left(\frac{1}{2}\right)\phantom{\rule{0ex}{0ex}}=26.4=30\xb0=\frac{\mathrm{\pi}}{6}$

(c) When the width,

*d =*$\frac{2mv}{qB}$

From the figure, it can be seen that the angle between the initial direction and final direction of velocity is π.

#### Answer:

Given:

Velocity of a narrow beam of singly-charged carbon ions, *v* = 6.0 × 10^{4} m s^{−1}

Strength of magnetic field *B* = 0.5 T

Separations between the two beams from the incident beam are 3.0 cm and 3.5 cm.

Mass of an ion = *A*(1.6 × 10^{−27}) kg

The radius of the curved path taken by the first beam,* ${r}_{1}=\frac{3}{2}=1.5\mathrm{cm}$*

The radius of the curved path taken by the second beam, ${r}_{2}=\frac{3.5}{2}\mathrm{cm}$

We know:

${r}_{1}=\frac{{m}_{1}v}{qB}$,

where *m*_{1} is the mass of the first isotope and *q* is the charge.

For the second beam:

${r}_{2}=\frac{{m}_{2}v}{qB}$,

where *m*_{2} is the mass of the first isotope and *q* is the charge.

$\frac{{r}_{1}}{{r}_{2}}=\frac{{m}_{1}}{{m}_{2}}\phantom{\rule{0ex}{0ex}}\frac{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{{\displaystyle \raisebox{1ex}{$3.5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=\frac{{A}_{1}\times 1.6\times {10}^{-27}}{{A}_{2}\times 1.6\times {10}^{-27}}\phantom{\rule{0ex}{0ex}}\frac{6}{7}=\frac{{A}_{1}}{{A}_{2}}$

As

${r}_{1}=\frac{{m}_{1}v}{qB}$

$\Rightarrow {m}_{1}=\frac{qB{r}_{1}}{v}\phantom{\rule{0ex}{0ex}}=\frac{1.6\times {10}^{-19}\times 0.5\times 0.015}{6\times {10}^{4}}\phantom{\rule{0ex}{0ex}}=20\times {10}^{-27}\mathrm{kg}\phantom{\rule{0ex}{0ex}}=\frac{20\times {10}^{-27}}{1.6\times {10}^{-27}}\mathrm{u}\phantom{\rule{0ex}{0ex}}=12.5\mathrm{u}$

Also,

${A}_{2}=\frac{7}{6}{A}_{1}\phantom{\rule{0ex}{0ex}}=\frac{7}{6}\times 12.5\phantom{\rule{0ex}{0ex}}=14.58\mathrm{u}$

So, the two isotopes of carbon used are ^{12}C_{6} and ^{14}C_{6}.

#### Page No 233:

#### Question 41:

Given:

Velocity of a narrow beam of singly-charged carbon ions, *v* = 6.0 × 10^{4} m s^{−1}

Strength of magnetic field *B* = 0.5 T

Separations between the two beams from the incident beam are 3.0 cm and 3.5 cm.

Mass of an ion = *A*(1.6 × 10^{−27}) kg

The radius of the curved path taken by the first beam,* ${r}_{1}=\frac{3}{2}=1.5\mathrm{cm}$*

The radius of the curved path taken by the second beam, ${r}_{2}=\frac{3.5}{2}\mathrm{cm}$

We know:

${r}_{1}=\frac{{m}_{1}v}{qB}$,

where *m*_{1} is the mass of the first isotope and *q* is the charge.

For the second beam:

${r}_{2}=\frac{{m}_{2}v}{qB}$,

where *m*_{2} is the mass of the first isotope and *q* is the charge.

$\frac{{r}_{1}}{{r}_{2}}=\frac{{m}_{1}}{{m}_{2}}\phantom{\rule{0ex}{0ex}}\frac{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{{\displaystyle \raisebox{1ex}{$3.5$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=\frac{{A}_{1}\times 1.6\times {10}^{-27}}{{A}_{2}\times 1.6\times {10}^{-27}}\phantom{\rule{0ex}{0ex}}\frac{6}{7}=\frac{{A}_{1}}{{A}_{2}}$

As

${r}_{1}=\frac{{m}_{1}v}{qB}$

$\Rightarrow {m}_{1}=\frac{qB{r}_{1}}{v}\phantom{\rule{0ex}{0ex}}=\frac{1.6\times {10}^{-19}\times 0.5\times 0.015}{6\times {10}^{4}}\phantom{\rule{0ex}{0ex}}=20\times {10}^{-27}\mathrm{kg}\phantom{\rule{0ex}{0ex}}=\frac{20\times {10}^{-27}}{1.6\times {10}^{-27}}\mathrm{u}\phantom{\rule{0ex}{0ex}}=12.5\mathrm{u}$

Also,

${A}_{2}=\frac{7}{6}{A}_{1}\phantom{\rule{0ex}{0ex}}=\frac{7}{6}\times 12.5\phantom{\rule{0ex}{0ex}}=14.58\mathrm{u}$

So, the two isotopes of carbon used are ^{12}C_{6} and ^{14}C_{6}.

#### Answer:

Given:

Potential difference through which the Fe^{+} ions are accelerated, *V =* 500 V

Strength of the homogeneous magnetic field, *B* = 20.0 mT = 20 × 10^{−3} T

Mass numbers of the two isotopes are 57 and 58.

Mass of an ion = *A* (1.6 × 10^{−27}) kg

We know that the radius of the circular path described by a particle in a magnetic field,

$r=\frac{mv}{qB}\phantom{\rule{0ex}{0ex}}\mathrm{For}\mathrm{isotope}1,\phantom{\rule{0ex}{0ex}}{r}_{1}=\frac{{m}_{1}{v}_{1}}{qB}\phantom{\rule{0ex}{0ex}}\mathrm{For}\mathrm{isotope}2,\phantom{\rule{0ex}{0ex}}{r}_{2}=\frac{{m}_{2}{v}_{2}}{qB}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{r}_{1}}{{r}_{2}}=\frac{{m}_{1}{v}_{1}}{{m}_{2}{v}_{2}}\phantom{\rule{0ex}{0ex}}$

As both the isotopes are accelerated via the same potential V, the K.E gained by the two particles will be same.

$qV=\frac{1}{2}{m}_{1}{{v}_{1}}^{2}=\frac{1}{2}{m}_{2}{{v}_{2}}^{2}\phantom{\rule{0ex}{0ex}}\frac{{m}_{1}}{{m}_{2}}=\frac{{{v}_{1}}^{2}}{{{v}_{2}}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{r}_{1}}{{r}_{2}}=(\frac{{m}_{1}}{{m}_{2}}{)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

$\mathrm{Also},{r}_{1}=\frac{m{v}_{1}}{qB}\phantom{\rule{0ex}{0ex}}=\frac{{m}_{1}\sqrt{\frac{2qV}{{m}_{1}}}}{qB}\phantom{\rule{0ex}{0ex}}=\frac{1}{B}\sqrt{\frac{2{m}_{1}V}{q}}\phantom{\rule{0ex}{0ex}}$

$=\frac{\sqrt{1000\times 57\times 1.6\times {10}^{-27}}}{\sqrt{1.6\times {10}^{-19}}\times 20\times {10}^{-3}}\phantom{\rule{0ex}{0ex}}=1.19\times {10}^{-2}\mathrm{m}=119\mathrm{cm}$

For the second isotope:

$\mathrm{As}\frac{{r}_{1}}{{r}_{2}}=(\frac{{m}_{1}}{{m}_{2}}{)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.},\phantom{\rule{0ex}{0ex}}{r}_{2}=(\frac{{m}_{2}}{{m}_{1}}{)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{r}_{1}\phantom{\rule{0ex}{0ex}}=(\frac{58}{57}{)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\times 119\mathrm{cm}\phantom{\rule{0ex}{0ex}}=120\mathrm{cm}$

#### Page No 233:

#### Question 42:

Given:

Potential difference through which the Fe^{+} ions are accelerated, *V =* 500 V

Strength of the homogeneous magnetic field, *B* = 20.0 mT = 20 × 10^{−3} T

Mass numbers of the two isotopes are 57 and 58.

Mass of an ion = *A* (1.6 × 10^{−27}) kg

We know that the radius of the circular path described by a particle in a magnetic field,

$r=\frac{mv}{qB}\phantom{\rule{0ex}{0ex}}\mathrm{For}\mathrm{isotope}1,\phantom{\rule{0ex}{0ex}}{r}_{1}=\frac{{m}_{1}{v}_{1}}{qB}\phantom{\rule{0ex}{0ex}}\mathrm{For}\mathrm{isotope}2,\phantom{\rule{0ex}{0ex}}{r}_{2}=\frac{{m}_{2}{v}_{2}}{qB}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{r}_{1}}{{r}_{2}}=\frac{{m}_{1}{v}_{1}}{{m}_{2}{v}_{2}}\phantom{\rule{0ex}{0ex}}$

As both the isotopes are accelerated via the same potential V, the K.E gained by the two particles will be same.

$qV=\frac{1}{2}{m}_{1}{{v}_{1}}^{2}=\frac{1}{2}{m}_{2}{{v}_{2}}^{2}\phantom{\rule{0ex}{0ex}}\frac{{m}_{1}}{{m}_{2}}=\frac{{{v}_{1}}^{2}}{{{v}_{2}}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{r}_{1}}{{r}_{2}}=(\frac{{m}_{1}}{{m}_{2}}{)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

$\mathrm{Also},{r}_{1}=\frac{m{v}_{1}}{qB}\phantom{\rule{0ex}{0ex}}=\frac{{m}_{1}\sqrt{\frac{2qV}{{m}_{1}}}}{qB}\phantom{\rule{0ex}{0ex}}=\frac{1}{B}\sqrt{\frac{2{m}_{1}V}{q}}\phantom{\rule{0ex}{0ex}}$

$=\frac{\sqrt{1000\times 57\times 1.6\times {10}^{-27}}}{\sqrt{1.6\times {10}^{-19}}\times 20\times {10}^{-3}}\phantom{\rule{0ex}{0ex}}=1.19\times {10}^{-2}\mathrm{m}=119\mathrm{cm}$

For the second isotope:

$\mathrm{As}\frac{{r}_{1}}{{r}_{2}}=(\frac{{m}_{1}}{{m}_{2}}{)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.},\phantom{\rule{0ex}{0ex}}{r}_{2}=(\frac{{m}_{2}}{{m}_{1}}{)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{r}_{1}\phantom{\rule{0ex}{0ex}}=(\frac{58}{57}{)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\times 119\mathrm{cm}\phantom{\rule{0ex}{0ex}}=120\mathrm{cm}$

#### Answer:

Kinetic energy of singly-charged potassium ions = 32 keV

Width of the magnetic region = 1.00 cm

Magnetic field's strength, *B =* 0.500 T

Distance between the screen and the region = 95.5 cm

Atomic weights of the two isotopes are 39 and 41.

Mass of a potassium ion = *A* (1.6 × 10^{−27}) kg

For a singly-charged potassium ion K-39 :

Mass of K-39 = 39 × 1.6 × 10^{−27} kg,

Charge, *q* = 1.6 × 10^{−19} C

As per the question, the narrow beam of singly-charged potassium ions is injected into a region of magnetic field.

As

K.E = 32 keV

$\frac{1}{2}m{v}^{2}=32\times {10}^{3}\times 1.6\times {10}^{-19}$

$\frac{1}{2}$ × 39 × (1.6 × 10^{−27}) × *v*^{2} = 32 × 10^{3} × 1.6 × 10^{−19}*v* = 4.05 × 10^{5}

We know that throughout the motion, the horizontal velocity remains constant.

So, the time taken to cross the magnetic field,*t* = $\frac{d}{v}=\frac{0.01}{4.05\times {10}^{5}}$

= 24.7 × 10^{−9} s

Now, the acceleration in the magnetic field region:*F* = *qvB = ma$a=\frac{qvB}{m}\phantom{\rule{0ex}{0ex}}=\frac{1.6\times {10}^{-19}\times 4.05\times {10}^{5}}{39\times 1.6\times {10}^{-27}}$*

= 5192 × 10

^{8}m/s

^{2}

Velocity in the vertical direction,

*v*

_{y}=

*at*

= 5193.53 × 10

^{8}× 24.7 × 10

^{−9}

= 12824.24 m/s

Time taken to reach the screen

$=\frac{d}{v}=\frac{0.955}{4.05\times {10}^{5}}$

= 0.000002358 s.

Distance moved vertically in this time

=

*v*

_{y}×

*t*

= 12824.24 × 2358 × 10

^{−9}

= 3023.95×10

^{-5}m

Vertical distance travelled by the particle inside magnetic field can be found out by using equaton of motion

*v*

^{2}

^{}= 2

*aS*

⇒ (12824.24)

^{2}= 2 × 5192 × 10

^{8}×

*S*

⇒ 15.83×10

^{−5}=

*S*

Net display from line

= 15.83×10

^{−5}+3023.95×10

^{-5}

= 3039.787×10

^{−5}m.

For the potassium ion K-41

*l*:

$\frac{1}{2}$ × 41 × 1.6 × 10

^{−27}

*v*

^{2}= 32 × 10

^{3}× 1.6 × 10

^{−9}

⇒

*v*= 3.94×10

^{5 }m/s

Similarly, acceleration,

*a*= 4805 × 10

^{8}m/s

^{2}

*t*= time taken for exiting the magnetic field

= 25.4 × 10

^{−9}sec.

*v*

_{y1}=

*at*(vertical velocity)

= 4805 × 10

^{8}× 25.4 × 10

^{−9}

= 12204.7× 10

^{−9}m/s

Time to reach the screen

= 2423× 10

^{−9}s.

Distance moved vertically

= 12204.7 × 2423 × 10

^{−9}

= 2957.1× 10

^{−5}

Now,

Vertical distance travelled by the particle inside magnetic field can be found out by using equaton of motion

*v*

^{2}

^{}= 2

*aS*

(12204.7)

^{2}= 2 × 4805 × 10

^{8}S

⇒ S = 15.49× 10

^{-5}m

Net distance travelled

= 15.49× 10

^{-5}+2957.1× 10

^{−5}

= 2972.68× 10

^{-5}m.

Net gap between K-39 and K-41

= 3039.787×10

^{−5}− 2972.68× 10

^{-5}

= 67 mm.

#### Page No 233:

#### Question 43:

Kinetic energy of singly-charged potassium ions = 32 keV

Width of the magnetic region = 1.00 cm

Magnetic field's strength, *B =* 0.500 T

Distance between the screen and the region = 95.5 cm

Atomic weights of the two isotopes are 39 and 41.

Mass of a potassium ion = *A* (1.6 × 10^{−27}) kg

For a singly-charged potassium ion K-39 :

Mass of K-39 = 39 × 1.6 × 10^{−27} kg,

Charge, *q* = 1.6 × 10^{−19} C

As per the question, the narrow beam of singly-charged potassium ions is injected into a region of magnetic field.

As

K.E = 32 keV

$\frac{1}{2}m{v}^{2}=32\times {10}^{3}\times 1.6\times {10}^{-19}$

$\frac{1}{2}$ × 39 × (1.6 × 10^{−27}) × *v*^{2} = 32 × 10^{3} × 1.6 × 10^{−19}*v* = 4.05 × 10^{5}

We know that throughout the motion, the horizontal velocity remains constant.

So, the time taken to cross the magnetic field,*t* = $\frac{d}{v}=\frac{0.01}{4.05\times {10}^{5}}$

= 24.7 × 10^{−9} s

Now, the acceleration in the magnetic field region:*F* = *qvB = ma$a=\frac{qvB}{m}\phantom{\rule{0ex}{0ex}}=\frac{1.6\times {10}^{-19}\times 4.05\times {10}^{5}}{39\times 1.6\times {10}^{-27}}$*

= 5192 × 10

^{8}m/s

^{2}

Velocity in the vertical direction,

*v*

_{y}=

*at*

= 5193.53 × 10

^{8}× 24.7 × 10

^{−9}

= 12824.24 m/s

Time taken to reach the screen

$=\frac{d}{v}=\frac{0.955}{4.05\times {10}^{5}}$

= 0.000002358 s.

Distance moved vertically in this time

=

*v*

_{y}×

*t*

= 12824.24 × 2358 × 10

^{−9}

= 3023.95×10

^{-5}m

Vertical distance travelled by the particle inside magnetic field can be found out by using equaton of motion

*v*

^{2}

^{}= 2

*aS*

⇒ (12824.24)

^{2}= 2 × 5192 × 10

^{8}×

*S*

⇒ 15.83×10

^{−5}=

*S*

Net display from line

= 15.83×10

^{−5}+3023.95×10

^{-5}

= 3039.787×10

^{−5}m.

For the potassium ion K-41

*l*:

$\frac{1}{2}$ × 41 × 1.6 × 10

^{−27}

*v*

^{2}= 32 × 10

^{3}× 1.6 × 10

^{−9}

⇒

*v*= 3.94×10

^{5 }m/s

Similarly, acceleration,

*a*= 4805 × 10

^{8}m/s

^{2}

*t*= time taken for exiting the magnetic field

= 25.4 × 10

^{−9}sec.

*v*

_{y1}=

*at*(vertical velocity)

= 4805 × 10

^{8}× 25.4 × 10

^{−9}

= 12204.7× 10

^{−9}m/s

Time to reach the screen

= 2423× 10

^{−9}s.

Distance moved vertically

= 12204.7 × 2423 × 10

^{−9}

= 2957.1× 10

^{−5}

Now,

Vertical distance travelled by the particle inside magnetic field can be found out by using equaton of motion

*v*

^{2}

^{}= 2

*aS*

(12204.7)

^{2}= 2 × 4805 × 10

^{8}S

⇒ S = 15.49× 10

^{-5}m

Net distance travelled

= 15.49× 10

^{-5}+2957.1× 10

^{−5}

= 2972.68× 10

^{-5}m.

Net gap between K-39 and K-41

= 3039.787×10

^{−5}− 2972.68× 10

^{-5}

= 67 mm.

#### Answer:

Given:

Focal length of the convex lens = 12 cm

Uniform magnetic field, *B =* 1.2 T

Charge of the particle, *q =* 2.0 × 10^{−3}^{}C

and mass, *m =* 2.0 × 10^{−5} kg

Speed of the particle, *v = *4.8 m s^{−1}

The distance of the particle from the lens = 18 cm

As per the question, the object is projected perpendicular to the plane of the paper.

Let the radius of the circle on which the object is moving be *r*.

We know:

$r=\frac{mv}{qB}$

$r=\frac{2\times {10}^{-5}\times 4.8}{2\times {10}^{-3}\times 1.2}\phantom{\rule{0ex}{0ex}}r=0.04m=4cm$

Here, object distance*, u* = -18 cm

Using the lens equation

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f},\phantom{\rule{0ex}{0ex}}\frac{1}{v}-\frac{1}{\left(-18\right)}=\frac{1}{12}$

Image distance, *v* = 36 cm.

Let the radius of the circular path of image be *r*'.

So, magnification:

$\frac{v}{u}=\frac{r\text{'}}{r}\phantom{\rule{0ex}{0ex}}r\text{'}=\frac{v}{u}\times r\phantom{\rule{0ex}{0ex}}=8\mathrm{cm}$

Therefore, the radius of the circular path in which the image moves is 8 cm.

#### Page No 233:

#### Question 44:

Given:

Focal length of the convex lens = 12 cm

Uniform magnetic field, *B =* 1.2 T

Charge of the particle, *q =* 2.0 × 10^{−3}^{}C

and mass, *m =* 2.0 × 10^{−5} kg

Speed of the particle, *v = *4.8 m s^{−1}

The distance of the particle from the lens = 18 cm

As per the question, the object is projected perpendicular to the plane of the paper.

Let the radius of the circle on which the object is moving be *r*.

We know:

$r=\frac{mv}{qB}$

$r=\frac{2\times {10}^{-5}\times 4.8}{2\times {10}^{-3}\times 1.2}\phantom{\rule{0ex}{0ex}}r=0.04m=4cm$

Here, object distance*, u* = -18 cm

Using the lens equation

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f},\phantom{\rule{0ex}{0ex}}\frac{1}{v}-\frac{1}{\left(-18\right)}=\frac{1}{12}$

Image distance, *v* = 36 cm.

Let the radius of the circular path of image be *r*'.

So, magnification:

$\frac{v}{u}=\frac{r\text{'}}{r}\phantom{\rule{0ex}{0ex}}r\text{'}=\frac{v}{u}\times r\phantom{\rule{0ex}{0ex}}=8\mathrm{cm}$

Therefore, the radius of the circular path in which the image moves is 8 cm.

#### Answer:

Given:

Electrons are accelerated through a potential difference = *V*

Let the mass of an electron be *m* and the charge of an electron be *e*.

We know:

Electric field, *E* = $\frac{V}{r}$

Force experienced by the electron, *F* = *eE*

Acceleration of the electron, *a* = $\frac{eV}{rm}$

Using the equation of motion*v*^{2} − *u*^{2}= 2 × *a* × *s*,*v*^{2} = 2 × *a* × *s (*As *u* = 0)

Here, *s* = *r*

$v=\sqrt{\frac{2eVr}{rm}}\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{2eV}{m}}$

Time taken by electron to cover the curved path,

${T}{=}\frac{2\pi m}{eB}$

As the acceleration of the electron is along the *y* axis only, it travels along the *x* axis with uniform velocity.

Velocity of the electron moving along the field remains *v*.

Therefore, the distance at which the beam is refocused, *d* = *v* × *T**$d=\sqrt{\frac{2eV}{m}}\times \frac{2\pi m}{eB}\phantom{\rule{0ex}{0ex}}d=\sqrt{\frac{8{\pi}^{2}mV}{e{B}^{2}}}$*

#### Page No 234:

#### Question 45:

Given:

Electrons are accelerated through a potential difference = *V*

Let the mass of an electron be *m* and the charge of an electron be *e*.

We know:

Electric field, *E* = $\frac{V}{r}$

Force experienced by the electron, *F* = *eE*

Acceleration of the electron, *a* = $\frac{eV}{rm}$

Using the equation of motion*v*^{2} − *u*^{2}= 2 × *a* × *s*,*v*^{2} = 2 × *a* × *s (*As *u* = 0)

Here, *s* = *r*

$v=\sqrt{\frac{2eVr}{rm}}\phantom{\rule{0ex}{0ex}}=\sqrt{\frac{2eV}{m}}$

Time taken by electron to cover the curved path,

${T}{=}\frac{2\pi m}{eB}$

As the acceleration of the electron is along the *y* axis only, it travels along the *x* axis with uniform velocity.

Velocity of the electron moving along the field remains *v*.

Therefore, the distance at which the beam is refocused, *d* = *v* × *T**$d=\sqrt{\frac{2eV}{m}}\times \frac{2\pi m}{eB}\phantom{\rule{0ex}{0ex}}d=\sqrt{\frac{8{\pi}^{2}mV}{e{B}^{2}}}$*

#### Answer:

Given,

Mass of two particles = *m*

Distance between them = *d*

Both the particles have equal charges in magnitude but opposite polarity equal to *q.*

As per the question, both the particles are projected towards each other with equal speed *v*.

It is assumed that Coulomb force between the charges is switched off.

(a) The maximum value *v _{m}* of the projection speed so that the two particles do not collide:

Both the particles will not collide if

*d*=

*r*

_{1}+

*r*

_{2}(where,

*r*

_{1}=

*r*

_{2}= radius of circular orbit described by the charged particles)

$\Rightarrow d=\frac{m{v}_{\mathrm{m}}}{qB}+\frac{m{v}_{\mathrm{m}}}{qB}=\frac{2m{\mathrm{v}}_{\mathrm{m}}}{qB}\phantom{\rule{0ex}{0ex}}\mathrm{So},{v}_{\mathrm{m}}=\frac{qBd}{2m}$

(b) The minimum and maximum separation between the particles if

*v*=

*v*/2:

_{m}Let the radius of the curved path taken by the particles, when they are projected with speed

*v*/2, be

_{m}*r*.

So, minimum separation between the particles = (

*d*$-$ 2

*r*)

$\Rightarrow (d-2r)=\frac{2m{v}_{\mathrm{m}}}{qB}-\frac{2mv}{qB}\phantom{\rule{0ex}{0ex}}\Rightarrow (d-2r)=\frac{m{v}_{\mathrm{m}}}{qB}\phantom{\rule{0ex}{0ex}}\Rightarrow (d-2r)=\frac{d}{2}\phantom{\rule{0ex}{0ex}}$

Maximum distance separation = (

*d*+ 2

*r*)

$\Rightarrow d+2r=d+\frac{d}{2}=\frac{3d}{2}$

(c) The instant at which the collision occurs between the particles when $v=2{v}_{\mathrm{m}}$:

The particles will collide at a distance

*d*/2 along the horizontal direction.

Let they collide after time

*t*.

Velocity of the particles along the horizontal direction will remain the same.

Therefore, $t=\frac{d/2}{2{v}_{m}}$

$\Rightarrow t=\frac{d}{4}\times \frac{2m}{qBd}\phantom{\rule{0ex}{0ex}}\Rightarrow t=\frac{m}{2qB}$

(d) The motion of the two particles after collision when the collision is completely inelastic:

*v*= 2

*v*

_{m}

Let the particles collide at point P.

And at point P, both the particles will have motion in upward direction.

As the collision is inelastic they stick together.

Distance between centres =

*d*

Velocity along the horizontal direction does not get affected due to the magnetic force.

At point P, velocities along the horizontal direction are equal and opposite. So, they cancel each other.

Velocity along the vertical direction (upward) will add up.

Magnetic force acting along the vertical direction,

$F=q\left(2{v}_{m}\right)B$

Acceleration along the vertical direction,

$a=\frac{F}{m}=\frac{2q{v}_{m}B}{m}$

Velocity of the combined mass at point P is along the vertical direction. So,

$v\text{'}=0+a\times t\phantom{\rule{0ex}{0ex}}v\text{'}=0+\left(\frac{2q{v}_{m}B}{m}\right)\times \left(\frac{m}{2qB}\right)\phantom{\rule{0ex}{0ex}}v\text{'}={v}_{m}$

Hence, both the particles will behave as a combined mass and move with velocity

*v*

_{m}.

#### Page No 234:

#### Question 46:

Given,

Mass of two particles = *m*

Distance between them = *d*

Both the particles have equal charges in magnitude but opposite polarity equal to *q.*

As per the question, both the particles are projected towards each other with equal speed *v*.

It is assumed that Coulomb force between the charges is switched off.

(a) The maximum value *v _{m}* of the projection speed so that the two particles do not collide:

Both the particles will not collide if

*d*=

*r*

_{1}+

*r*

_{2}(where,

*r*

_{1}=

*r*

_{2}= radius of circular orbit described by the charged particles)

$\Rightarrow d=\frac{m{v}_{\mathrm{m}}}{qB}+\frac{m{v}_{\mathrm{m}}}{qB}=\frac{2m{\mathrm{v}}_{\mathrm{m}}}{qB}\phantom{\rule{0ex}{0ex}}\mathrm{So},{v}_{\mathrm{m}}=\frac{qBd}{2m}$

(b) The minimum and maximum separation between the particles if

*v*=

*v*/2:

_{m}Let the radius of the curved path taken by the particles, when they are projected with speed

*v*/2, be

_{m}*r*.

So, minimum separation between the particles = (

*d*$-$ 2

*r*)

$\Rightarrow (d-2r)=\frac{2m{v}_{\mathrm{m}}}{qB}-\frac{2mv}{qB}\phantom{\rule{0ex}{0ex}}\Rightarrow (d-2r)=\frac{m{v}_{\mathrm{m}}}{qB}\phantom{\rule{0ex}{0ex}}\Rightarrow (d-2r)=\frac{d}{2}\phantom{\rule{0ex}{0ex}}$

Maximum distance separation = (

*d*+ 2

*r*)

$\Rightarrow d+2r=d+\frac{d}{2}=\frac{3d}{2}$

(c) The instant at which the collision occurs between the particles when $v=2{v}_{\mathrm{m}}$:

The particles will collide at a distance

*d*/2 along the horizontal direction.

Let they collide after time

*t*.

Velocity of the particles along the horizontal direction will remain the same.

Therefore, $t=\frac{d/2}{2{v}_{m}}$

$\Rightarrow t=\frac{d}{4}\times \frac{2m}{qBd}\phantom{\rule{0ex}{0ex}}\Rightarrow t=\frac{m}{2qB}$

(d) The motion of the two particles after collision when the collision is completely inelastic:

*v*= 2

*v*

_{m}

Let the particles collide at point P.

And at point P, both the particles will have motion in upward direction.

As the collision is inelastic they stick together.

Distance between centres =

*d*

Velocity along the horizontal direction does not get affected due to the magnetic force.

At point P, velocities along the horizontal direction are equal and opposite. So, they cancel each other.

Velocity along the vertical direction (upward) will add up.

Magnetic force acting along the vertical direction,

$F=q\left(2{v}_{m}\right)B$

Acceleration along the vertical direction,

$a=\frac{F}{m}=\frac{2q{v}_{m}B}{m}$

Velocity of the combined mass at point P is along the vertical direction. So,

$v\text{'}=0+a\times t\phantom{\rule{0ex}{0ex}}v\text{'}=0+\left(\frac{2q{v}_{m}B}{m}\right)\times \left(\frac{m}{2qB}\right)\phantom{\rule{0ex}{0ex}}v\text{'}={v}_{m}$

Hence, both the particles will behave as a combined mass and move with velocity

*v*

_{m}.

#### Answer:

Given:

Uniform magnetic field, *B* = 0.20 T

Mass of the particle, *m* = 0.010 g = 1 × 10^{−5} kg

Charge of the particle, *q =* 1.0 × 10^{−5} C

As per the question, if the particle has to move with uniform velocity in the region of the applied field,

gravitational force experienced by the particle should be equal to the magnetic force experienced by the particle.

So, *qvB* = *mg*, where* v* is the uniform velocity and *g* is the acceleration due to gravity.

⇒ 1 × 10^{−5} × *v* × 2 × 10^{−1} = 1 × 10^{−5} × 9.8

⇒ *v* = 4.9 × 10 = 49 m/s

#### Page No 234:

#### Question 47:

Given:

Uniform magnetic field, *B* = 0.20 T

Mass of the particle, *m* = 0.010 g = 1 × 10^{−5} kg

Charge of the particle, *q =* 1.0 × 10^{−5} C

As per the question, if the particle has to move with uniform velocity in the region of the applied field,

gravitational force experienced by the particle should be equal to the magnetic force experienced by the particle.

So, *qvB* = *mg*, where* v* is the uniform velocity and *g* is the acceleration due to gravity.

⇒ 1 × 10^{−5} × *v* × 2 × 10^{−1} = 1 × 10^{−5} × 9.8

⇒ *v* = 4.9 × 10 = 49 m/s

#### Answer:

Given:

Diameter of the circle = 1.0 cm

Thus, radius of circle, *r *= = 0.5 × 10^{−2} m,

Magnetic field, *B* = 0.40 T

Electric field, *E* = 200 V m^{−1}.

As per the question, the particle is moving in a circle under the action of a magnetic field. But when an electric field is applied on the particle, it moves in a straight line.

So, we can write:*F*_{e} = *F*_{m}*qE* = qvB, where *q* is the charge and *v* is the velocity of the particle.

$\Rightarrow v=\frac{E}{B}=\frac{200}{0.4}=500\mathrm{m}/\mathrm{s}.\phantom{\rule{0ex}{0ex}}\mathrm{As}r=\frac{mv}{qB},\phantom{\rule{0ex}{0ex}}\frac{q}{m}=\frac{v}{rB}\phantom{\rule{0ex}{0ex}}=\frac{500}{0.5\times {10}^{-2}\times 0.4}\phantom{\rule{0ex}{0ex}}=2.5\times {10}^{5}\mathrm{C}/\mathrm{kg}$

#### Page No 234:

#### Question 48:

Given:

Diameter of the circle = 1.0 cm

Thus, radius of circle, *r *= = 0.5 × 10^{−2} m,

Magnetic field, *B* = 0.40 T

Electric field, *E* = 200 V m^{−1}.

As per the question, the particle is moving in a circle under the action of a magnetic field. But when an electric field is applied on the particle, it moves in a straight line.

So, we can write:*F*_{e} = *F*_{m}*qE* = qvB, where *q* is the charge and *v* is the velocity of the particle.

$\Rightarrow v=\frac{E}{B}=\frac{200}{0.4}=500\mathrm{m}/\mathrm{s}.\phantom{\rule{0ex}{0ex}}\mathrm{As}r=\frac{mv}{qB},\phantom{\rule{0ex}{0ex}}\frac{q}{m}=\frac{v}{rB}\phantom{\rule{0ex}{0ex}}=\frac{500}{0.5\times {10}^{-2}\times 0.4}\phantom{\rule{0ex}{0ex}}=2.5\times {10}^{5}\mathrm{C}/\mathrm{kg}$

#### Answer:

Given:

Mass of the proton, *m* = 1.6 × 10^{−27} kg

Speed of the proton inside the crossed electric and magnetic field, *v *= 2.0 × 10^{5} ms^{−1}

As per the question, the proton is not deflected under the combined action of the electric and magnetic fields. Thus, the forces applied by both the fields are equal and opposite.

That is, *qE = qvB *

⇒ *E = vB* ...(1)

But when the electric field is switched off, the proton moves in a circle due to the force of the magnetic field.

Radius of the circle, *r* = 4.0 cm = 4 × 10^{−2} m

We know: *r* = $\frac{mv}{qB}$

$\Rightarrow B=\frac{mv}{qr}\phantom{\rule{0ex}{0ex}}=\frac{1.6\times {10}^{-27}\times 2\times {10}^{5}}{1.6\times {10}^{-19}\times 4\times {10}^{-2}}$* = *0.5 × 10^{−1} = 0.05 T

Putting the value of *B* in equation (1), we get:*E *= 2 × 10^{5} × 0.05

= 1 × 10^{4} N/c

#### Page No 234:

#### Question 49:

Given:

Mass of the proton, *m* = 1.6 × 10^{−27} kg

Speed of the proton inside the crossed electric and magnetic field, *v *= 2.0 × 10^{5} ms^{−1}

As per the question, the proton is not deflected under the combined action of the electric and magnetic fields. Thus, the forces applied by both the fields are equal and opposite.

That is, *qE = qvB *

⇒ *E = vB* ...(1)

But when the electric field is switched off, the proton moves in a circle due to the force of the magnetic field.

Radius of the circle, *r* = 4.0 cm = 4 × 10^{−2} m

We know: *r* = $\frac{mv}{qB}$

$\Rightarrow B=\frac{mv}{qr}\phantom{\rule{0ex}{0ex}}=\frac{1.6\times {10}^{-27}\times 2\times {10}^{5}}{1.6\times {10}^{-19}\times 4\times {10}^{-2}}$* = *0.5 × 10^{−1} = 0.05 T

Putting the value of *B* in equation (1), we get:*E *= 2 × 10^{5} × 0.05

= 1 × 10^{4} N/c

#### Answer:

Given:

Charge of the particle*, q* = 5 µC = 5 × 10^{−6} C

Magnetic field intensity, *B *= 5 × 10^{−3} T

Mass of the particle, *m* = 5 × 10^{−12} kg

Velocity of projection, *v* =* *1 Km/s = 10^{3}* *m/s

Angle between the magnetic field and velocity, $\theta $ = sin^{−1}(0.9)

Component of velocity perpendicular to the magnetic field, ${v}_{\perp}=v\mathrm{sin}\theta $

Component of velocity in the direction of magnetic field, ${v}_{\parallel}=v\mathrm{cos}\theta $

Since there are no forces in the horizontal direction (the direction of magnetic field), the particle moves with uniform velocity.

The velocity has a vertical component along which it accelerates with an acceleration *a *and moves in a circular cross-section. Thus, it moves in a helix.

$\frac{m{{v}_{\perp}}^{2}}{r}$ = *q${v}_{\perp}$B*

$\Rightarrow r=\frac{mv\mathrm{sin\theta}}{qB}\phantom{\rule{0ex}{0ex}}=\frac{5\times {10}^{-12}\times {10}^{3}\times 0.90}{5\times {10}^{-6}\times 5\times {10}^{-3}}\phantom{\rule{0ex}{0ex}}=0.18\mathrm{m}$

Hence, diameter of the helix, 2*r* = 0.36 m = 36 cm

Pitch,

$P=\frac{2\mathrm{\pi}r}{v\mathrm{sin\theta}}\times v\mathrm{cos\theta}\phantom{\rule{0ex}{0ex}}=\frac{2\times 3.14\times 0.18}{0.90}\times \sqrt{1-0.81}\phantom{\rule{0ex}{0ex}}=0.55\mathrm{m}=55\mathrm{cm}$

#### Page No 234:

#### Question 50:

Given:

Charge of the particle*, q* = 5 µC = 5 × 10^{−6} C

Magnetic field intensity, *B *= 5 × 10^{−3} T

Mass of the particle, *m* = 5 × 10^{−12} kg

Velocity of projection, *v* =* *1 Km/s = 10^{3}* *m/s

Angle between the magnetic field and velocity, $\theta $ = sin^{−1}(0.9)

Component of velocity perpendicular to the magnetic field, ${v}_{\perp}=v\mathrm{sin}\theta $

Component of velocity in the direction of magnetic field, ${v}_{\parallel}=v\mathrm{cos}\theta $

Since there are no forces in the horizontal direction (the direction of magnetic field), the particle moves with uniform velocity.

The velocity has a vertical component along which it accelerates with an acceleration *a *and moves in a circular cross-section. Thus, it moves in a helix.

$\frac{m{{v}_{\perp}}^{2}}{r}$ = *q${v}_{\perp}$B*

$\Rightarrow r=\frac{mv\mathrm{sin\theta}}{qB}\phantom{\rule{0ex}{0ex}}=\frac{5\times {10}^{-12}\times {10}^{3}\times 0.90}{5\times {10}^{-6}\times 5\times {10}^{-3}}\phantom{\rule{0ex}{0ex}}=0.18\mathrm{m}$

Hence, diameter of the helix, 2*r* = 0.36 m = 36 cm

Pitch,

$P=\frac{2\mathrm{\pi}r}{v\mathrm{sin\theta}}\times v\mathrm{cos\theta}\phantom{\rule{0ex}{0ex}}=\frac{2\times 3.14\times 0.18}{0.90}\times \sqrt{1-0.81}\phantom{\rule{0ex}{0ex}}=0.55\mathrm{m}=55\mathrm{cm}$

#### Answer:

Mass of the proton, *m _{p}* = 1.6 × 10

^{−27}kg

Magnetic field intensity,

*B*= 0.02 T

Radius of the helical path,

*r*= 5 cm = 5 × 10

^{−2}m

Pitch of the helical path,

*p*= 20cm = 2 × 10

^{−1}m

We know that for a helical path, the velocity of the proton has two components, ${v}_{\parallel}\mathrm{and}{v}_{\perp}$.

Now, $\frac{m{{v}_{\perp}}^{2}}{r}$ =

*$q{v}_{\perp}B$*

$\Rightarrow r=\frac{m{v}_{\perp}}{qB}\phantom{\rule{0ex}{0ex}}\Rightarrow 5\times {10}^{-2}=\frac{1.6\times {10}^{-27}\times {\mathrm{v}}_{\perp}}{1.6\times {10}^{-19}\times 0.02}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{v}}_{\perp}={10}^{5}\mathrm{m}/\mathrm{s}$

$\mathrm{Pitch}=\frac{{v}_{\parallel}2\mathrm{\pi}r}{{v}_{\perp}}\phantom{\rule{0ex}{0ex}}{v}_{\parallel}=\frac{{v}_{\perp}P}{2\mathrm{\pi}r}\phantom{\rule{0ex}{0ex}}=\frac{{10}^{5}\times 0.2}{2\times 3.14\times 5\times {10}^{-2}}\phantom{\rule{0ex}{0ex}}=0.6369\times {10}^{5}\phantom{\rule{0ex}{0ex}}=6.4\times {10}^{4}\mathrm{m}/\mathrm{s}$

#### Page No 234:

#### Question 51:

Mass of the proton, *m _{p}* = 1.6 × 10

^{−27}kg

Magnetic field intensity,

*B*= 0.02 T

Radius of the helical path,

*r*= 5 cm = 5 × 10

^{−2}m

Pitch of the helical path,

*p*= 20cm = 2 × 10

^{−1}m

We know that for a helical path, the velocity of the proton has two components, ${v}_{\parallel}\mathrm{and}{v}_{\perp}$.

Now, $\frac{m{{v}_{\perp}}^{2}}{r}$ =

*$q{v}_{\perp}B$*

$\Rightarrow r=\frac{m{v}_{\perp}}{qB}\phantom{\rule{0ex}{0ex}}\Rightarrow 5\times {10}^{-2}=\frac{1.6\times {10}^{-27}\times {\mathrm{v}}_{\perp}}{1.6\times {10}^{-19}\times 0.02}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{v}}_{\perp}={10}^{5}\mathrm{m}/\mathrm{s}$

$\mathrm{Pitch}=\frac{{v}_{\parallel}2\mathrm{\pi}r}{{v}_{\perp}}\phantom{\rule{0ex}{0ex}}{v}_{\parallel}=\frac{{v}_{\perp}P}{2\mathrm{\pi}r}\phantom{\rule{0ex}{0ex}}=\frac{{10}^{5}\times 0.2}{2\times 3.14\times 5\times {10}^{-2}}\phantom{\rule{0ex}{0ex}}=0.6369\times {10}^{5}\phantom{\rule{0ex}{0ex}}=6.4\times {10}^{4}\mathrm{m}/\mathrm{s}$

#### Answer:

Given:

Mass of the particle = *m*

Charge of the particle = *q*

Electric field and magnetic field are given by

$\overrightarrow{B}=-{B}_{0}\overrightarrow{j}\mathrm{and}\overrightarrow{E}={E}_{0}\overrightarrow{k}$

Velocity, $v={v}_{\mathrm{x}}\hat{i}+{v}_{\mathrm{y}}\hat{j}+{v}_{\mathrm{z}}\hat{\mathit{k}}$

So, total force on the particle,*F = q (E + v × B)*

*$=q\left[{E}_{0}\hat{k}-\left({v}_{\mathrm{x}}\hat{i}+{v}_{\mathrm{y}}\hat{j}+{v}_{\mathrm{z}}\hat{\mathit{k}}\right)\times {B}_{\mathit{0}}\hat{\mathit{j}}\right]\phantom{\rule{0ex}{0ex}}=q{E}_{0}\hat{k}-{v}_{\mathrm{x}}{B}_{0}\hat{k}+{v}_{\mathrm{z}}{B}_{0}\hat{i}$*

Since *v*_{x} = 0, *F*_{z}* = qE _{0}*

$\mathrm{So},{a}_{\mathrm{z}}=\frac{q{E}_{0}}{m}\phantom{\rule{0ex}{0ex}}{v}^{2}={u}^{2}+2as=2\frac{q{E}_{0}}{m}z\phantom{\rule{0ex}{0ex}}\mathrm{So},v=\sqrt{\frac{2q{E}_{0}z}{m}}$

Here,

*z*is the distance along the

*z*-direction.

#### Page No 234:

#### Question 52:

Given:

Mass of the particle = *m*

Charge of the particle = *q*

Electric field and magnetic field are given by

$\overrightarrow{B}=-{B}_{0}\overrightarrow{j}\mathrm{and}\overrightarrow{E}={E}_{0}\overrightarrow{k}$

Velocity, $v={v}_{\mathrm{x}}\hat{i}+{v}_{\mathrm{y}}\hat{j}+{v}_{\mathrm{z}}\hat{\mathit{k}}$

So, total force on the particle,*F = q (E + v × B)*

*$=q\left[{E}_{0}\hat{k}-\left({v}_{\mathrm{x}}\hat{i}+{v}_{\mathrm{y}}\hat{j}+{v}_{\mathrm{z}}\hat{\mathit{k}}\right)\times {B}_{\mathit{0}}\hat{\mathit{j}}\right]\phantom{\rule{0ex}{0ex}}=q{E}_{0}\hat{k}-{v}_{\mathrm{x}}{B}_{0}\hat{k}+{v}_{\mathrm{z}}{B}_{0}\hat{i}$*

Since *v*_{x} = 0, *F*_{z}* = qE _{0}*

$\mathrm{So},{a}_{\mathrm{z}}=\frac{q{E}_{0}}{m}\phantom{\rule{0ex}{0ex}}{v}^{2}={u}^{2}+2as=2\frac{q{E}_{0}}{m}z\phantom{\rule{0ex}{0ex}}\mathrm{So},v=\sqrt{\frac{2q{E}_{0}z}{m}}$

Here,

*z*is the distance along the

*z*-direction.

#### Answer:

Given:

Potential difference across the plates of the capacitor = *V*

Separation between the plates = *d*

Magnetic field intensity = *B*

The electric field set up between the plates of a capacitor, $E=\frac{V}{d}$

The force experienced by the electron due to this electric field, $F=\frac{eV}{d}$

$\Rightarrow a=\frac{F}{m}=\frac{eV}{{m}_{\mathrm{e}}d}$,

where *e* = charge of the electron and *m _{e}* = mass of the electron

Using

*v*

^{2}=

*u*

^{2}

^{}+ 2

*as*and substituting the value of

*a*, we get:

${v}^{2}=2\times \left(\frac{eV}{{m}_{\mathrm{e}}d}\right)\times d\phantom{\rule{0ex}{0ex}}\Rightarrow v=\sqrt{\frac{2eV}{{m}_{\mathrm{e}}}}$

The electron will move in a circular path due to the given magnetic field. Radius of the circular path,

$r=\frac{{m}_{\mathrm{e}}v}{eB}$

And the electron will fail to strike the upper plate only when the radius of the circular path will be less than

*d*,

$\mathrm{i}.\mathrm{e}.dr\phantom{\rule{0ex}{0ex}}\Rightarrow d\frac{{m}_{\mathrm{e}}}{eB}\times \sqrt{\frac{2eV}{{m}_{\mathrm{e}}}}\phantom{\rule{0ex}{0ex}}\Rightarrow d\sqrt{\frac{2{m}_{\mathrm{e}}V}{e{B}^{2}}}\phantom{\rule{0ex}{0ex}}$

Thus, $d>{\left(\frac{2{m}_{e}V}{e{{B}_{0}}^{2}}\right)}^{\frac{1}{2}}$

#### Page No 234:

#### Question 53:

Given:

Potential difference across the plates of the capacitor = *V*

Separation between the plates = *d*

Magnetic field intensity = *B*

The electric field set up between the plates of a capacitor, $E=\frac{V}{d}$

The force experienced by the electron due to this electric field, $F=\frac{eV}{d}$

$\Rightarrow a=\frac{F}{m}=\frac{eV}{{m}_{\mathrm{e}}d}$,

where *e* = charge of the electron and *m _{e}* = mass of the electron

Using

*v*

^{2}=

*u*

^{2}

^{}+ 2

*as*and substituting the value of

*a*, we get:

${v}^{2}=2\times \left(\frac{eV}{{m}_{\mathrm{e}}d}\right)\times d\phantom{\rule{0ex}{0ex}}\Rightarrow v=\sqrt{\frac{2eV}{{m}_{\mathrm{e}}}}$

The electron will move in a circular path due to the given magnetic field. Radius of the circular path,

$r=\frac{{m}_{\mathrm{e}}v}{eB}$

And the electron will fail to strike the upper plate only when the radius of the circular path will be less than

*d*,

$\mathrm{i}.\mathrm{e}.dr\phantom{\rule{0ex}{0ex}}\Rightarrow d\frac{{m}_{\mathrm{e}}}{eB}\times \sqrt{\frac{2eV}{{m}_{\mathrm{e}}}}\phantom{\rule{0ex}{0ex}}\Rightarrow d\sqrt{\frac{2{m}_{\mathrm{e}}V}{e{B}^{2}}}\phantom{\rule{0ex}{0ex}}$

Thus, $d>{\left(\frac{2{m}_{e}V}{e{{B}_{0}}^{2}}\right)}^{\frac{1}{2}}$

#### Answer:

Given:

No. of turns in the coil, n = 100

Area of the coil, *A = *5 × 4 cm^{2} = 20 × 10^{−4} m^{2}

Magnitude of current = 2 A

Torque acting on the coil, *τ* = 0.2 N m^{−1}

τ = *niA* × *B*

⇒*τ* = *ni*B*A* sin 90°

⇒ 0.2 = 100 × 2 ×20 × 10^{−4} × *B*

⇒ *B* = 0.5 T

#### Page No 234:

#### Question 54:

Given:

No. of turns in the coil, n = 100

Area of the coil, *A = *5 × 4 cm^{2} = 20 × 10^{−4} m^{2}

Magnitude of current = 2 A

Torque acting on the coil, *τ* = 0.2 N m^{−1}

τ = *niA* × *B*

⇒*τ* = *ni*B*A* sin 90°

⇒ 0.2 = 100 × 2 ×20 × 10^{−4} × *B*

⇒ *B* = 0.5 T

#### Answer:

Given:

No. of turns of the coil, n = 50

Magnetic field intensity, *B = *0.20 T = 2 × 10^{−1} T

Radius of the coil*, **r* = 0.02 m = 2 × 10^{−2} m

Magnitude of current =5 A

Torque acting on the coil,

τ = *niA**B*sin*θ*

Here,* A* is the area of the coil and *θ is *the angle between the area vector and the magnetic field.

τ is maximum when *θ *= 90°.

τ_{max} = *niAB*sin90°

= 50 × 5 × 3.14 × 4 × 10^{−4} × 2 × 10^{−1}

= 6.28 × 10^{−2} N-m

$\mathrm{Given},\tau =\frac{1}{2}\times {\tau}_{max}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{sin\theta}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \theta \mathit{}=30\xb0$

So, the angle between the magnetic field and the plane of the coil = 90° − 30° = 60°

#### Page No 234:

#### Question 55:

Given:

No. of turns of the coil, n = 50

Magnetic field intensity, *B = *0.20 T = 2 × 10^{−1} T

Radius of the coil*, **r* = 0.02 m = 2 × 10^{−2} m

Magnitude of current =5 A

Torque acting on the coil,

τ = *niA**B*sin*θ*

Here,* A* is the area of the coil and *θ is *the angle between the area vector and the magnetic field.

τ is maximum when *θ *= 90°.

τ_{max} = *niAB*sin90°

= 50 × 5 × 3.14 × 4 × 10^{−4} × 2 × 10^{−1}

= 6.28 × 10^{−2} N-m

$\mathrm{Given},\tau =\frac{1}{2}\times {\tau}_{max}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{sin\theta}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \theta \mathit{}=30\xb0$

So, the angle between the magnetic field and the plane of the coil = 90° − 30° = 60°

#### Answer:

Let ABCD be the rectangular loop.

Given:

No. of turns of the coil, *n =* 50

Magnetic field intensity, *B = *0.20 T = 2 × 10^{−1} T

Magnitude of current, *I* = 5 A

Length of the loop,* l *= 20 cm = 20 × 10^{−2} m,

Breadth of the loop, *w** *= 10 cm = 10 × 10^{−2} m,

So, area of the loop, *A *= *lw* = 0.02 m^{2},

fig

(a) There is no force on the sides *ab* and *cd*, as they are along the magnetic field.

But the force on the sides *ad* and *bc* are equal and opposite; so, they cancel each other.

Hence, the net force on the loop is zero.

(b) Torque acting on the coil,

τ = *niA**B*sin*θ*

Here,* A* is the area of the coil and *θ *is the angle between the area vector and magnetic field.

τ = *niA*Bsin90°

= 1 × 5 × 0.02 × 0.2

= 0.02 N-m

So, the torque acting on the loop is 0.02 N-m and is parallel to the shorter side.

#### Page No 234:

#### Question 56:

Let ABCD be the rectangular loop.

Given:

No. of turns of the coil, *n =* 50

Magnetic field intensity, *B = *0.20 T = 2 × 10^{−1} T

Magnitude of current, *I* = 5 A

Length of the loop,* l *= 20 cm = 20 × 10^{−2} m,

Breadth of the loop, *w** *= 10 cm = 10 × 10^{−2} m,

So, area of the loop, *A *= *lw* = 0.02 m^{2},

fig

(a) There is no force on the sides *ab* and *cd*, as they are along the magnetic field.

But the force on the sides *ad* and *bc* are equal and opposite; so, they cancel each other.

Hence, the net force on the loop is zero.

(b) Torque acting on the coil,

τ = *niA**B*sin*θ*

Here,* A* is the area of the coil and *θ *is the angle between the area vector and magnetic field.

τ = *niA*Bsin90°

= 1 × 5 × 0.02 × 0.2

= 0.02 N-m

So, the torque acting on the loop is 0.02 N-m and is parallel to the shorter side.

#### Answer:

Given:

No. of turns of the coil, *n *= 500

Magnetic field intensity, *B = *0.40 T = 4 × 10^{−1} T

Radius of the coil*, **r* = 2 cm = 2 × 10^{−2} m

Magnitude of current, *i =* 1 A

Angle between the area vector and magnetic field, *θ = *30°

Torque acting on the coil,

τ = *niA**B*sin*θ*

Here,* A* is the area of the coil.

τ = 500 × 1 × 3.14 × 4 × 10^{−4} ×4×10^{−1} × $\frac{1}{2}$

= 12.56 × 10^{−2}

= 0.1256 = 0.13 N-m

#### Page No 234:

#### Question 57:

Given:

No. of turns of the coil, *n *= 500

Magnetic field intensity, *B = *0.40 T = 4 × 10^{−1} T

Radius of the coil*, **r* = 2 cm = 2 × 10^{−2} m

Magnitude of current, *i =* 1 A

Angle between the area vector and magnetic field, *θ = *30°

Torque acting on the coil,

τ = *niA**B*sin*θ*

Here,* A* is the area of the coil.

τ = 500 × 1 × 3.14 × 4 × 10^{−4} ×4×10^{−1} × $\frac{1}{2}$

= 12.56 × 10^{−2}

= 0.1256 = 0.13 N-m

#### Answer:

Given,

Magnetic field intensity = *B *

Circumference, *L* = 2π*r*, where *r* is the radius of the coil*.*

So,* *area of the coil, * A* $=\frac{{L}^{2}}{4\mathrm{\pi}}$

Magnitude of current =* i*

Torque acting on the coil,

τ = *niAB*sin*θ**, *where *θ *is the angle between the area vector and the magnetic field.

$\tau =niAB=\frac{i{L}^{2}B}{4\mathrm{\pi}}$

(b) Let *s* be the length of the square loop.

$4s=L\phantom{\rule{0ex}{0ex}}\Rightarrow s=\frac{L}{4}\phantom{\rule{0ex}{0ex}}A={\left(\frac{L}{4}\right)}^{2}=\left(\frac{{L}^{2}}{16}\right)\phantom{\rule{0ex}{0ex}}\tau =niAB=\frac{i{L}^{2}B}{16}$

So, the torque on the circular loop is larger.

#### Page No 234:

#### Question 58:

Given,

Magnetic field intensity = *B *

Circumference, *L* = 2π*r*, where *r* is the radius of the coil*.*

So,* *area of the coil, * A* $=\frac{{L}^{2}}{4\mathrm{\pi}}$

Magnitude of current =* i*

Torque acting on the coil,

τ = *niAB*sin*θ**, *where *θ *is the angle between the area vector and the magnetic field.

$\tau =niAB=\frac{i{L}^{2}B}{4\mathrm{\pi}}$

(b) Let *s* be the length of the square loop.

$4s=L\phantom{\rule{0ex}{0ex}}\Rightarrow s=\frac{L}{4}\phantom{\rule{0ex}{0ex}}A={\left(\frac{L}{4}\right)}^{2}=\left(\frac{{L}^{2}}{16}\right)\phantom{\rule{0ex}{0ex}}\tau =niAB=\frac{i{L}^{2}B}{16}$

So, the torque on the circular loop is larger.

#### Answer:

Given:

Number of turns in the coil = *n*

Edge of the square loop = *l*

Magnetic field intensity = *B *

Magnitude of current = *i *

Angle between area vector and magnetic field, *θ = *90°

Torque acting on the coil due to magnetic field,*τ* = *niA**B*sin*θ*

Here,* A* is the area of the coil.*τ* = n*il*^{2}*B*sin90° = n*il*^{2}*B*

Torque produced due to weight, *τ*_{weight} = $\frac{mgl}{2}$

For the coil to start tipping over,*τ* ≥ *τ*_{weight}

For minimum value of *B,**τ* = *τ*_{weight}

$\Rightarrow ni{l}^{2}B=\frac{mgl}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow B=\frac{mg}{2nil}$

#### Page No 234:

#### Question 59:

Given:

Number of turns in the coil = *n*

Edge of the square loop = *l*

Magnetic field intensity = *B *

Magnitude of current = *i *

Angle between area vector and magnetic field, *θ = *90°

Torque acting on the coil due to magnetic field,*τ* = *niA**B*sin*θ*

Here,* A* is the area of the coil.*τ* = n*il*^{2}*B*sin90° = n*il*^{2}*B*

Torque produced due to weight, *τ*_{weight} = $\frac{mgl}{2}$

For the coil to start tipping over,*τ* ≥ *τ*_{weight}

For minimum value of *B,**τ* = *τ*_{weight}

$\Rightarrow ni{l}^{2}B=\frac{mgl}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow B=\frac{mg}{2nil}$

#### Answer:

Given:

Radius of the ring* = * *r*

Mass of the ring* = * *m*

Total charge of the ring* = * *q*

(a) Angular speed, $\omega =\frac{2\mathrm{\pi}}{T}\Rightarrow T=\frac{2\mathrm{\pi}}{\omega}$

Current in the ring, $i=\frac{q}{T}=\frac{q\omega}{2\mathrm{\pi}}$

(b) For a ring of area *A *with current* i*, magnetic moment,

µ = *niA* = *ia* [*n* = 1]

$=\frac{q\omega}{2\mathrm{\pi}}\times \mathrm{\pi}{r}^{2}\mathit{}\mathit{=}\mathit{}\frac{q\omega {r}^{2}}{2}$ ...(i)

(c) Angular momentum, *l = $I\omega $*,

where *I* is moment of inertia of the ring about its axis of rotation.

$I=m{r}^{2}$

So, $I=m{r}^{2}\omega $

$\Rightarrow \omega {r}^{2}=\frac{l}{m}$

Putting this value in equation (i), we get:

$\mu =\frac{ql}{2m}$

#### Page No 235:

#### Question 60:

Given:

Radius of the ring* = * *r*

Mass of the ring* = * *m*

Total charge of the ring* = * *q*

(a) Angular speed, $\omega =\frac{2\mathrm{\pi}}{T}\Rightarrow T=\frac{2\mathrm{\pi}}{\omega}$

Current in the ring, $i=\frac{q}{T}=\frac{q\omega}{2\mathrm{\pi}}$

(b) For a ring of area *A *with current* i*, magnetic moment,

µ = *niA* = *ia* [*n* = 1]

$=\frac{q\omega}{2\mathrm{\pi}}\times \mathrm{\pi}{r}^{2}\mathit{}\mathit{=}\mathit{}\frac{q\omega {r}^{2}}{2}$ ...(i)

(c) Angular momentum, *l = $I\omega $*,

where *I* is moment of inertia of the ring about its axis of rotation.

$I=m{r}^{2}$

So, $I=m{r}^{2}\omega $

$\Rightarrow \omega {r}^{2}=\frac{l}{m}$

Putting this value in equation (i), we get:

$\mu =\frac{ql}{2m}$

#### Answer:

Given:

Radius of the ring* = * *r*

Mass of the ring* = * *m*

Total charge of the ring* = * *q*

Angular speed, $\omega =\frac{2\pi}{T}\Rightarrow T=\frac{2\pi}{\omega}$

Current in the ring, $i=\frac{q}{T}=\frac{q\omega}{2\pi}$

For the ring of area *A *with current* i, *magnetic moment,

µ = *niA* = *ia* [*n* = 1]

$=\frac{q\omega}{2\mathrm{\pi}}\times \mathrm{\pi}{r}^{2}\mathit{=}\frac{q\omega {r}^{2}}{2}$ ...(i)

Angular momentum, *l = $I\omega $*,

where *I* is the moment of inertia of the ring about its axis of rotation.

$I=m{r}^{2}$

So, $l=m{r}^{2}\omega $

$\Rightarrow \omega {r}^{2}=\frac{l}{m}$

Putting this value in equation (i), we get:

$\mu =\frac{ql}{2m}$

#### Page No 235:

#### Question 61:

Given:

Radius of the ring* = * *r*

Mass of the ring* = * *m*

Total charge of the ring* = * *q*

Angular speed, $\omega =\frac{2\pi}{T}\Rightarrow T=\frac{2\pi}{\omega}$

Current in the ring, $i=\frac{q}{T}=\frac{q\omega}{2\pi}$

For the ring of area *A *with current* i, *magnetic moment,

µ = *niA* = *ia* [*n* = 1]

$=\frac{q\omega}{2\mathrm{\pi}}\times \mathrm{\pi}{r}^{2}\mathit{=}\frac{q\omega {r}^{2}}{2}$ ...(i)

Angular momentum, *l = $I\omega $*,

where *I* is the moment of inertia of the ring about its axis of rotation.

$I=m{r}^{2}$

So, $l=m{r}^{2}\omega $

$\Rightarrow \omega {r}^{2}=\frac{l}{m}$

Putting this value in equation (i), we get:

$\mu =\frac{ql}{2m}$

#### Answer:

Considering the strip of width *dx* at a distance *x* from the centre of the sphere.

Small area of the strip is given as,*da* = 4π *x* *dx$i=\frac{dq}{dt}=\frac{q\omega}{2\mathrm{\pi}}\phantom{\rule{0ex}{0ex}}\mathrm{\mu}=ia=\frac{q\omega}{2\mathrm{\pi}}4\mathrm{\pi}\int x\mathit{}dx\phantom{\rule{0ex}{0ex}}=q.\omega {r}^{2}=\frac{q}{m}l$*

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