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#### Page No 248:

#### Answer:

According to the right-hand thumb rule, if the thumb of our right hand points in the direction of the current flowing, then the curling of the fingers will show the direction of the magnetic field developed due to it and vice versa. Let us consider the case where an electric current flows north to south in a wire.

According to the right-hand thumb rule,

(a) for any point in the east of the wire, the magnetic field will come out of the plane of paper

(b) for a point in the west of the wire, the magnetic field will enter the plane of paper

(c) for any point vertically above the wire, the magnetic field will be from right to left

(d) for any point vertically below the wire, the magnetic field will be from left to right

#### Page No 248:

#### Answer:

The magnetic field due to a long, straight wire is given by

$B=\frac{{\mathrm{\mu}}_{\mathrm{o}}i}{2\mathrm{\pi}d}\phantom{\rule{0ex}{0ex}}\because \mathrm{Speed}\mathrm{of}\mathrm{light},\mathrm{c}=\frac{1}{\sqrt{{\mathrm{\mu}}_{\mathrm{o}}{\mathrm{\epsilon}}_{\mathrm{o}}}}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{\mu}}_{\mathrm{o}}=\frac{1}{{\mathrm{c}}^{2}{\mathrm{\epsilon}}_{\mathrm{o}}}\phantom{\rule{0ex}{0ex}}\Rightarrow B=\frac{i}{2{\mathrm{\pi c}}^{2}{\mathrm{\epsilon}}_{\mathrm{o}}d}$

(In terms of ε_{0}, *c*, *i* and *d*)

#### Page No 248:

#### Answer:

According to the right-hand thumb rule, if we curl the fingers of our right hand in the direction of the current flowing, then the thumb will point in the direction of the magnetic field developed due to it and vice versa. Therefore, in this case, the field at the centre is going away from us.

#### Page No 248:

#### Answer:

In Ampere's law $\oint \overrightarrow{B}.\overrightarrow{dl}={\mathrm{\mu}}_{\mathrm{o}}i$, *i* is the total current crossing the area bounded by the closed curve. The magnetic field *B* on the left-hand side is the resultant field due to all existing currents.

#### Page No 248:

#### Answer:

The magnetic field due to a long solenoid is given as *B *= *µ*_{0}*ni**,* where *n* is the number of loops per unit length. So, if we add more loops at the ends of the solenoid, there will be an increase in the number of loops and an increase in the length, due to which the ratio *n *will remain unvaried, thereby leading to not a considerable effect on the field inside the solenoid.

#### Page No 248:

#### Answer:

Ampere's law is valid for a loop that is not circular. However, it should have some charge distribution in the area enclosed so as to have a constant electric field in the region and a non-zero magnetic field. Even if the loop defined does not have its own charge distribution but has electric influence of some other charge distribution, it can have some constant magnetic field ($\oint \overrightarrow{B}.d\overrightarrow{l}={\mathrm{\mu}}_{\mathrm{o}}{i}_{\mathrm{enclosed}}$).

#### Page No 248:

#### Answer:

The magnetic force on a wire carrying an electric current *i* is $\overrightarrow{F}=i.(\overrightarrow{l}\times \overrightarrow{B})$, where *l* is the length of the wire and *B* is the magnetic field acting on it. If a uniformly charged ring starts rotating around a straight wire, then according to the right-hand thumb rule, the magnetic field due to the ring on the current carrying straight wire placed at its axis will be parallel to it. So, the cross product will be

$(\overrightarrow{l}\times \overrightarrow{B})=0\phantom{\rule{0ex}{0ex}}\Rightarrow \overrightarrow{F}=0$

Therefore, no magnetic force will act on the wire.

#### Page No 248:

#### Answer:

The magnetic force on a wire carrying an electric current *i* is $\overrightarrow{F}=i.(\overrightarrow{l}\times \overrightarrow{B})$, where *l* is the length of the wire and *B* is the magnetic field acting on it. Suppose we have one wire in the horizontal direction (fixed) and other wire in the vertical direction (free to move). If the horizontal wire is carrying current from right to left is held fixed perpendicular to the vertical wire, which is free to move, the upper portion of the free wire will tend to move in the left direction and the lower portion of the wire will tend to move in the right direction, according to Fleming's left hand rule, as the magnetic field acting on the wire due to the fixed wire will point into the plane of paper above the wire and come out of the paper below the horizontal wire and the current will flow in upward direction in the free wire. Thus, the free wire will tend to become parallel to the fixed wire so as to experience maximum attractive force.

#### Page No 248:

#### Answer:

Two proton beams going in the same direction repel each other, as they are like charges and we know that like charges repel each other.

When a charge is in motion then a magnetic field is associated with it. Two wires carrying currents in the same direction produce their fields (acting on each other) in opposite directions so the resulting magnetic force acting on them is attractive. Due to the magnetic force, these two wires attract each other.

But when a charge is at rest then only an electric field is associated with it and no magnetic fiels is produced by it. So at rest, it repels a like charge by exerting a electric force on it.

Charge in motion can produce both electric field and magnetic field.

The attractive force between two current carrying wires is due to the magnetic field and repulsive force is due to the electric field.

#### Page No 248:

#### Answer:

We can obtain a magnetic field due to a straight, long wire using Ampere's law by mentioning the current flowing in the wire, without emphasising on the source of the current in the wire. To apply Ampere's circuital law, we need to have a constant current flowing in the wire, irrespective of its source.

#### Page No 248:

#### Answer:

Connecting wires carrying currents in opposite directions are twisted together in using electrical appliances.If the wire is twisted, then the resultant fields from consecutive twists are in opposite directions. So the cumulative effect over a long length of wire is roughly zero.

Let at any point in between the two wires, B_{1} and B_{2} be the magnetic field due to wire 1 and wire 2 respectively.

From the diagram, we can see that the net magnetic field due to first turn is into the paper and due to second twisted turn is out of the plane of paper so these fields will cancel each other. Hence if the wire is twisted, then the resultant fields from consecutive twists are in opposite directions. So the cumulative effect over a long length of wire is roughly zero.

#### Page No 248:

#### Answer:

Magnetic field can not do any work and hence can never speed up or down a particle.

Consider 2 wires carrying current in upward direction.

Magnetic field due to current in wire 1 produces a magnetic field out of the plane of paper at the position of wire 2. Due to this magnetic field, a force is exerted on wire 2. Wire 2 electron, moving in downward direction, move in circular paths due to this magnetic force. As these electrons can not come out of the wire so while describing circular path,they hit the edges of the wire and tranfer a momentum to the wire. Due to this change in momentum, wire starts moving and gains kinetic energy.

#### Page No 248:

#### Answer:

(c) upwards

A vertical wire is carrying current in upward direction, so the magnetic field produced will be anticlockwise (according to the right-hand thumb rule). As the electron beam is sent horizontally towards the wire, the direction of the current will be horizontally away from the wire (direction of conventional current is opposite to the direction of the negative charge). According to Fleming's left-hand rule, the force will act in upward direction, deflecting the beam in the same direction.

#### Page No 248:

#### Answer:

(c) will not exert any force on the circular loop

The magnetic force on a wire carrying an electric current *i* is given as $\overrightarrow{F}=i.(\overrightarrow{l}\times \overrightarrow{B})$, where *l* is the length of the wire and *B* is the magnetic field acting on it. If a current-carrying straight wire is kept along the axis of a circular loop carrying current, then according to the right-hand thumb rule, the magnetic field due to the wire on the current-carrying loop will be along its circumference, which contains a current element $id\overrightarrow{l}$.

So, the cross product will be

$(\overrightarrow{l}\times \overrightarrow{B})=0\phantom{\rule{0ex}{0ex}}\Rightarrow \overrightarrow{F}=0$

Thus, the straight wire will not exert any force on the loop.

#### Page No 248:

#### Answer:

(a) towards the proton beam

A proton beam is going from north to south, so the direction of the current due to the beam will also be from north to south. Also, an electron beam is going from south to north, so the direction of the current due to the beam will also be from north to south. The direction of conventional current is along the direction of the flow of the positive charge and opposite to the flow of the negative charge. The magnetic field generated due to them will enter the plane of paper in the west and come out of the plane of paper in the east, according to the right-hand thumb rule. Since both the beams have currents in the same direction, they will apply equal and opposite forces on each other and, hence, will attract each other. Thus, the electron beam will be deflected towards the proton beam.

#### Page No 248:

#### Answer:

(d) is towards west at both A and B

According to the right-hand thumb rule, if we curl the fingers of our right hand in the direction of the current flowing, then the stretching of the thumb will show the direction of the magnetic field developed due to it and vice versa.

Let north-south is along x axis and east-west is along y axis. Circular wire is in xz plane. Then point A will lie on positive y axis and B on negative y axis. On looking from point B, current is flowing in anticlockwise direction so the magnetic field will point from right to left. Hence, the magnetic field due to the loop will be towards west at both A and B.

#### Page No 248:

#### Answer:

(b) move towards the wire

Force acting on the wire per unit length carrying current *i*_{2} due to the wire carrying current *i*_{1} placed at a distance *d* is given by

$F=\frac{{\mathrm{\mu}}_{\mathrm{o}}{i}_{1}{i}_{2}}{2\mathrm{\pi}d}$

So, forces per unit length acting on sides *AB* and *CD *are as follows:

${F}_{\mathrm{AB}}=\frac{{\mathrm{\mu}}_{\mathrm{o}}{i}_{1}{i}_{2}}{2\mathrm{\pi}d}(\mathrm{Towards}\mathrm{the}\mathrm{wire})\phantom{\rule{0ex}{0ex}}{F}_{\mathrm{CD}}=\frac{{\mathrm{\mu}}_{\mathrm{o}}{i}_{1}{i}_{2}}{2\mathrm{\pi}(d+a)}(\mathrm{Away}\mathrm{from}\mathrm{the}\mathrm{wire})$

Here, *F*_{AB}* > **F*_{CD}_{ }because force is inversly proportional to the distance from the wire and wire AB is closer to the wire carrying current *i*_{1.}

The forces per unit length acting on sides *BC* and *DA* will be equal and opposite, as they are equally away from the wire carrying current *i*_{1}, with current *i*_{2}_{ }flowing in the opposite direction.

∴ *F*_{BC}_{ }*= $-$F*_{DA}

Now,

Net force:

$F={F}_{\mathrm{AB}}\mathit{+}{F}_{\mathrm{BC}}\mathit{+}{F}_{\mathrm{CD}}\mathit{+}{F}_{\mathrm{DA}}\phantom{\rule{0ex}{0ex}}\Rightarrow F=\frac{{\mathrm{\mu}}_{\mathrm{o}}{i}_{1}{i}_{2}}{2\mathrm{\pi}d}+{F}_{\mathrm{BC}}-\frac{{\mathrm{\mu}}_{\mathrm{o}}{i}_{1}{i}_{2}}{2\mathrm{\pi}(d+a)}-{F}_{\mathrm{BC}}\phantom{\rule{0ex}{0ex}}\Rightarrow F=\frac{{\mathrm{\mu}}_{\mathrm{o}}{i}_{1}{i}_{2}}{2\mathrm{\pi}}\left(\frac{1}{d}-\frac{1}{d+a}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow F=\frac{{\mathrm{\mu}}_{\mathrm{o}}{i}_{1}{i}_{2}a}{2\mathrm{\pi}d(d+a)}$

(Towards the wire)

Therefore, the loop will move towards the wire.

#### Page No 248:

#### Answer:

(d) zero

The force on a charged particle *q* moving with velocity *v* in a magnetic field *B* is given by

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

As the charge is moving along the magnetic line of force, the velocity and magnetic field vectors will point in the same direction, making a cross product.

$(\overrightarrow{v}\times \overrightarrow{B})=0\phantom{\rule{0ex}{0ex}}\Rightarrow \overrightarrow{F}=0$

So, the magnetic force on the particle will be zero.

#### Page No 248:

#### Answer:

(c) both of them

Because of the presence of a charge, a particle produces an electric field. Also, because of its motion, that is, the flow of charge or current, there is generation of a magnetic field.

#### Page No 249:

#### Answer:

(c) the kinetic energy

When a particle of mass *m* carrying charge* q* is projected with speed* v* in a plane perpendicular to a uniform magnetic field *B*, the field tends to deflect the particle in a circular path of radius *r*.

$\therefore \frac{m{v}^{2}}{r}=qvB\phantom{\rule{0ex}{0ex}}\Rightarrow r=\frac{mv}{qB}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}\mathrm{Area},A=\mathrm{\pi}{r}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow A=\mathrm{\pi}{\left(\frac{mv}{qB}\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow A=k{v}^{2}\phantom{\rule{0ex}{0ex}}\mathrm{Here},\phantom{\rule{0ex}{0ex}}k=\mathrm{\pi}{\left(\frac{m}{qB}\right)}^{2}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Kinetic energy of the particle, $E=\frac{1}{2}m{v}^{2}$

Therefore, the area bounded is proportional to the kinetic energy.

#### Page No 249:

#### Answer:

(c)$\frac{{{R}_{1}}^{2}}{{{R}_{2}}^{2}}$

Particles *X* and *Y* of respective masses *m*_{1} and *m*_{2} are carrying charge *q* describing circular paths with respective radii *R*_{1} and *R*_{2} such that

${R}_{1}=\frac{{m}_{1}{v}_{1}}{qB}\phantom{\rule{0ex}{0ex}}{R}_{2}=\frac{{m}_{2}{v}_{2}}{qB}$

Since both the particles are accelerated through the same potential difference, both will have the same kinetic energy.

$\therefore \frac{1}{2}{m}_{1}{{v}_{1}}^{2}=\frac{1}{2}{m}_{2}{{v}_{2}}^{2}\phantom{\rule{0ex}{0ex}}\because {R}_{1}=\frac{{m}_{1}{v}_{1}}{qB}\Rightarrow {v}_{1}=\frac{{R}_{1}qB}{{m}_{1}}\phantom{\rule{0ex}{0ex}}\mathrm{And},\phantom{\rule{0ex}{0ex}}{R}_{2}=\frac{{m}_{2}{v}_{2}}{qB}\Rightarrow {v}_{2}=\frac{{R}_{2}qB}{{m}_{2}}\phantom{\rule{0ex}{0ex}}\therefore {m}_{1}{\left(\frac{{R}_{1}qB}{{m}_{1}}\right)}^{2}={m}_{2}{\left(\frac{{R}_{2}qB}{{m}_{2}}\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{m}_{1}}{{m}_{2}}=\frac{{{R}_{1}}^{2}}{{{R}_{2}}^{2}}\phantom{\rule{0ex}{0ex}}$

#### Page No 249:

#### Answer:

(b) towards 40 A

According to Fleming's left-hand rule, if the forefinger and middle finger of our left hand point towards the magnetic field acting on a wire and the current flowing in the wire, respectively, then the thumb will point towards the direction in which the force will act (keeping all three perpendicular). Direction of force can be determined using Fleming's left-hand rule.

In the figure, dotted circle shows the magnetic filed lines due to both the wires.

Magnetic field at any point on the middle wire will be acting along the tangent to the masgnetic field lines at that point.

Therefore, the wire will experience a magnetic field pointing towards the 40 A wire.

Due to AB, the force will be towards right and due to CD, the force on the wire will be towards right. So, both the forces will add to give a resultant force, which will be towards right, that is, towards the 40 A current-carrying wire.

#### Page No 249:

#### Answer:

(c) 2

The magnetic field due to the current-carrying long, straight wire at point *a* is given by

$B=\frac{{\mathrm{\mu}}_{\mathrm{o}}i}{2\mathrm{\pi}a}$

When both the wires carry currents *i*_{1} and *i*_{2}_{ }_{}in the same direction, they produce magnetic fields in opposite directions at any point in between the wires.

$B\text{'}=\frac{{\mathrm{\mu}}_{\mathrm{o}}{i}_{1}}{2\mathrm{\pi}a}-\frac{{\mathrm{\mu}}_{\mathrm{o}}{i}_{2}}{2\mathrm{\pi}a}=10\mathrm{\mu T}..\left(1\right)$

Here, *a *is the distance of the midpoint from both the wires.

When both the wires carry currents in opposite directions, they produce fields in the same direction at the midpoint of the two wires.

$B\text{'}\text{'}=\frac{{\mathrm{\mu}}_{\mathrm{o}}{i}_{1}}{2\mathrm{\pi}a}+\frac{{\mathrm{\mu}}_{\mathrm{o}}{i}_{2}}{2\mathrm{\pi}a}=30\mathrm{\mu T}..\left(2\right)$

On solving eqs. (1) and (2), we get

${i}_{1}-{i}_{2}=10\phantom{\rule{0ex}{0ex}}{i}_{1}+{i}_{2}=30\phantom{\rule{0ex}{0ex}}\Rightarrow {i}_{1}=20,\mathit{}{i}_{2}=10\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{i}_{1}}{{i}_{2}}=\frac{2}{1}=2$

#### Page No 249:

#### Answer:

(a) $\frac{{\mathrm{\mu}}_{o}i}{2\mathrm{\pi}r}$

Magnetic field will be independent of the motion of the observer because the velocity with which the observer is moving is comparable to drift velocity of electron which is very small as compared to the speed of flow of current from one end of wire to other end. So it can be neglected and hence, magnetic field due to the wire w.r.t the observer will be

*B* = $\frac{{\mathrm{\mu}}_{o}i}{2\mathrm{\pi}r}$

^{ }

#### Page No 249:

#### Answer:

(a) $\frac{{\mathrm{\mu}}_{0}i}{4\pi}\frac{d\overrightarrow{l}\times \overrightarrow{r}}{{r}^{3}}$

(b) $-\frac{{\mathrm{\mu}}_{0}i}{4\pi}\frac{\overrightarrow{r}\times d\overrightarrow{l}}{{r}^{3}}$

The magnetic field at the origin due to current element $id\overrightarrow{l}$ placed at a position $\overrightarrow{r}$ is given by

$d\overrightarrow{B}=\frac{{\mathrm{\mu}}_{\mathrm{o}}}{4\mathrm{\pi}}i\frac{\overrightarrow{r}\times d\overrightarrow{l}}{{r}^{3}}\phantom{\rule{0ex}{0ex}}$

According to the cross product property,

$\overrightarrow{A}\times \overrightarrow{B}=-\overrightarrow{B}\times \overrightarrow{A}\phantom{\rule{0ex}{0ex}}\Rightarrow d\overrightarrow{B}=-\frac{{\mathrm{\mu}}_{\mathrm{o}}}{4\mathrm{\pi}}i\frac{\overrightarrow{r}\times \overrightarrow{dl}}{{r}^{3}}$

#### Page No 249:

#### Answer:

(a) *x*, *y* have the same dimensions.

(b) *y*, *z* have the same dimensions.

(c) *z*, *x* have the same dimensions.

Lorentz Force:

$qvB=qE\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{Dimensions}\mathrm{of}x=\left[v\right]=\left[\frac{E}{B}\right]=\left[{\mathrm{LT}}^{-1}\right]$

$y=\frac{1}{\sqrt{{\mathrm{\mu}}_{\mathrm{o}}{\mathrm{\epsilon}}_{\mathrm{o}}}}=\sqrt{\frac{4\mathrm{\pi}}{{\mathrm{\mu}}_{\mathrm{o}}}\times \frac{1}{4{\mathrm{\pi \epsilon}}_{\mathrm{o}}}}=\sqrt{\frac{9\times {10}^{9}}{{10}^{-7}}}=3\times {10}^{8}=\mathrm{c}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{Dimensions}\mathrm{of}\mathrm{y}=\left[\mathrm{c}\right]=\left[{\mathrm{LT}}^{-1}\right]$

Time constant of *RC *circuit = *RC* so dimensionally [*RC]* = [T]

$\Rightarrow z=\left[\frac{l}{\mathrm{RC}}\right]\Rightarrow \left[z\right]=\left[{\mathrm{LT}}^{-1}\right]$

Therefore, *x*, *y* and *z* have the same dimensions.

#### Page No 249:

#### Answer:

(b) the directions of the magnetic fields are the same

(c) the magnitudes of the magnetic fields are equal

(d) the field at one point is opposite to that at the other point.

Consider a current carrying wire lying along x axis.

At any two points on z axis which are at equal distance from the wire,one above the wire and one below the wire,the magnitude of magnetic field will be same and their directions will be opposite to each other.

At any two points on z axis which are at different distances from the wire,one above the wire and other also above the wire,the magnitude of magnetic field will be different and their directions will be same to each other.

#### Page No 249:

#### Answer:

(b) minimum at the axis of the wire

(c) maximum at the surface of the wire

A long, straight wire of radius *R* is carrying current *i*, which is uniformly distributed over its cross section. According to Ampere's law,

$\oint \overrightarrow{B}.d\overrightarrow{l}={\mathrm{\mu}}_{\mathrm{o}}{i}_{\mathrm{inside}}\phantom{\rule{0ex}{0ex}}\mathrm{At}\mathrm{surface},\phantom{\rule{0ex}{0ex}}B\times 2\mathrm{\pi}R={\mathrm{\mu}}_{\mathrm{o}}i\phantom{\rule{0ex}{0ex}}\Rightarrow {B}_{\mathrm{surface}}=\frac{{\mathrm{\mu}}_{\mathrm{o}}i}{2\mathrm{\pi}R}\phantom{\rule{0ex}{0ex}}\mathrm{Inside},B\times 2\mathrm{\pi}r={\mathrm{\mu}}_{\mathrm{o}}i\text{'}\mathrm{for}rR\phantom{\rule{0ex}{0ex}}$

Here *i ^{,} *is the current enclosed by the amperian loop drawn inside the wire.

*B*

_{inside}will be proportional to the distance from the axis.

On the axis

*B*=0

The magnetic fields from points on the cross section will point in opposite directions and will cancel each other at the centre.

$\mathrm{Outside},B\times 2\mathrm{\pi}r={\mathrm{\mu}}_{\mathrm{o}}i\phantom{\rule{0ex}{0ex}}\Rightarrow {B}_{\mathrm{outside}}=\frac{{\mathrm{\mu}}_{\mathrm{o}}i}{2\mathrm{\pi}r},rR$

Therefore, the magnitude of the magnetic field is maximum at the surface of the wire and minimum at the axis of the wire.

#### Page No 249:

#### Answer:

(b) is constant inside the tube

(c) is zero at the axis

A hollow tube is carrying uniform electric current along its length, so the current enclosed inside the tube is zero.

According to Ampere's law,

$\oint \overrightarrow{B}.d\overrightarrow{l}={\mathrm{\mu}}_{\mathrm{o}}{i}_{\mathrm{inside}}\phantom{\rule{0ex}{0ex}}\mathrm{Inside}\mathrm{the}\mathrm{tube},\phantom{\rule{0ex}{0ex}}\oint \overrightarrow{B}.d\overrightarrow{l}=0,rR\phantom{\rule{0ex}{0ex}}\Rightarrow {B}_{\mathrm{inside}}=\mathrm{Constant}\phantom{\rule{0ex}{0ex}}\Rightarrow {B}_{\mathrm{axis}}=0\phantom{\rule{0ex}{0ex}}$

The magnetic fields from points on the circular surface will point in opposite directions and cancel each other.

$\mathrm{Outside}\mathrm{the}\mathrm{tube},\phantom{\rule{0ex}{0ex}}B\times 2\mathrm{\pi}r={\mathrm{\mu}}_{\mathrm{o}}i\phantom{\rule{0ex}{0ex}}\Rightarrow {B}_{\mathrm{outside}}=\frac{{\mathrm{\mu}}_{\mathrm{o}}i}{2\mathrm{\pi}r},rR$

#### Page No 249:

#### Answer:

(a) outside the cable

(b) inside the inner conductor

According to Ampere's law, in a coaxial, straight cable carrying currents *i* in the inner conductor and -*i* (equally in the opposite direction) in the outside conductor.

Inside the inner conductor

$\oint \overrightarrow{B}.d\overrightarrow{l}={\mathrm{\mu}}_{\mathrm{o}}{i}_{\mathrm{inside}}\phantom{\rule{0ex}{0ex}}\oint \overrightarrow{B}.d\overrightarrow{l}=0\phantom{\rule{0ex}{0ex}}\Rightarrow B.l=0\phantom{\rule{0ex}{0ex}}\Rightarrow B=0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

In between the 2 conductors

$\oint \overrightarrow{B}.d\overrightarrow{l}={\mathrm{\mu}}_{\mathrm{o}}i\phantom{\rule{0ex}{0ex}}\Rightarrow B=\frac{{\mathrm{\mu}}_{\mathrm{o}}i}{2\mathrm{\pi}r}\phantom{\rule{0ex}{0ex}}$

Outside the outer conductor

$\oint \overrightarrow{B}.d\overrightarrow{l}={\mathrm{\mu}}_{\mathrm{o}}(i-i)\phantom{\rule{0ex}{0ex}}\Rightarrow B=0$

Therefore, the magnetic field is zero outside the cable.

#### Page No 249:

#### Answer:

(b) The magnetic field at the axis of the conductor is zero.

(c) The electric field in the vicinity of the conductor is zero.

As the current is flowing through a conductor so it it is distributed only on the surface of the conductor not in the volume of the cylindrical conductor. It is equivalent to charge distribution on a cylindrical sheet for which electric field inside a conducting cylindrical sheet is always zero.

Magnetic fields at any point inside the conducting cylinder is proportional to the distance from the axis of the cylinder.

At the axis, *r* = 0. This implies that field will be zero at the axis.

#### Page No 249:

#### Answer:

Using the relation $\overrightarrow{F}=q\stackrel{\rightharpoonup}{v}\times \overrightarrow{B}$, we get

$B=\frac{F}{qv}\phantom{\rule{0ex}{0ex}}=\frac{F}{Itv}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Units of

Force (*F*) = N

Current (*I*) = A

Time (*T*) = s

Velocity (*v*) = m/s

$\Rightarrow B=\mathrm{N}/\mathrm{A}-\mathrm{m}$

Now, using the relation $B=\frac{{\mu}_{0}i}{2\mathrm{\pi}r}$, we get

$\Rightarrow {\mu}_{0}=B\frac{2\mathrm{\pi}r}{i}=\frac{\mathrm{N}}{\mathrm{A}-\mathrm{m}}\times \frac{\mathrm{m}}{\mathrm{A}}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mu}_{0}\mathit{=}\mathrm{N}/{\mathrm{A}}^{2}$

#### Page No 249:

#### Answer:

Given:

Magnitude of current, *I* = 10 A

Separation of the point from the wire, *d* = 1 m

The magnetic field $\overrightarrow{B}$ at point (1 m, 0, 0) is given by

$B=\frac{{\mathrm{\mu}}_{0}i}{2\mathrm{\pi}d}=\frac{4\mathrm{\pi}\times {10}^{-7}\times 10}{2\mathrm{\pi}\times 1}\phantom{\rule{0ex}{0ex}}\Rightarrow B=2\times {10}^{-6}\mathrm{T}$

(Along the +ve *y*-axis by the right-hand thumb rule)

#### Page No 250:

#### Answer:

Given:

Magnitude of current, *I* = 10 A

Diameter of the wire, *d* = 1.6 × 10^{−3} m

∴ Radius of the wire = 0.8 × 10^{−3} m

The magnetic field intensity is given by

$\mathrm{B}=\frac{{\mu}_{0}i}{2\mathrm{\pi}r}$

$=\frac{2\times {10}^{-7}\times 20}{0.8\times {10}^{-3}}=5\mathrm{mT}$

#### Page No 250:

#### Answer:

Given:

Magnitude of current,* I *= 100 A

Separation of the road from the wire, *d* = 8 m

Thus, the magnetic field is given by

$B=\frac{{\mu}_{0}i}{2\mathrm{\pi}d}=\frac{2\times {10}^{-7}\times 100}{8}=2.5\mathrm{mT}$

#### Page No 250:

#### Answer:

Given:

Uniform magnetic field, *B*_{0} = 1.0 × 10^{−5} T (Vertically upwards)

Separation of the point from the wire, *d* = 2 cm = 0.02 m

The magnetic field due to the wire is given by

${B}_{\mathrm{w}}=\frac{{\mu}_{0}i}{2\mathrm{\pi}d}=\frac{4\mathrm{\pi}\times {10}^{-7}\times 1}{2\pi \times 0.02}\phantom{\rule{0ex}{0ex}}\Rightarrow {B}_{\mathrm{w}}=1\times {10}^{-5}\mathrm{T}$

Now,

Net magnetic field at point P:

*B*_{P} = *B*_{w}_{ }*+ **B*_{0} = 2 × 10^{−5} T

Net magnetic field at point Q:

*B*_{Q} = *B*_{w}_{ }−* **B*_{0} = 0

#### Page No 250:

#### Answer:

(a) As the wire in question is carrying current, so it will also generate a magnetic field around it. And for a long straight wire it will be maximum at the mid-point called P.

Now,

Magnetic field generated by the current carrying wire$=\frac{{\mathrm{\mu}}_{\mathrm{o}}i}{2\pi r}$

Net magnetic field = $\mathrm{B}+\frac{{\mathrm{\mu}}_{0}i}{2\mathrm{\pi}r}$

(b) Magnetic field B = 0

when $r<\frac{{\mathrm{\mu}}_{0}i}{2\mathrm{\pi B}}$

Clearly,

B = 0

when $r=\frac{{\mathrm{\mu}}_{0}i}{2\mathrm{\pi B}}$

But when $r>\frac{{\mathrm{\mu}}_{0}i}{2\mathrm{\pi B}}$,

Net magnetic field = $\mathrm{B}-\frac{{\mathrm{\mu}}_{0}i}{2\mathrm{\pi}r}$

#### Page No 250:

#### Answer:

Given:

Uniform magnetic field, *B*_{0} = 4.0 × 10^{−4} T

Magnitude of current, *I = *30 A

Separation of the point from the wire, *d* = 0.02 m

Thus, the magnetic field due to current in the wire is given by

$B=\frac{{\mu}_{0}I}{2\mathrm{\pi}d}$

$=\frac{2\times {10}^{-7}\times 30}{0.02}\phantom{\rule{0ex}{0ex}}=3\times {10}^{-4}\mathrm{T}$

*B*_{0} is perpendicular to B (as shown in the figure).

∴ Resultant magnetic field

${B}_{\mathrm{net}}=\sqrt{{B}^{2}+{{\mathrm{B}}_{0}}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{(4\times {10}^{-4}{)}^{2}+(3\times {10}^{-4}{)}^{2}}\phantom{\rule{0ex}{0ex}}=5\times {10}^{-4}\mathrm{T}$

#### Page No 250:

#### Answer:

Given:

Uniform magnetic field, *B*_{0} =^{ }2.0 × 10^{−3} T (From south to north)

To make the resultant magnetic field zero, the magnetic field due to the wire should be of the same magnitude as *B*_{0} and in the direction north to south.

The above condition will be satisfied when the required point will be placed in the west w.r.t. the wire.

Let the separation of the point from the wire be *d.*

The magnetic field due to current in the wire is given by

$B=\frac{{\mu}_{0}I}{2\pi d}$

From the question, *B = **B*_{0}*.*

$\Rightarrow 2.0\times {10}^{-3}=\frac{{\mu}_{0}I}{2\mathrm{\pi}d}$

$\Rightarrow 2.0\times {10}^{-3}=\frac{2\times {10}^{-7}\times 10}{d}\phantom{\rule{0ex}{0ex}}\Rightarrow d={10}^{-3}\mathrm{m}=1\mathrm{mm}$

#### Page No 250:

#### Answer:

For point A_{1},

Magnitude of current in wires, *I* = 10 A

Separation of point A_{1}_{ }from the wire on the left side, *d* = 2 cm

Separation of point A_{1}_{ }from the wire on the right side, *d*' = 6 cm

In the figure

Red and blue arrow denotes the direction of magnetic field due to the wire marked as red and blue respectively.

P (marked red) denotes the wire carrying current in a plane going into the paper.

Q (marked blue) denotes the wire carrying current in a plane coming out of the paper.

Also from the figure, we can see that

$P{A}_{4}=Q{A}_{4}\phantom{\rule{0ex}{0ex}}\angle {A}_{4}{A}_{3}P=\angle {A}_{4}{A}_{3}Q=90\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle {A}_{4}P{A}_{3}=\angle {A}_{4}Q{A}_{3}=45\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle P{A}_{4}{A}_{3}=\angle Q{A}_{4}{A}_{3}=45\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle P{A}_{4}Q=90\xb0$

The magnetic field at A_{1} due to current in the wires is given by

$B=\frac{{\mu}_{0}I}{2\mathrm{\pi}d}-\frac{{\mu}_{0}I}{2\mathrm{\pi}d\text{'}}$ ...(1)

$\Rightarrow B=\frac{2\times {10}^{-7}\times 10}{2\times {10}^{-2}}-\frac{2\times {10}^{-7}\times 10}{6\times {10}^{-2}}\phantom{\rule{0ex}{0ex}}=\left(1-\frac{1}{3}\right)\times {10}^{-4}\phantom{\rule{0ex}{0ex}}=0.67\times {10}^{-4}\mathrm{T}$

Similarly, we get the magnetic field at A_{2} using eq. (1).

$B=\frac{2\times {10}^{-7}\times 10}{1\times {10}^{-2}}+\frac{2\times {10}^{-7}\times 10}{3\times {10}^{-2}}\phantom{\rule{0ex}{0ex}}=\frac{8}{3}\times {10}^{-4}\mathrm{T}\phantom{\rule{0ex}{0ex}}=2.67\times {10}^{-4}\mathrm{T}$

Now,

Magnetic field at A_{3}:

$B=\frac{2\times {10}^{-7}\times 10}{2\times {10}^{-2}}+\frac{2\times {10}^{-7}\times 10}{2\times {10}^{-2}}\phantom{\rule{0ex}{0ex}}=2\times {10}^{-4}\mathrm{T}$

Magnetic field at A_{4}:

Separation of point A_{4}_{ }from the wire on the left side, *d* = $\sqrt{{2}^{2}+{2}^{2}}=2\sqrt{2}\mathrm{cm}$

Separation of point A_{4}_{ }from the wire on the right side, *d*' =$\sqrt{{2}^{2}+{2}^{2}}=2\sqrt{2}\mathrm{cm}$

Thus, the magnetic field at A_{4} due to current in the wires is given by

$B=\sqrt{{\left(\frac{2\times {10}^{-7}\times 10}{2\sqrt{2}\times {10}^{-2}}\right)}^{2}+{\left(\frac{2\times {10}^{-7}\times 10}{2\sqrt{2}\times {10}^{-2}}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=1\times {10}^{-4}\mathrm{T}$

#### Page No 250:

#### Answer:

Given:

Magnitude of currents, *I*_{1}_{ }= *I*_{2} = 10 A

Separation of the point from the wires, *d* = 2 cm

Thus, the magnetic field due to current in the wire is given by

${B}_{1}={B}_{2}=\frac{{\mu}_{0}I}{2\mathrm{\pi}d}$

In the figure, dotted circle shows the magnetic field lines due to current carrying wire placed in a plane perpendicular to the plane of the paper.

From the figure, we can see that $\u2206\mathrm{P}{I}_{1}{I}_{2}$ is an equilateral triangle.

$\angle {I}_{1}\mathrm{P}{I}_{\mathit{2}}\mathit{}\mathit{=}\mathit{}{60}^{\xb0}$

Angle between the magnetic fields due to current in the wire, *θ* = $60\xb0$

∴ Required magnetic field at P

${B}_{\mathrm{net}}=\sqrt{{{B}_{1}}^{2}+{{B}_{2}}^{2}+2{B}_{1}{B}_{2}\mathrm{cos}\theta}$

$=\sqrt{{\left(\frac{2\times {10}^{-7}\times 10}{2\times {10}^{-2}}\right)}^{2}+{\left(\frac{2\times {10}^{-7}\times 10}{2\times {10}^{-2}}\right)}^{2}+\left(\frac{2\times {10}^{-7}\times 10}{2\times {10}^{-2}}\right)+\left(\frac{2\times {10}^{-7}\times 10}{2\times {10}^{-2}}\right)\mathrm{cos}60\xb0}$

$=\sqrt{\left({10}^{-4}\right)+({10}^{-4}{)}^{2}+2({10}^{-4}\left)\right({10}^{-4})\times \frac{1}{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{3}\times {10}^{-4}\mathrm{T}\phantom{\rule{0ex}{0ex}}=1.732\times {10}^{-4}\mathrm{T}$

#### Page No 250:

#### Answer:

Given:

Magnitude of current, *I* = 5 A

Separation of the point from the wire, *d* = 1 m

Thus, the magnitude of magnetic field due to current in the wires is given by

${B}_{1}={B}_{2}=\frac{{\mu}_{0}I}{2\pi d}$

(a) At point (1 m, 1 m), the magnetic fields due to the wires are the same in magnitude, but they are opposite in direction.

Hence, the net magnetic field is zero.

(b) At point (−1 m, 1 m), the magnetic fields due to the wires are in upward direction.

$\Rightarrow {B}_{\mathrm{net}}={B}_{1}+{B}_{2}\phantom{\rule{0ex}{0ex}}=\left(\frac{2\times {10}^{-7}\times 5}{1}+\frac{2\times {10}^{-7}\times 5}{1}\right)\phantom{\rule{0ex}{0ex}}$

= 2 × 10^{−6} T (Along the *z*-axis)

(c) At point (−1 m, −1 m), the magnetic fields due to the wires are the same in magnitude, but they are opposite in direction.

Hence, the net magnetic field is zero.

(d) At point (1 m, −1 m), the magnetic fields due to the wires are in upward direction.

$\Rightarrow {B}_{\mathrm{net}}={B}_{1}+{B}_{2}\phantom{\rule{0ex}{0ex}}=\left(\frac{2\times {10}^{-7}\times 5}{1}+\frac{2\times {10}^{-7}\times 5}{1}\right)\phantom{\rule{0ex}{0ex}}$

= 2 × 10^{−6} T (Along the negative *z*-axis)

#### Page No 250:

#### Answer:

Given:

Let the horizontal wires placed at the bottom and top are denoted as W_{1} and W_{2} respectively.

Let the vertical wires placed at the right and left to point P are denoted as W_{3} and W_{4} respectively.

Magnitude of current, *I* = 5 A

(a) Consider point P.

Magnetic fields due to wires W_{1} and W_{2}_{ }are the same in magnitude, but they are opposite in direction.

Magnetic fields due to wires W_{3} and W_{4}_{ }are the same in magnitude, but they are opposite in direction.

Hence, the net magnetic field is zero.

Net magnetic field at P due to these four wires = 0

(b) Consider point Q_{1}.

Due to wire W_{1}, separation of point Q_{1}_{ }from the wire (*d*) is 7.5 cm.

So, the magnetic field due to current in the wire is given by

${B}_{{W}_{1}}=\frac{{\mu}_{0}I}{2\pi d}$

* *= 4 × 10^{−5}^{ }T (In upward direction)

Due to wire W_{2}, separation of point Q_{1}_{ }from the wire (*d*) is 2.5 cm.

So, the magnetic field due to current in the wire is given by

${B}_{{\mathrm{W}}_{2}}=\frac{4}{3}\times {10}^{-5}\mathrm{T}$ (In upward direction)

Due to wire W_{3}, separation of point Q_{1}_{ }from the wire (*d*) is 7.5 cm.

So, the magnetic field due to current in the wire is given by

*B*_{W}_{3} = 4 × 10^{−5} T (In upward direction)

Due to wire W_{4}, separation of point Q_{1}_{ }from the wire (*d*) is 2.5 cm.

So, the magnetic field due to current in the wire is given by

${B}_{{\mathrm{W}}_{4}}=\frac{4}{3}\times {10}^{-5}\mathrm{T}$ (In upward direction)

∴ Net magnetic field at point Q_{1}

${B}_{{\mathrm{Q}}_{1}}=\left(4+\frac{4}{3}+4+\frac{4}{3}\right)\times {10}^{-5}\phantom{\rule{0ex}{0ex}}=\frac{32}{3}\times {10}^{-5}\phantom{\rule{0ex}{0ex}}=1.06\times {10}^{-4}\mathrm{T}(\mathrm{In}\mathrm{upward}\mathrm{direction})$

At point Q_{2},

Magnetic field due to wire W_{1}:

*B*_{W}_{1}_{ }= 4 × 10^{−5} T (In upward direction)

Magnetic field due to wire W_{2}:

${B}_{{\mathrm{W}}_{2}}=\frac{4}{3}\times {10}^{-5}\mathrm{T}$ (In upward direction)

Magnetic field due to wire W_{3}:

${B}_{{\mathrm{W}}_{3}}=\frac{4}{3}\times {10}^{-5}\mathrm{T}$ (In downward direction)

Magnetic field due to wire W_{4}:

${\mathrm{B}}_{{\mathrm{W}}_{4}}=4\times {10}^{-5}\mathrm{T}$ (In downward direction)

∴ Net magnetic field at point Q_{2}_{,
${B}_{{Q}_{2}}=0$}

Similarly, at point Q_{3}, the magnetic field is 1.1 × 10^{−4}^{ }T (in downward direction) and at point Q_{4}, the magnetic field is zero.

#### Page No 250:

#### Answer:

As shown in the figure, points P, Q, R and S lie on a circle of radius *d*.

Let the wires be named W_{1} and W_{2}.

Now,

At point P, the magnetic field due to wire W_{1} is given by

*B*_{1} = 0

At point P, the magnetic field due to wire W_{2} is given by

${B}_{2}=\frac{{\mathrm{\mu}}_{0}i}{4\mathrm{\pi}d}$ (Perpendicular to the plane in outward direction)

$\Rightarrow {B}_{\mathrm{net}}=\frac{{\mathrm{\mu}}_{0}i}{4\mathrm{\pi}d}$ (Perpendicular to the plane in outward direction)

At point Q, the magnetic field due to wire W_{1} is given by

${B}_{1}=\frac{{\mathrm{\mu}}_{0}i}{4\mathrm{\pi}d}$ (Perpendicular to the plane in inward direction)

At point Q, the magnetic field due to wire W_{2} is given by

*B*_{2} = 0

$\Rightarrow {B}_{\mathrm{net}}=\frac{{\mathrm{\mu}}_{0}i}{4\mathrm{\pi}d}$ (Perpendicular to the plane in inward direction)

At point R, the magnetic field due to wire W_{1} is given by

*B*_{1} = 0

At point R, the magnetic field due to wire W_{2} is given by

${B}_{2}=\frac{{\mathrm{\mu}}_{0}i}{4\mathrm{\pi}d}$ (Perpendicular to the plane in inward direction)

$\Rightarrow {B}_{\mathrm{net}}=\frac{{\mathrm{\mu}}_{0}i}{4\mathrm{\pi}d}$ (Perpendicular to the plane in inward direction)

At point S, the magnetic field due to wire W_{1} is given by

${B}_{1}=\frac{{\mathrm{\mu}}_{0}i}{4\mathrm{\pi}d}$ (Perpendicular to the plane in outward direction)

At point S, the magnetic field due to wire W_{2} is given by

*B*_{2} = 0

$\Rightarrow {B}_{\mathrm{net}}=\frac{{\mathrm{\mu}}_{0}i}{4\mathrm{\pi}d}$ (Perpendicular to the plane in outward direction)

Hence, the magnitude of the magnetic field at points P, Q, R and S is $\frac{{\mathrm{\mu}}_{0}i}{4\mathrm{\pi}d}$.

#### Page No 250:

#### Answer:

Let AB be the wire of length *x* with midpoint O.

Given:

Magnitude of current = *i*

Separation of the point from the wire = *d*

Now,

The magnetic field on a perpendicular bisector is given by

$B=\frac{{\mathrm{\mu}}_{0}i}{4\mathrm{\pi}d}(\mathrm{sin}\theta +\mathrm{sin}\theta )$

$B=\frac{{\mathrm{\mu}}_{0}i}{4\mathrm{\pi}d}\frac{2x}{\sqrt{{x}^{2}+4{d}^{2}}}$

So, if *d* > > *x* (neglecting* x*), then

$B=\frac{{\mathrm{\mu}}_{0}i}{4\mathrm{\pi}d}\frac{2x}{2d}\phantom{\rule{0ex}{0ex}}\Rightarrow B\propto \frac{1}{{d}^{2}}$

And, if *d *< <* x* (neglecting* d*), then

$B=\frac{{\mathrm{\mu}}_{0}i}{4\mathrm{\pi}d}\frac{2x}{x}\phantom{\rule{0ex}{0ex}}\Rightarrow B\propto \frac{1}{d}$

#### Page No 250:

#### Answer:

Let AB be the wire of length 10* *cm and P be the required point.

Given:

Magnitude of current, *i* = 10 A

The angles made by points A and B with point P are ${\theta}_{1}=30\xb0\mathrm{and}{\theta}_{2}=30\xb0$, respectively.

∴ Separation of the point from the wire, *d* = $\frac{\sqrt{3}}{2}a=\frac{\sqrt{3}}{2}\times 10=5\sqrt{3}\mathrm{cm}$

Thus, the magnetic field due to current in the wire is given by

$B=\frac{{\mathrm{\mu}}_{0}i}{4\mathrm{\pi}d}(\mathrm{sin}{\theta}_{1}+\mathrm{sin}{\theta}_{2})\phantom{\rule{0ex}{0ex}}=\frac{{10}^{-7}\times 10}{5\sqrt{3}\times {10}^{-2}}\left(\frac{1}{2}+\frac{1}{2}\right)\phantom{\rule{0ex}{0ex}}=11.54\times {10}^{-6}\mathrm{T}\phantom{\rule{0ex}{0ex}}$

#### Page No 250:

#### Answer:

Given:

Magnitude of current* = i*

Separation of the point from the wire = *d*

Thus, the magnetic field due to current in the long wire is given by

${B}_{1}=\frac{{\mu}_{0}i}{2\mathrm{\pi}d}\phantom{\rule{0ex}{0ex}}$

Also, the magnetic field due to a section of length *l* on a perpendicular bisector is given by

${B}_{2}=\frac{{\mu}_{0}i}{4\pi d}\frac{2l}{\sqrt{{l}^{2}+4{d}^{2}}}$

$\Rightarrow \frac{{\mu}_{0}i\mathit{}l}{4\mathrm{\pi}d}\frac{2}{d\sqrt{{\displaystyle \frac{{l}^{2}}{{d}^{2}}}+4}}\phantom{\rule{0ex}{0ex}}\mathrm{Neglecting}\frac{{l}^{2}}{{d}^{2}}\left(\mathrm{very}\mathrm{small}\right),\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}{B}_{2}=\frac{{\mu}_{0}i\mathit{}l}{4\mathrm{\pi}{d}^{2}}\times \frac{2}{\sqrt{2}}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{2}{\mu}_{0}i\mathit{}l}{4\mathrm{\pi}{d}^{2}}\phantom{\rule{0ex}{0ex}}$

Now,

B_{1} > B_{2}

According to the question,

$\frac{{B}_{1}\mathit{-}{B}_{2}}{{B}_{1}}\mathit{=}\frac{1}{100}\phantom{\rule{0ex}{0ex}}\Rightarrow {B}_{2}=0.99{B}_{1}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\sqrt{2}{\mu}_{0}il}{4\pi {d}^{2}}=0.99\times \frac{{\mu}_{0}i}{2\pi d}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{d}{l}=\frac{1.414}{1.98}=0.71$

#### Page No 250:

#### Answer:

Let the currents in wires ABC and ADC be *i*_{1} and *i*_{2}, respectively.

The resistances in wires ABC and ADC are *r* and 2*r*, respectively.

$\therefore \frac{{i}_{1}}{{i}_{2}}=\frac{2}{1}\phantom{\rule{0ex}{0ex}}\Rightarrow {i}_{1}-2{i}_{2}=0...\left(1\right)$

And,

${i}_{1}+{i}_{2}=i...\left(2\right)$

Using (1) and (2), we get

${i}_{1}=\frac{2i}{3}$ and ${i}_{2}=\frac{i}{3}$

The angles made by points A and D with point O are ${\theta}_{1}=45\xb0\mathrm{and}{\theta}_{2}=45\xb0$, respectively.

Separation of the point from the wire, *d* =* a*/2

Now,

The magnetic field due to current in wire AD is given by

$B=\frac{{\mathrm{\mu}}_{0}{i}_{2}}{4\mathrm{\pi}d}(\mathrm{sin}{\theta}_{1}+\mathrm{sin}{\theta}_{2})\phantom{\rule{0ex}{0ex}}\Rightarrow B=\frac{{\mathrm{\mu}}_{0}{\displaystyle \frac{i}{3}}}{4\mathrm{\pi}{\displaystyle \frac{a}{2}}}(\mathrm{sin}45+\mathrm{sin}45)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

The magnetic field at centre due to wire ADC is given by

$B\text{'}=2B=\frac{{\mathrm{\mu}}_{0}}{4\mathrm{\pi}}\frac{i}{3}\frac{a}{{a}^{2}}\times 4\times \sqrt{2}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{2}{\mathrm{\mu}}_{0}i}{3\mathrm{\pi}a}\phantom{\rule{0ex}{0ex}}$

(Perpendicular to the plane in outward direction)

The magnetic field at centre due to wire ABC is given by

$B\text{'}\text{'}=\frac{\mathrm{\mu}}{4\mathrm{\pi}}\frac{2i}{3}\frac{a}{{a}^{2}}\times 4\times \sqrt{2}\phantom{\rule{0ex}{0ex}}=\frac{2\sqrt{2}{\mathrm{\mu}}_{0}i}{3\mathrm{\pi}a}$

(Perpendicular to the plane in inward direction)

$\therefore {B}_{\mathrm{net}}=B\text{'}\text{'}-B\text{'}=\frac{\sqrt{2}{\mathrm{\mu}}_{0}i}{3\mathrm{\pi}a}$

(Perpendicular to the plane in inward direction)

#### Page No 250:

#### Answer:

B at P due to AD = $\frac{{\mathrm{\mu}}_{0}}{4\pi}.\frac{i}{2}.\frac{4}{{d}^{2}}.a\left[\frac{\left({\displaystyle \frac{a}{2}}\right)}{\sqrt{{\left({\displaystyle \frac{a}{2}}\right)}^{2}+{\left({\displaystyle \frac{a}{4}}\right)}^{2}}}+\frac{\left({\displaystyle \frac{a}{2}}\right)}{\sqrt{{\left({\displaystyle \frac{a}{2}}\right)}^{2}+{\left({\displaystyle \frac{3a}{4}}\right)}^{2}}}\right]$along•

$\frac{{\mathrm{\mu}}_{0}i}{4\pi a}\left[\frac{\left({\displaystyle \frac{a}{2}}\right)}{\sqrt{{\left({\displaystyle \frac{a}{2}}\right)}^{2}+{\left({\displaystyle \frac{a}{4}}\right)}^{2}}}+\frac{\left({\displaystyle \frac{a}{2}}\right)}{\sqrt{{\left({\displaystyle \frac{a}{2}}\right)}^{2}+{\left({\displaystyle \frac{3a}{4}}\right)}^{2}}}\right]$along•

B at P due to AC=$\frac{{\mathrm{\mu}}_{0}}{4\pi}.\frac{i}{2}.\frac{16}{9{a}^{2}}.a.2\left[\frac{\left({\displaystyle \frac{3a}{4}}\right)}{\sqrt{{\left({\displaystyle \frac{3a}{4}}\right)}^{2}+{\left({\displaystyle \frac{a}{2}}\right)}^{2}}}\right]$

$=\frac{4{\mathrm{\mu}}_{0}i}{9\pi a}\left[\frac{\left({\displaystyle \frac{3a}{4}}\right)}{\sqrt{{\left({\displaystyle \frac{a}{4}}\right)}^{2}+{\left({\displaystyle \frac{3a}{2}}\right)}^{2}}}\right]$along•

B at P due to AB = $\frac{{\mathrm{\mu}}_{0}}{4\pi}.\frac{i}{2}.\frac{16}{9{a}^{2}}.a.2\left[\frac{\left({\displaystyle \frac{3a}{4}}\right)}{\sqrt{{\left({\displaystyle \frac{3a}{4}}\right)}^{2}+{\left({\displaystyle \frac{a}{2}}\right)}^{2}}}\right]$along•

B at P due to BC = $\frac{{\mathrm{\mu}}_{0}}{4\pi}.\frac{i}{2}.\frac{4}{{a}^{2}}.a.\left[\frac{\left({\displaystyle \frac{a}{2}}\right)}{\sqrt{{\left({\displaystyle \frac{a}{2}}\right)}^{2}+{\left({\displaystyle \frac{a}{4}}\right)}^{2}}}+\frac{\left({\displaystyle \frac{a}{2}}\right)}{\sqrt{{\left({\displaystyle \frac{a}{2}}\right)}^{2}+{\left({\displaystyle \frac{3a}{4}}\right)}^{2}}}\right]$ along•

$=\frac{{\mathrm{\mu}}_{0}i}{2\pi a}\left[\frac{\left({\displaystyle \frac{a}{2}}\right)}{\sqrt{{\left({\displaystyle \frac{a}{2}}\right)}^{2}+{\left({\displaystyle \frac{a}{4}}\right)}^{2}}}+\frac{\left({\displaystyle \frac{a}{2}}\right)}{\sqrt{{\left({\displaystyle \frac{a}{2}}\right)}^{2}+{\left({\displaystyle \frac{3a}{4}}\right)}^{2}}}\right]\mathrm{along}\otimes $

So, net magnetic field at point P.

$\mathrm{B}=\frac{4{\mathrm{\mu}}_{0}i}{\mathrm{\pi}a}\left[\frac{\left({\displaystyle \frac{\mathrm{a}}{4}}\right)}{\sqrt{{\left({\displaystyle \frac{\mathrm{a}}{2}}\right)}^{2}+{\left({\displaystyle \frac{\mathrm{a}}{4}}\right)}^{2}}}\right]-\frac{4{\mathrm{\mu}}_{0}i}{9\mathrm{\pi}a}\left[\frac{\left({\displaystyle \frac{3\mathrm{a}}{4}}\right)}{\sqrt{{\left({\displaystyle \frac{\mathrm{a}}{2}}\right)}^{2}+{\left({\displaystyle \frac{3\mathrm{a}}{4}}\right)}^{2}}}\right]\phantom{\rule{0ex}{0ex}}=\frac{4{\mathrm{\mu}}_{0}i}{\mathrm{\pi}a}\frac{1}{4}\left[\frac{1}{\sqrt{{\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{16}}}}\right]-\frac{4{\mathrm{\mu}}_{0}i}{9\mathrm{\pi}a}.\frac{3}{4}\left[\frac{1}{\sqrt{{\displaystyle \frac{1}{4}}+{\displaystyle \frac{9}{16}}}}\right]$

$=\frac{4{\mathrm{\mu}}_{0}i}{4\pi a}\left[\frac{4}{\sqrt{5}}\right]-\frac{{\mathrm{\mu}}_{0}i}{3\pi a}\left[\frac{4}{\sqrt{13}}\right]\left[\frac{4}{\sqrt{13}}\right]\phantom{\rule{0ex}{0ex}}$ along•

$=\frac{4{\mathrm{\mu}}_{0}i}{2\pi a}\left[\frac{1}{\sqrt{5}}-\frac{1}{3\sqrt{13}}\right]$ along•

$=\frac{2{\mathrm{\mu}}_{0}i}{\pi a}\left[\frac{1}{\sqrt{5}}-\frac{1}{3\sqrt{13}}\right]$ along•

#### Page No 251:

#### Answer:

The loop ABCD can be considered as a circuit with two resistances in parallel, one along branch AB and other along branch ADC.

As, the sides of the loop are identical, their resistances are also same.

Let the resistance of each side be *r*.

The resistance of branch AB = *r*

The resistance of branch ADC = 3*r*

The current in the branches are calculated as:

${i}_{\mathrm{AB}}=i\times \frac{3r}{3r+r}=\frac{3i}{4}$

${i}_{\mathrm{ADC}}=i\times \frac{r}{3r+r}=\frac{i}{4}$

As current follow the least resistive path so

Current in branch AB = $\frac{3i}{4}$

Current in branch ADC = $\frac{i}{4}$

At the centre of the loop:

Magnetic field due to wire AD, DC and CB will be into the plane of paper according to right hand thumb rule.

Magnetic field due to wire AB will be out of the plane of paper according to right hand thumb rule.

Net magnetic field at the centre = B_{AD} +B_{DC} +B_{CB} − B_{AB}_{ }which will be out of the plane of paper.

As, perpendicular distance of the centre from every wire will be equal to $\frac{a}{\sqrt{2}}$ and angle made by corner points of each side at the centre is $45\xb0$.

${B}_{AD}=\frac{{\mu}_{0}\left({\displaystyle \frac{i}{4}}\right)}{4\mathrm{\pi}\left({\displaystyle \frac{\mathrm{a}}{\sqrt{2}}}\right)}(\mathrm{sin}45\xb0+\mathrm{sin}45\xb0)\phantom{\rule{0ex}{0ex}}=\frac{{\mu}_{0}{\displaystyle i}}{8\mathrm{\pi a}}\phantom{\rule{0ex}{0ex}}{B}_{AD}={B}_{DC}={B}_{CB}\phantom{\rule{0ex}{0ex}}{B}_{AD}+{B}_{DC}+{B}_{CB}={B}^{,}=\frac{3{\mu}_{0}{\displaystyle i}}{8\mathrm{\pi a}}\phantom{\rule{0ex}{0ex}}{B}_{AB}=\frac{{\mu}_{0}\left({\displaystyle \frac{3i}{4}}\right)}{4\mathrm{\pi}\left({\displaystyle \frac{\mathrm{a}}{\sqrt{2}}}\right)}(\mathrm{sin}45\xb0+\mathrm{sin}45\xb0)\phantom{\rule{0ex}{0ex}}=\frac{3{\mu}_{0}{\displaystyle i}}{8\mathrm{\pi a}}\phantom{\rule{0ex}{0ex}}{B}_{net}={B}^{,}-{B}_{AB}\phantom{\rule{0ex}{0ex}}=0\phantom{\rule{0ex}{0ex}}$

#### Page No 251:

#### Answer:

Let current* *2*I** *enter the circuit.

Since the wire is uniform, the current will be equally divided at point A (as shown in the figure).

Now,

Magnetic field at P due to wire AB = *B* (say)

(Perpendicular to the plane in outward direction)

Magnetic field at P due to wire BD = *B' *(say)

(Perpendicular to the plane in outward direction)

Magnetic field at P due to wire AC = Magnetic field at P due to wire AB = *B*

(Perpendicular to the plane in inward direction)

Magnetic field at P due to wire CD = Magnetic field at P due to wire BD = *B*'

(Perpendicular to the plane in inward direction)

∴ Net magnetic field at P =* B + B' − B − B'* = 0

#### Page No 251:

#### Answer:

Let ABC be the equilateral triangle with side *l*/3 and centre M.

(a)

$\mathrm{In}\u2206\mathrm{AOB},\phantom{\rule{0ex}{0ex}}\mathrm{AO}=\sqrt{{\left(\frac{l}{3}\right)}^{2}-{\left(\frac{I}{6}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=l\sqrt{\frac{1}{9}-\frac{1}{36}}=l\sqrt{\frac{4-1}{36}}=l\sqrt{\frac{1}{12}}\phantom{\rule{0ex}{0ex}}\therefore \mathrm{MO}=\frac{1}{3}\times l\sqrt{\frac{1}{12}}=\frac{l}{6\sqrt{3}}$

The angles made by points B and C with centre M are ${\theta}_{1}=60\xb0\mathrm{and}{\theta}_{2}=60\xb0$, respectively.

Separation of the point from the wire, *d* = MO *=* $\frac{l}{6\sqrt{3}}$

Thus, the magnetic field due to current in wire BC is given by

$B=\frac{{\mathrm{\mu}}_{0}i}{4\mathrm{\pi}d}(\mathrm{sin}{\theta}_{1}+\mathrm{sin}{\theta}_{2})\phantom{\rule{0ex}{0ex}}\Rightarrow B=\frac{{\mathrm{\mu}}_{0}{\displaystyle i}}{4\mathrm{\pi}{\displaystyle \frac{\mathrm{l}}{6\sqrt{3}}}}(\mathrm{sin}60+\mathrm{sin}60)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

$\Rightarrow B=\frac{{\mathrm{\mu}}_{0}i}{4\pi l}6\sqrt{3}\times \sqrt{3}$

Now,

Net magnetic field at M = Magnetic field due to wire BC + Magnetic field due to wire CA + Magnetic field due to wire AB

Since all wires are the same,

${B}_{\mathrm{net}}=3B$$=\frac{27{\mathrm{\mu}}_{0}i}{\mathrm{\pi l}}$

It is perpendicular to the plane in outward direction if the current is anticlockwise and perpendicular to the plane in inward direction if the current is clockwise.

(b)

The angles made by points B and C with centre M are ${\theta}_{1}=45\xb0\mathrm{and}{\theta}_{2}=45\xb0$, respectively.

Separation of the point from the wire, *d* =* l*/8

Thus, the magnetic field due to current in wire BC is given by

$B=\frac{{\mathrm{\mu}}_{0}i}{4\mathrm{\pi}d}(\mathrm{sin}{\theta}_{1}+\mathrm{sin}{\theta}_{2})\phantom{\rule{0ex}{0ex}}\Rightarrow B=\frac{{\mathrm{\mu}}_{0}{\displaystyle \mathrm{i}}}{4\mathrm{\pi}{\displaystyle \frac{\mathrm{l}}{8}}}(\mathrm{sin}45+\mathrm{sin}45)\phantom{\rule{0ex}{0ex}}=\frac{2\sqrt{2}{\mathrm{\mu}}_{0}{\displaystyle i}}{\mathrm{\pi}{\displaystyle l}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Since all wires are the same,

Net magnetic field at M = 4 × Magnetic field due to wire BC

$\Rightarrow {B}_{\mathrm{net}}=4B=$$\frac{8\sqrt{2}{\mathrm{\mu}}_{0}{\displaystyle i}}{\mathrm{\pi}{\displaystyle l}}$

#### Page No 251:

#### Answer:

Let CAB be the wire making an angle α, P be the point on the bisector of this angle situated at a distance *x* from the vertex A and *d* be the perpendicular distance of AC and AB from P.

From the figure,

$\mathrm{sin}\left(\frac{\alpha}{2}\right)=\frac{d}{x}$

$d=x\mathrm{sin}\left(\frac{\alpha}{2}\right)$

The angles made by points A and C with point P are ${\theta}_{1}=90-\frac{\alpha}{2}\mathrm{and}{\theta}_{2}=90\xb0$, respectively.

Separation of the point from the wire, $d=x\mathrm{sin}\left(\frac{\alpha}{2}\right)$

Thus, the magnetic field due to current in wire AC is given by

$B=\frac{{\mathrm{\mu}}_{0}i}{4\mathrm{\pi}d}(\mathrm{sin}{\theta}_{1}+\mathrm{sin}{\theta}_{2})\phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{\mu}}_{0}{\displaystyle i}}{4\mathrm{\pi}{\displaystyle x\mathrm{sin}\frac{{\displaystyle \alpha}}{2}}}\left[\mathrm{sin}\left(90-\frac{{\displaystyle \alpha}}{2}\right)+\mathrm{sin}90\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

$\frac{{\mathrm{\mu}}_{0}i}{4\mathrm{\pi}x\mathrm{sin}\left({\displaystyle \frac{\alpha}{2}}\right)}\left[\mathrm{cos}\frac{\alpha}{2}+1\right]\phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{\mu}}_{0}i2{\mathrm{cos}}^{2}\left({\displaystyle \frac{\alpha}{4}}\right)}{4\mathrm{\pi}x2\mathrm{sin}\left({\displaystyle \frac{\alpha}{4}}\right)\mathrm{cos}\left({\displaystyle \frac{\alpha}{4}}\right)}=\frac{{\mathrm{\mu}}_{0}i\mathrm{cot}\left({\displaystyle \frac{\alpha}{4}}\right)}{4\mathrm{\pi}x}$

Now, the magnetic field due to wires AC and AB is given by

${B}_{\mathrm{net}}=2B=\frac{{\mathrm{\mu}}_{0}i\mathrm{cot}\left({\displaystyle \frac{\alpha}{4}}\right)}{2\mathrm{\pi}x}$

#### Page No 251:

#### Answer:

Let the angles made by points A and B with point O be ${\theta}_{1}\mathrm{and}{\theta}_{2}$, respectively.

And,

${\theta}_{1}={\theta}_{2}=\theta $

Separation of the point from the wire, *d* =* b*/2

$\mathrm{In}\u2206\mathrm{AOM},\phantom{\rule{0ex}{0ex}}\mathrm{AO}=\sqrt{{\left(\frac{b}{2}\right)}^{2}+{\left(\frac{l}{2}\right)}^{2}}=\frac{1}{2}\sqrt{{b}^{2}+{l}^{2}}$

And,

$\mathrm{sin}\theta =\frac{{\displaystyle \frac{l}{2}}}{{\displaystyle \frac{1}{2}1}\sqrt{{b}^{2}+{l}^{2}}}=\frac{l}{\sqrt{{b}^{2}+{l}^{2}}}$

Thus, the magnetic field due to current in wire AB is given by

$B=\frac{{\mathrm{\mu}}_{0}i}{4\mathrm{\pi}d}(\mathrm{sin}{\theta}_{1}+\mathrm{sin}{\theta}_{2})\phantom{\rule{0ex}{0ex}}\Rightarrow B=\frac{{\mathrm{\mu}}_{0}i}{4\mathrm{\pi}d}\times 2\mathrm{sin}\theta \phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

$\Rightarrow B=\frac{{\mathrm{\mu}}_{0}i}{4\mathrm{\pi}\left({\displaystyle \frac{b}{2}}\right)}\times \frac{2l}{\sqrt{{l}^{2}+{b}^{2}}}\phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{\mu}}_{0}i}{\mathrm{\pi}b}\frac{l}{\sqrt{{l}^{2}+{b}^{2}}}$

Similarly, the magnetic field due to current in wire BC is given by

$B\text{'}=\frac{{\mathrm{\mu}}_{0}i}{\mathrm{\pi}l}\frac{b}{\sqrt{{l}^{2}+{b}^{2}}}$

Now,

Magnetic field due to current in wire CD = Magnetic field due to current in wire AB = *B*

And,

Magnetic field due to current in wire DA = Magnetic field due to current in wire BC = *B*'

$\therefore {B}_{net}=2\left[B+B\text{'}\right]\phantom{\rule{0ex}{0ex}}=2\left[\frac{{\mathrm{\mu}}_{0}i}{\mathrm{\pi}b}\frac{l}{\sqrt{{l}^{2}+{b}^{2}}}+\frac{{\mathrm{\mu}}_{0}i}{\mathrm{\pi}l}\frac{b}{\sqrt{{l}^{2}+{b}^{2}}}\right]\phantom{\rule{0ex}{0ex}}=\frac{2{\mathrm{\mu}}_{0}i}{\mathrm{\pi}\sqrt{{l}^{2}+{b}^{2}}}\left[\frac{l}{b}+\frac{b}{l}\right]\phantom{\rule{0ex}{0ex}}=\frac{2{\mathrm{\mu}}_{0}i\sqrt{{b}^{2}+{l}^{2}}}{\mathrm{\pi}bl}$

#### Page No 251:

#### Answer:

(a)

Using figure,

$\mathrm{For}\mathrm{a}\mathrm{regure}\mathrm{polygon}\mathrm{of}\mathrm{n}-\mathrm{sides},\mathrm{the}\mathrm{angle}\mathrm{subtended}\mathrm{at}\mathrm{the}\mathrm{centre}\mathrm{is}\frac{2\mathrm{\pi}}{n}\phantom{\rule{0ex}{0ex}}\mathrm{tan}\theta =\frac{l}{2x}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{l}{2\mathrm{tan}\theta}\phantom{\rule{0ex}{0ex}}\mathrm{considering}\mathrm{angle}\mathrm{to}\mathrm{be}\mathrm{small}\phantom{\rule{0ex}{0ex}}\frac{l}{2}=\frac{\mathrm{\pi r}}{n}\phantom{\rule{0ex}{0ex}}\mathrm{Using}\mathrm{Biot}-\mathrm{Savart}\mathrm{law}\mathrm{for}\mathrm{one}\mathrm{side}\phantom{\rule{0ex}{0ex}}B=\frac{{\mathrm{\mu}}_{0}}{4\pi}\frac{idl\mathrm{sin}\theta}{{x}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow B=\frac{{\mathrm{\mu}}_{0}}{4\pi}\frac{i(\mathrm{sin}\theta +\mathrm{sin}\theta )}{{x}^{2}}=\frac{{\mathrm{\mu}}_{0}i2\left(\mathrm{tan}\theta \right)\left(2\mathrm{sin}\theta \right)}{4\pi l}\left(\mathrm{Putting}\mathrm{value}\mathrm{of}r\right)\phantom{\rule{0ex}{0ex}}\Rightarrow B=\frac{{\mathrm{\mu}}_{0}i2n\left(\mathrm{tan}\left({\displaystyle \frac{\mathrm{\pi}}{n}}\right)\right)\left(2\mathrm{sin}\left({\displaystyle \frac{\mathrm{\pi}}{n}}\right)\right)}{4\mathrm{\pi}\left(2\mathrm{\pi r}\right)}\left(\mathrm{Putting}\mathrm{value}\mathrm{of}l\right)\phantom{\rule{0ex}{0ex}}\mathrm{For}\mathrm{n}-\mathrm{sided}\mathrm{polygon}\phantom{\rule{0ex}{0ex}}B\text{'}=nB\phantom{\rule{0ex}{0ex}}\Rightarrow B\text{'}=\frac{{\mathrm{\mu}}_{0}i{n}^{2}\mathrm{tan}\left({\displaystyle \frac{\mathrm{\pi}}{n}}\right)\mathrm{sin}\left({\displaystyle \frac{\mathrm{\pi}}{n}}\right)}{2{\pi}^{2}r}\phantom{\rule{0ex}{0ex}}$

(b)

when *n*→ ∞, polygon becomes a circle with radius *r*

and magnetic field will become

$B=\frac{{\mathrm{\mu}}_{0}i}{2r}$

#### Page No 251:

#### Answer:

By appluing Kirchoff voltagee law,we can see that the current in the circuit is zero.

Net current in the circuit = 0

As

magnetic field is always proportional to the current flowing in the circuit hence,

Net magnetic field at point P = 0

Field at point P is independent of the values of the resistances in the circuit.

#### Page No 251:

#### Answer:

Given:

Magnitude of current in both wires, *i*_{1}* = **i*_{2} = 20 A

Force acting on 0.1 m of the second wire,* F* = 2.0 × 10^{−5} N

∴ Force per unit length = $\frac{2\times {10}^{-5}}{0.1}$ = 2.0 × 10^{−4} N/m

Now,

Let the separation between the two wires be *d*.

Thus, the force per unit length is given by

$\frac{F}{l}=\frac{{\mathrm{\mu}}_{0}{i}_{1}{i}_{2}}{2\mathrm{\pi}d}\phantom{\rule{0ex}{0ex}}\Rightarrow 2.0\times {10}^{-4}=\frac{2\times {10}^{-7}\times 20\times 20}{d}\phantom{\rule{0ex}{0ex}}\Rightarrow d=0.4\mathrm{m}=40\mathrm{cm}$

#### Page No 251:

#### Answer:

Let wires W_{1}, W_{2} and W_{3} be arranged as shown in the figure.

Given:

Magnitude of current in each wire,* i*_{1}_{ }= *i*_{2} =* **i*_{3} = 10 A

The magnetic force per unit length on a wire due to a parallel current-carrying wire is given by

$\frac{F}{l}=\frac{{\mathrm{\mu}}_{0}{i}_{1}{i}_{2}}{2\mathrm{\pi}d}$

So, for wire W_{1},

$\frac{F}{l}=\frac{F}{l}\mathrm{by}\mathrm{wire}{W}_{2}+\frac{F}{l}\mathrm{by}\mathrm{wire}{W}_{3}\phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{\mu}}_{0}\times 10\times 10}{2\mathrm{\pi}\times 5\times {10}^{-2}}+\frac{{\mathrm{\mu}}_{0}\times 10\times 10}{2\mathrm{\pi}\times 10\times {10}^{-2}}\phantom{\rule{0ex}{0ex}}=\frac{2\times {10}^{-7}\times {10}^{2}}{5\times {10}^{-2}}+\frac{2\times {10}^{-7}\times {10}^{2}}{10\times {10}^{-2}}\phantom{\rule{0ex}{0ex}}=4\times {10}^{-4}+2\times {10}^{-4}\phantom{\rule{0ex}{0ex}}=6\times {10}^{-4}\mathrm{N}$

For wire W_{2},

$\frac{F}{l}=\frac{F}{l}\mathrm{by}\mathrm{wire}{\mathrm{W}}_{1}-\frac{F}{l}\mathrm{by}\mathrm{wire}{\mathrm{W}}_{3}\phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{\mu}}_{0}\times 10\times 10}{2\mathrm{\pi}\times 5\times {10}^{-2}}-\frac{{\mathrm{\mu}}_{0}\times 10\times 10}{2\mathrm{\pi}\times 5\times {10}^{-2}}=0$

For wire W_{3},

$\frac{F}{l}=\frac{F}{l}\mathrm{by}\mathrm{wire}{\mathrm{W}}_{1}+\frac{F}{l}\mathrm{by}\mathrm{wire}{\mathrm{W}}_{2}\phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{\mu}}_{0}\times 10\times 10}{2\mathrm{\pi}\times 5\times {10}^{-2}}+\frac{{\mathrm{\mu}}_{0}\times 10\times 10}{2\mathrm{\pi}\times 10\times {10}^{-2}}$

= 6 × 10^{−4} N

#### Page No 251:

#### Answer:

Let the third wire W_{3} having current* i* in upward direction be placed *x* cm from the 10 A current wire.

The magnetic force per unit length on a wire due to a parallel current-carrying wire is given by

$\frac{F}{l}=\frac{{\mathrm{\mu}}_{0}{i}_{1}{i}_{2}}{2\mathrm{\pi}d}$

According to the question, wire W_{3}_{ }experiences no magnetic force.

∴ $\frac{F}{l}\mathrm{due}\mathrm{to}\mathrm{wire}{\mathrm{W}}_{1}=\frac{F}{l}\mathrm{due}\mathrm{to}\mathrm{wire}{\mathrm{W}}_{2}$

$\Rightarrow \frac{{\mathrm{\mu}}_{0}10i}{2\mathrm{\pi}x}=\frac{{\mathrm{\mu}}_{0}40i}{2\mathrm{\pi}(10-x)}\phantom{\rule{0ex}{0ex}}\Rightarrow 10-x=4x\phantom{\rule{0ex}{0ex}}\Rightarrow x=2\mathrm{cm}$

Thus, wire W_{3} is placed 2 cm from the 10 A current wire.

#### Page No 251:

#### Answer:

Since wires AB, CD and EF have identical resistance, the current (30 A) gets equally distributed in them, that is, 10 A in each wire.

The magnetic force per unit length on a wire due to a parallel current-carrying wire is given by

$\frac{F}{l}=\frac{{\mathrm{\mu}}_{0}{i}_{1}{i}_{2}}{2\mathrm{\pi}d}$

∴ Magnetic force per unit length of AB = Force due to current in CD + Force due to current in EF

$\frac{F}{l}=\frac{{\mathrm{\mu}}_{0}\times 10\times 10}{2\mathrm{\pi}\times 1\times {10}^{-2}}+\frac{{\mathrm{\mu}}_{0}\times 10\times 10}{2\mathrm{\pi}\times 2\times {10}^{-2}}\phantom{\rule{0ex}{0ex}}=\frac{2\times {10}^{-7}\times {10}^{2}}{{10}^{-2}}+\frac{2\times {10}^{-7}\times {10}^{2}}{2\times {10}^{-2}}\phantom{\rule{0ex}{0ex}}=2\times {10}^{-3}+{10}^{-3}\phantom{\rule{0ex}{0ex}}=3\times {10}^{-3}\mathrm{N}/\mathrm{m}$

Similarly,

Magnetic force per unit length of CD = Force due to current in AB − Force due to current in EF

∵ Force on CD due to current in AB = Force due to current in EF

∴ Magnetic force per unit length of CD = 0

#### Page No 251:

#### Answer:

Given:

Magnitude of current, *i*_{1} = 10 A

Separation between two wires, *d* = 5 mm

Linear mass density of the second wire,* λ* = 1.0 × 10^{−4} kgm^{−1}

Now,

Let *i*_{2}_{ }be the current in the second wire in opposite direction.

Thus, the magnetic force per unit length on the wire due to a parallel current-carrying wire is given by

$\frac{{F}_{\mathrm{m}}}{l}=\frac{{\mathrm{\mu}}_{0}{i}_{1}{i}_{2}}{2\mathrm{\pi}d}$ (upwards)

Also,

Weight of the second wire, *W* = *mg*

Weight per unit length of the second wire, *$\frac{W}{l}=\lambda g$* (downwards)

Now, according to the question,

$\frac{{F}_{\mathrm{m}}}{l}=\frac{W}{l}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{\mathrm{\mu}}_{0}{i}_{1}{i}_{2}}{2\mathrm{\pi}d}=\lambda \mathrm{g}$

$\Rightarrow \frac{2\times {10}^{-7}\times 50\times {i}_{2}}{5\times {10}^{-3}}=1\times {10}^{-4}\times 9.8\phantom{\rule{0ex}{0ex}}\Rightarrow {i}_{2}=\frac{9.8\times {10}^{-7}}{20\times {10}^{-7}}=0.49\mathrm{A}$

#### Page No 251:

#### Answer:

Given:

Current in the loop, *i*_{1} = 6 A

Current in the wire, *i*_{2} = 10 A

Now, consider an element on PQ of width *dx* at a distance *x* from the wire.

Force on the element is given by

$dF=\frac{{\mu}_{0}{i}_{1}{i}_{2}}{2\pi x}dx\phantom{\rule{0ex}{0ex}}$

Force acting on part PQ is given by

${F}_{\mathrm{PQ}}=\frac{{\mathrm{\mu}}_{0}{i}_{1}{i}_{2}}{2\mathrm{\pi}}\underset{1}{\overset{3}{\int}}\frac{dx}{x}\phantom{\rule{0ex}{0ex}}=2\times {10}^{-7}\times 6\times 10[\mathrm{ln}x{{]}_{1}}^{3}\phantom{\rule{0ex}{0ex}}=120\times {10}^{-7}\mathrm{ln}\left(3\right)\mathrm{N}\phantom{\rule{0ex}{0ex}}\mathrm{Similarly},\phantom{\rule{0ex}{0ex}}{F}_{\mathrm{RS}}=\frac{{\mu}_{0}{i}_{1}{i}_{2}}{2\mathrm{\pi}}\underset{3}{\overset{1}{\int}}\frac{dx}{x}\phantom{\rule{0ex}{0ex}}=120\times {10}^{-7}\mathrm{ln}\left(\frac{1}{3}\right)\phantom{\rule{0ex}{0ex}}=-120\times {10}^{-7}\mathrm{ln}\left(3\right)\mathrm{N}$

Both forces are equal in magnitude, but they are opposite in direction.

(b) The magnetic field intensity due to wire on SP is given by

$B=\frac{{\mu}_{0}{i}_{2}}{2\mathrm{\pi}r}$

Force on part SP is given by

${F}_{\mathrm{SP}}={i}_{1}Bl\phantom{\rule{0ex}{0ex}}=\frac{{\mu}_{0}{i}_{1}{i}_{2}}{2\mathrm{\pi}}\left(\frac{l}{r}\right)\phantom{\rule{0ex}{0ex}}=\frac{{\mu}_{0}{i}_{1}{i}_{2}}{2\mathrm{\pi}}\left(\frac{2}{1}\right)$

(Towards right)

Force on part RQ is given by

${F}_{\mathrm{RQ}}={i}_{1}Bl\phantom{\rule{0ex}{0ex}}=\frac{{\mu}_{0}{i}_{1}{i}_{2}}{2\mathrm{\pi}}\left(\frac{l}{r}\right)\phantom{\rule{0ex}{0ex}}=\frac{{\mu}_{0}{i}_{1}{i}_{2}}{2\mathrm{\pi}}\left(\frac{2}{3}\right)$

(Towards left)

Thus, the net force on the loop is given by

${F}_{\mathrm{net}}={F}_{\mathrm{SP}}-{F}_{\mathrm{RQ}}\phantom{\rule{0ex}{0ex}}=\frac{{\mu}_{0}{i}_{1}{i}_{2}}{2\mathrm{\pi}}\left(\frac{2}{1}-\frac{2}{3}\right)\phantom{\rule{0ex}{0ex}}=2\times {10}^{-7}\times 6\times 10\times \frac{4}{3}\phantom{\rule{0ex}{0ex}}=16\times {10}^{-6}\mathrm{N}$

(Towards right)

#### Page No 251:

#### Answer:

Given:

No. of turns, *n* = 1

Magnitude of current, *i* = 5.00 A

Now, let the radius of the loop be *r*.

Thus, the magnetic field at the centre due to the current in the loop is given by

$B=\frac{{\mathrm{\mu}}_{0}i}{2r}$

$\Rightarrow 0.2\times {10}^{-3}=\frac{4\mathrm{\pi}\times {10}^{-7}\times 5}{2r}\phantom{\rule{0ex}{0ex}}\Rightarrow r=1.57\times {10}^{-2}\mathrm{m}=1.57\mathrm{cm}$

#### Page No 251:

#### Answer:

Given:

No. of turns, *n* = 100

Radius of the loop, *r* = 5 cm = 0.05 m

Magnetic field intensity, *B =* 6.0 × 10^{−5} T

Now, let the magnitude of current be *i.*

$\mathrm{Using}B=\frac{{\mathrm{\mu}}_{0}\mathrm{ni}}{2\mathrm{r}},\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}6.0\times {10}^{-7}=\frac{4\mathrm{\pi}\times {10}^{-7}\times {10}^{2}\times i}{2\times 0.05}\phantom{\rule{0ex}{0ex}}\Rightarrow i=\frac{60\times {10}^{-7}}{4\mathrm{\pi}\times {10}^{-7}\times {10}^{2}}\phantom{\rule{0ex}{0ex}}=4.777\times {10}^{-2}\approx 48\mathrm{mA}$

#### Page No 251:

#### Answer:

Given:

Frequency of the electron = 3 × 10^{5}

Time taken by the electron to complete one revolution,* T = *$\frac{1}{\mathrm{Frequency}}$

Current in the circle, *i *= $\frac{q}{t}$

Radius of the loop, *r = * 0.5 ${\mathrm{A}}^{\xb0}$ = $0.5\times {10}^{-10}\mathrm{m}$

Thus, the magnetic field at the centre due to the current in the loop is given by

$B=\frac{{\mathrm{\mu}}_{0}I}{2r}$

$=\frac{4\mathrm{\pi}\times {10}^{-7}\times {\displaystyle \frac{q}{T}}}{2\times 0.5\times {10}^{-10}}\phantom{\rule{0ex}{0ex}}=\frac{4\mathrm{\pi}\times {10}^{-7}\times {\displaystyle 1.6\times {10}^{-19}}}{2\times 0.5\times {10}^{-10}\times 3\times {10}^{5}}\phantom{\rule{0ex}{0ex}}=6\times {10}^{-10}\mathrm{T}$

#### Page No 251:

#### Answer:

As the centre of the loop, that is, point O, lies on the same line of two long, straight wires, the magnetic field at O due to each straight wire is zero.

Since wires ABC and ADC are identical, the current gets equally distributed in two parts at point A. So, the magnetic field due to ABC and ADC at O are equal in magnitude but are opposite in directions. (as shown in the figure).

∴ Net magnetic field at O = 0

#### Page No 252:

#### Answer:

No. of turns: *n*_{1} = 50 and *n*_{2} = 100

Magnitude of currents: *i*_{1} = *i*_{2} = 2 A

Radii of loops: *r*_{1} = 5 cm and *r*_{2} = 10 cm

(a) In the same sense:

The magnetic field intensity at the centre is given by

$B=\frac{{\mathrm{\mu}}_{0}{n}_{1}{i}_{1}}{2{r}_{1}}+\frac{{\mathrm{\mu}}_{0}{n}_{2}{i}_{2}}{2{r}_{2}}\phantom{\rule{0ex}{0ex}}=\frac{4\mathrm{\pi}\times {10}^{-7}\times 50\times 2}{2\times 5\times {10}^{-2}}+\frac{4\mathrm{\pi}\times {10}^{-7}\times 100\times 2}{2\times 10\times {10}^{-2}}\phantom{\rule{0ex}{0ex}}=4\mathrm{\pi}\times {10}^{-4}+4\mathrm{\pi}\times {10}^{-4}\phantom{\rule{0ex}{0ex}}=2\times 4\mathrm{\pi}\times {10}^{-6}\phantom{\rule{0ex}{0ex}}=8\mathrm{\pi}\times {10}^{-3}\mathrm{T}$

(b) In the opposite sense:

The magnetic field intensity at the centre is given by

$B=\frac{{\mathrm{\mu}}_{0}{n}_{1}{i}_{1}}{2{r}_{1}}-\frac{{\mathrm{\mu}}_{0}{n}_{2}{i}_{2}}{2{r}_{2}}\phantom{\rule{0ex}{0ex}}=\frac{4\mathrm{\pi}\times {10}^{-7}\times 50\times 2}{2\times 5\times {10}^{-2}}-\frac{4\mathrm{\pi}\times {10}^{-7}\times 100\times 2}{2\times 10\times {10}^{-2}}\phantom{\rule{0ex}{0ex}}=0$

#### Page No 252:

#### Answer:

Given:

No. of turns: *n*_{1} = 50 and *n*_{2} = 100

Magnitude of currents: *i*_{1} = *i*_{2} = 2 A

Radii of loops: *r*_{1} = 5 cm and *r*_{2} = 10 cm

(a) In the same sense:

The magnetic field intensity at the centre due to C_{1} is given by

${B}_{1}=\frac{{\mathrm{\mu}}_{0}{n}_{1}{i}_{1}}{2{r}_{1}}\phantom{\rule{0ex}{0ex}}=\frac{4\mathrm{\pi}\times {10}^{-7}\times 50\times 2}{2\times 5\times {10}^{-2}}\phantom{\rule{0ex}{0ex}}=4\mathrm{\pi}\times {10}^{-4}\mathrm{T}$

(In the plane of paper in upward direction)

The magnetic field intensity at the centre due to C_{2} is given by

${B}_{2}=\frac{{\mathrm{\mu}}_{0}{n}_{2}{i}_{2}}{2{r}_{2}}\phantom{\rule{0ex}{0ex}}=\frac{4\mathrm{\pi}\times {10}^{-7}\times 100\times 2}{2\times 10\times {10}^{-2}}\phantom{\rule{0ex}{0ex}}=4\mathrm{\pi}\times {10}^{-4}\mathrm{T}$

(In the plane of paper in upward direction)

In this case, magnetic fields due to C_{1} and C_{2} at the centre are along the same direction.

Thus, the net magnetic field is given by

${B}_{\mathrm{net}}={B}_{1}+{B}_{2}\phantom{\rule{0ex}{0ex}}=(4\mathrm{\pi}\times {10}^{-4})+(4\mathrm{\pi}\times {10}^{-4})\phantom{\rule{0ex}{0ex}}=8\mathrm{\pi}\times {10}^{-4}\mathrm{T}\phantom{\rule{0ex}{0ex}}=25.12\mathrm{mT}$

(b) When the direction of current in the two coils is opposite to each other then the magnetic fields will also point in opposite directions as shown in the figure. Hence, the net magnetic field will be obtained by the subtraction of the two magnetic fields.

${B}_{\mathrm{net}}={B}_{1}-{B}_{2}\phantom{\rule{0ex}{0ex}}=(4\mathrm{\pi}\times {10}^{-4})-(4\mathrm{\pi}\times {10}^{-4})\phantom{\rule{0ex}{0ex}}=0$

#### Page No 252:

#### Answer:

Given:

Magnitude of current in the loop, *I* = 10 A

Radius of the loop, *r* = 20 cm = 20 × 10^{−2} m

Thus, the magnetic field intensity at the centre is given by

$B=\frac{{\mathrm{\mu}}_{0}I}{2r}$

Now,

Velocity of the electron, *v* = 2 × 10^{6} m/s

Angle between the velocity and the magnetic field intensity, *θ *= 30°

Thus, the magnetic force on the electron is given by

$F=evB\mathrm{sin}\theta \phantom{\rule{0ex}{0ex}}=1.6\times {10}^{-19}\times 2\times {10}^{6}\times \frac{{\mathrm{\mu}}_{0}i}{2R}\mathrm{sin}30\xb0\phantom{\rule{0ex}{0ex}}=1.6\times {10}^{-19}\times 2\times {10}^{6}\times \frac{4\mathrm{\pi}\times {10}^{-7}\times 10}{2\times 20\times {10}^{-2}}\times \frac{1}{2}\phantom{\rule{0ex}{0ex}}=16\mathrm{\pi}\times {10}^{-19}\mathrm{N}$

#### Page No 252:

#### Answer:

Given:

For the outer loop,

Magnitude of current = *I*

Radius of the loop = *R*

Thus, the magnetic field at the centre due to the larger loop is given by

$B=\frac{{\mu}_{0}I}{2R}$

Let* A *be the area of the smaller loop and let current *i *pass through it.

Now,

Angle between the area vector of the smaller loop and the magnetic field due to the larger loop = 90°

Thus, the required torque is given by

$\Gamma =i(\overrightarrow{A}\times \overrightarrow{B})$

= *iAB*sin 90°

$=i\mathrm{\pi}{r}^{2}\frac{{\mathrm{\mu}}_{0}I}{2R}\phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{\mu}}_{0}\mathrm{\pi}{r}^{2}Ii}{2R}$

#### Page No 252:

#### Answer:

Given:

For the outer loop,

Magnitude of current = *I*

Radius of the loop = *R*

Thus, the magnetic field at the centre due to the larger loop is given by

$B=\frac{{\mathrm{\mu}}_{0}I}{2R}$

Let* A *be the area of the smaller loop and let current *i *pass through it.

Now,

Angle between the area vector of the smaller loop and the magnetic field due to the larger loop = 30°

Thus, the torque on the smaller loop is given by

$\Gamma =i(\overrightarrow{A}\times \overrightarrow{B})$

= *iAB*sin 30°

$=i\mathrm{\pi}{r}^{2}\frac{{\mathrm{\mu}}_{0}I}{4R}\phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{\mu}}_{0}\mathrm{\pi}{r}^{2}Ii}{4R}$

If the smaller loop is held fixed in its position, then

Torque due to the magnetic field = Torque due to the external force at its periphery

$\Rightarrow Fr=\frac{{\mathrm{\mu}}_{0}\mathrm{\pi}{r}^{2}Ii}{4R}\phantom{\rule{0ex}{0ex}}\Rightarrow F=\frac{{\mathrm{\mu}}_{0}\mathrm{\pi}Iir}{4R}$

This is the minimum magnitude of force to balance the given condition.

#### Page No 252:

#### Answer:

Given:

Magnitude of current, *I* = 5 A

Radius of the semi-circular wire, *r* = 10 cm

∴ Required magnetic field at the centre of curvature

$B=\frac{1}{2}\times \frac{{\mathrm{\mu}}_{0}i}{2r}\phantom{\rule{0ex}{0ex}}={10}^{-7}\times \frac{5}{10\times {10}^{-2}}\phantom{\rule{0ex}{0ex}}=5\mathrm{\pi}\times {10}^{-6}\phantom{\rule{0ex}{0ex}}=1.6\times {10}^{-5}\mathrm{T}$

#### Page No 252:

#### Answer:

Given:

Magnitude of current, *I* = 6 A

Radius of the semi-circular wire, *r* = 10 cm

Angle subtended at the centre, *θ* = 120° = $\frac{2\mathrm{\pi}}{3}$

∴ Required magnetic field at the centre of curvature

$B=\frac{{\mathrm{\mu}}_{0}i}{2r}\frac{\theta}{2\mathrm{\pi}}\phantom{\rule{0ex}{0ex}}=\frac{4\times {10}^{-7}\times 5}{2\times 10\times {10}^{-2}}\times \frac{2\mathrm{\pi}}{3\times 2\mathrm{\pi}}\phantom{\rule{0ex}{0ex}}=4\mathrm{\pi}\times {10}^{-6}\phantom{\rule{0ex}{0ex}}=1.26\times {10}^{-5}\mathrm{T}$

#### Page No 252:

#### Answer:

Given:

Magnitude of current = *i*

Radius of the loop = *r*

Magnetic field due to the loop at its centre, $B{}_{\mathrm{l}}=\frac{{\mathrm{\mu}}_{0}i}{2r}$

Let a straight wire carrying 4*i* current be placed at a distance *x* from the centre such that the magnetic fields of the loop and the wire are of equal magnitude but in opposite direction at O.

Magnetic field due to the wire at the centre of the loop, ${B}_{\mathrm{w}}=\frac{{\mathrm{\mu}}_{0}4i}{2\mathrm{\pi}x}$

According to the question,

${B}_{\mathrm{l}}={B}_{\mathrm{w}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{\mathrm{\mu}}_{0}i}{2r}=\frac{{\mathrm{\mu}}_{0}4i}{2\mathrm{\pi}x}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\frac{8r}{2\mathrm{\pi}}=\frac{4r}{\mathrm{\pi}}$

This means that the wire is placed $\frac{4r}{\mathrm{\pi}}$ from the centre of the loop (as shown in the figure).

#### Page No 252:

#### Answer:

Number of turns, *n *= 200

Radius of the coil, *r* = 10 cm

Current in the coil, *i* = 2A

(a) Let the magnetic field at the centre of the coil is *B*.

As the relation for magnetic field at the centre of a circular coil is given by

$\mathrm{B}=\frac{{\mathrm{n\mu}}_{0}\mathrm{i}}{2\mathrm{r}}\phantom{\rule{0ex}{0ex}}=\frac{200\times 4\mathrm{\pi}\times {10}^{-7}\times 2}{2\times 10\times {10}^{-2}}\phantom{\rule{0ex}{0ex}}=2.51\times {10}^{-3}\phantom{\rule{0ex}{0ex}}=12.56\mathrm{mT}$

(b) As magnetic field at any point *P* (say) on the axis of the circular coil is given by

${B}_{P}=n\frac{{\mu}_{0}i{r}^{2}}{2({x}^{2}+{r}^{2}{)}^{{\displaystyle \frac{3}{2}}}}\phantom{\rule{0ex}{0ex}}$

Where x is the distance of the point from the centre of the coil.

As per the question

$\frac{1}{2}{B}_{centre}={B}_{P}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{2}\frac{n{\mu}_{0}i}{2r}=\frac{n{\mu}_{0}i{r}^{2}}{2({x}^{2}+{r}^{2}{)}^{{\displaystyle \frac{3}{2}}}}\phantom{\rule{0ex}{0ex}}\Rightarrow ({x}^{2}+{r}^{2}{)}^{\frac{3}{2}}=2{r}^{3}\phantom{\rule{0ex}{0ex}}\Rightarrow ({x}^{2}+{r}^{2})={4}^{1/3}{r}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}+{r}^{2}=1.58{r}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow x=0.766\left|r\right|\phantom{\rule{0ex}{0ex}}\Rightarrow x=\pm 7.66cm$

Magnetic field will drop to half of its value at the centre if the distance of that point from the centre of the coil along the axis of coil is equal to 7.66 cm.

#### Page No 252:

#### Answer:

Given:

Magnitude of current, *I* = 5.0 A

Radius of the loop, *r* = 4.0 cm

(a) The magnetic field intensity *B* on point O at a distance* x* on the axial line is given by

$B=\frac{{\mathrm{\mu}}_{0}}{2}\frac{i{r}^{2}}{({x}^{2}+{r}^{2}{)}^{3/2}}$

$=\frac{4\mathrm{\pi}\times {10}^{-7}\times 5\times 16\times {10}^{-4}}{2\left[\right(9\times 16)\times {10}^{-4}{]}^{3/2}}\phantom{\rule{0ex}{0ex}}=\frac{4\mathrm{\pi}\times 80\times {10}^{-11}}{2\times 125\times {10}^{-6}}\phantom{\rule{0ex}{0ex}}=4.019\times {10}^{-5}\mathrm{T}(\mathrm{in}\mathrm{downward}\mathrm{direction})$

(b) The magnetic field intensity *B* on point O' at a distance* x* on the axial line is given by

$B=\frac{{\mathrm{\mu}}_{0}}{2}\frac{i{r}^{2}}{({x}^{2}+{r}^{2}{)}^{3/2}}$

$=\frac{4\mathrm{\pi}\times {10}^{-7}\times 5\times 16\times {10}^{-4}}{2\left[\right(9\times 16)\times {10}^{-4}{]}^{3/2}}\phantom{\rule{0ex}{0ex}}=\frac{4\mathrm{\pi}\times 80\times {10}^{-11}}{2\times 125\times {10}^{-6}}\phantom{\rule{0ex}{0ex}}=4.019\times {10}^{-5}\mathrm{T}(\mathrm{in}\mathrm{downward}\mathrm{direction})$

#### Page No 252:

#### Answer:

Given:

Magnitude of charges, *q* = 3.14 × 10^{−6} C

Radius of the ring, *$r=20\mathrm{cm}=20\times {10}^{-2}\mathrm{m}$*

Angular velocity of the ring, $\omega =60\mathrm{rad}/\mathrm{s}$

Time for 1 revolution = $\frac{2\mathrm{\pi}}{60}$

$\therefore \mathrm{Current},i=\frac{q}{t}=\frac{3.14\times {10}^{-6}\times 60}{2\mathrm{\pi}}\phantom{\rule{0ex}{0ex}}=30\times {10}^{-6}\mathrm{A}$

In the figure, *E*_{1} and* **E*_{2} denotes the electric field at a point on the axis at a distance of 5.00 cm from the centre due to small element 1 and 2 of the ring respectively.

*E *is the resultant electric field due to the entire ring at a point on the axis at a distance of 5.00 cm from the centre.

The electric field at a point on the axis at a distance *x* from the centre is given by

$E=\frac{xq}{4\mathrm{\pi}{\epsilon}_{0}({x}^{2}+{r}^{2}{)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}\phantom{\rule{0ex}{0ex}}$

The magnetic field at a point on the axis at a distance *x* from the centre is given by

$B=\frac{{\mathrm{\mu}}_{0}}{2}\frac{i{r}^{2}}{({x}^{2}+{r}^{2}{)}^{3/2}}\phantom{\rule{0ex}{0ex}}\frac{E}{B}=\frac{\frac{xq}{4\mathrm{\pi}{\epsilon}_{0}({x}^{2}+{r}^{2}{)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}}{\frac{{\mathrm{\mu}}_{0}}{2}\frac{i{r}^{2}}{({x}^{2}+{r}^{2}{)}^{3/2}}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}=\frac{9\times {10}^{9}\times 3.14\times {10}^{-6}\times 2\times (20.6{)}^{3}\times {10}^{-6}}{25\times {10}^{-4}\times 4\mathrm{\pi}\times {10}^{-14}\times 12}\phantom{\rule{0ex}{0ex}}=\frac{9\times 3.14\times 2\times (20.6{)}^{3}}{25\times 4\mathrm{\pi}\times 12}\phantom{\rule{0ex}{0ex}}=1.88\times {10}^{15}\mathrm{m}/\mathrm{s}$

#### Page No 252:

#### Answer:

(a) The magnetic field inside any conducting tube is always zero.

∴ Magnetic field inside the tube at a distance *r*/2 from the surface = 0

(b) Let the point outside the tube with distance $\frac{r}{2}$ be P.

∴ Net distance from centre, *r*' = $r+\frac{r}{2}=\frac{3r}{2}$

Consider an Amperian loop, as shown in the figure.

Length of the loop, *l* = $2\mathrm{\pi}\times \frac{3}{2}r=3\mathrm{\pi}r$

Current enclosed in the loop = *i*

On applying Ampere's law, we get

$\int B.dl={\mu}_{0}i\phantom{\rule{0ex}{0ex}}\Rightarrow B\times 3\mathrm{\pi}r={\mu}_{0}i\phantom{\rule{0ex}{0ex}}\Rightarrow B=\frac{{\mu}_{0}i}{3\mathrm{\pi}r}\phantom{\rule{0ex}{0ex}}$

#### Page No 252:

#### Answer:

a) The magnetic field inside any conducting tube is always zero.

∴ Magnetic field just inside the tube is zero.

(b) Let the point outside the tube with distance *b *be P.

Consider an Amperian loop, as shown in the figure.

Length of the loop, *l* = $2\mathrm{\pi}\times b=2\mathrm{\pi}b$

Current enclosed in the loop = *i*

On applying Ampere's law, we get

$\int B.dl={\mu}_{0}i\phantom{\rule{0ex}{0ex}}\Rightarrow B\times 2\mathrm{\pi}b={\mu}_{0}i\phantom{\rule{0ex}{0ex}}\Rightarrow B=\frac{{\mu}_{0}i}{2\mathrm{\pi}b}\phantom{\rule{0ex}{0ex}}$

#### Page No 252:

#### Answer:

Given:

Magnitude of current *= i *

Radius of the wire = *b*

For a point at a distance *a* from the axis,

Current enclosed, $i\text{'}=\frac{i}{\mathrm{\pi}{b}^{2}}\times \mathrm{\pi}{a}^{2}$

By Ampere's circuital law,

$\oint B.dl={\mathrm{\mu}}_{0}i\text{'}$

For the given conditions,

$B\times 2\mathrm{\pi}a={\mathrm{\mu}}_{0}\frac{i}{\mathrm{\pi}{b}^{2}}\times \mathrm{\pi}{a}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow B=\frac{{\mathrm{\mu}}_{0}ia}{2\mathrm{\pi}{b}^{2}}$

#### Page No 252:

#### Answer:

Given:

Magnitude of current,* i =* 5 A

Radius of the wire, *b*$=10\mathrm{cm}=10\times {10}^{-2}\mathrm{m}$

For a point at a distance *a* from the axis,

Current enclosed, $i\text{'}=\frac{i}{\mathrm{\pi}{b}^{2}}\times \mathrm{\pi}{a}^{2}$

By Ampere's circuital law,

$\oint B.dl={\mathrm{\mu}}_{0}i\text{'}$

For the given conditions,

$B\times 2\mathrm{\pi}a={\mathrm{\mu}}_{0}\frac{i}{\mathrm{\pi}{b}^{2}}\times \mathrm{\pi}{a}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow B=\frac{{\mathrm{\mu}}_{0}ia}{2\mathrm{\pi}{b}^{2}}\dots \left(1\right)$

$\left(\mathrm{a}\right)a=2\mathrm{cm}=2\times {10}^{-2}\mathrm{m}$

Again, using the circuital law, we get

$B=\frac{4\mathrm{\pi}\times {10}^{-7}\times 5\times 2\times {10}^{-2}}{2\mathrm{\pi}\times {10}^{-2}}\phantom{\rule{0ex}{0ex}}=2\times {10}^{-6}\mathrm{T}=2\mathrm{\mu T}$

(b) On putting $a=10\mathrm{cm}=10\times {10}^{-2}\mathrm{m}$ in (1), we get

*B* = 10 $\mathrm{\mu T}$

(c)Using the circuital law, we get

$\oint B.dl={\mathrm{\mu}}_{0}i\phantom{\rule{0ex}{0ex}}B=\frac{{\mathrm{\mu}}_{0}i}{2\mathrm{\pi}a}=\frac{2\times {10}^{-7}\times 5}{20\times {10}^{-2}}\phantom{\rule{0ex}{0ex}}=5\times {10}^{-6}\mathrm{T}=5\mathrm{\mu T}$

#### Page No 252:

#### Answer:

Half of the loop PQRS is in the region of magnetic field and half in the area of zero magnetic field.

Let us consider a current carrying circular wire, due to which there is uniform magnetic field in the region.

Take a point A inside the loop PQRS in the region where *B* = 0

According to Ampere's circuital law,

$\int \mathrm{B}.dl={\mu}_{0}i$

If there is current enclosed by the loop PQRS, then magnetic field *B* cannot be 0.

Whereas, we have taken the magnetic field at point A to be zero.

Thus, such a field is not possible.

#### Page No 252:

#### Answer:

At point P,

Current, *i* = 0

∴ Magnetic field B = 0

At point *Q*,

Applying Ampere's law, we get

$\int \mathrm{B}.dl={\mu}_{0}i\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{B}.dl={\mu}_{0}kdl\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{B}={\mu}_{0}k$

At point R,

Current, *i* = 0

∴ B = 0

#### Page No 252:

#### Answer:

Given:

Charge = *q*

Mass = *m*

Radius = *r*

We know that the radius described by a charged particle in a magnetic field is given by

$r=\frac{mv}{q\mathrm{B}}\phantom{\rule{0ex}{0ex}}$

$\mathrm{Using}\mathrm{Ampere}\mathrm{circuital}\mathrm{law}\phantom{\rule{0ex}{0ex}}\int \mathrm{B}.dl={\mu}_{0}i\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{B}.dl={\mu}_{0}kdl\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{B}={\mu}_{0}k$

$\Rightarrow v=\frac{\mathrm{B}qr}{m}=\frac{{\mu}_{0}kqr}{m}$

#### Page No 252:

#### Answer:

Given:

Magnitude of current, *i* = 5 A

Magnetic field intensity, *B =* 3.14 × 10^{−2} T

We know that the magnetic field inside a long solenoid having *n* turns per unit length is given by

$B={\mathrm{\mu}}_{0}ni\phantom{\rule{0ex}{0ex}}3.14\times {10}^{-2}=4\mathrm{\pi}\times {10}^{-7}\times \mathrm{n}\times 5\phantom{\rule{0ex}{0ex}}\Rightarrow n=\frac{{10}^{-2}}{20\times {10}^{-7}}\phantom{\rule{0ex}{0ex}}=5\times {10}^{3}=5000\mathrm{turns}/\mathrm{m}$

#### Page No 252:

#### Answer:

Given:

Radius of the wire, *r* = 0.5 mm

Width of each turn, = diameter of the wire, 2*r* = 1 mm = 1 × 10^{−3} m

∴ Total number of turns in 1 m solenoid

$n=\frac{1}{1\times {10}^{-3}}={10}^{3}\phantom{\rule{0ex}{0ex}}\mathrm{Magnitude}\mathrm{of}\mathrm{current},\mathit{}i=5\mathrm{A}\phantom{\rule{0ex}{0ex}}\mathrm{Using}B={\mathrm{\mu}}_{0}ni,\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}B=4\mathrm{\pi}\times {10}^{-7}\times {10}^{3}\times 5\phantom{\rule{0ex}{0ex}}B=2\mathrm{\pi}\times {10}^{-3}\mathrm{T}\phantom{\rule{0ex}{0ex}}$

#### Page No 252:

#### Answer:

Given:

Resistance per unit length of the wire, $\frac{R}{l}$ = 0.01 Ω/m

Radius of the wire, *r* = 1.0 cm = 0.01 m

Total no. of turns, *N* = 400

Magnetic field intensity, *B* = 1.0 × 10^{−2}^{ }T

Now,

Let *E* be the emf of the battery and *R*_{0} be the total resistance of the wire.

$\therefore i=\frac{E}{{R}_{0}}=\frac{E}{0.01\times 2\mathrm{\pi}r\times 400}\phantom{\rule{0ex}{0ex}}=\frac{E}{0.01\times 2\times \mathrm{\pi}\times 0.01\times 400}$

The magnetic field near the centre of the solenoid is given by

$B={\mathrm{\mu}}_{0}ni\phantom{\rule{0ex}{0ex}}=1\times {10}^{-2}=4\mathrm{\pi}\times {10}^{-7}\times \frac{400}{20\times {10}^{-2}}\times \frac{E}{2\mathrm{\pi}\times 4\times {10}^{-2}}\phantom{\rule{0ex}{0ex}}\Rightarrow E=\frac{{10}^{-2}\times 20\times {10}^{-2}\times 2\times {10}^{-2}}{{10}^{-7}\times 4\times {10}^{2}}\phantom{\rule{0ex}{0ex}}=1\mathrm{V}$

#### Page No 253:

#### Answer:

(a) Given:

Current in the loop or circular current = *indx*

Radius of the loop having circular current = *r*

Distance of the centre of the solenoid from the circular current = $\frac{l}{2}-x$

Magnetic field at the centre due to the circular loop,

$B=\frac{{\mu}_{0}}{2}\frac{i{r}^{2}}{({x}^{2}+{r}^{2}{)}^{3/2}}$

$\mathrm{B}=\int d\mathrm{B}\phantom{\rule{0ex}{0ex}}=\underset{0}{\overset{1}{\int}}\frac{{\mathrm{\mu}}_{0}{a}^{2}nidx}{4\pi {\left[{a}^{2}+(l-2x{)}^{2}\right]}^{3/2}}\phantom{\rule{0ex}{0ex}}=\underset{0}{\overset{1}{\int}}\frac{{\mathrm{\mu}}_{0}ni{a}^{2}dx}{4\pi {a}^{3}{\left[1+{\left({\displaystyle \frac{l-2x}{a}}\right)}^{2}\right]}^{3/2}}=\frac{{\mu}_{0}ni}{4\pi a}\underset{0}{\overset{1}{\int}}\frac{dx}{{\left[1+{\left({\displaystyle \frac{l-2x}{a}}\right)}^{2}\right]}^{3/2}}=\frac{{\mu}_{0}ni}{4\pi a}.\frac{4\pi a}{\sqrt{1+\left({\displaystyle \frac{2a}{l}}\right)}}=\frac{{\mu}_{0}ni}{\sqrt{1+{\left({\displaystyle \frac{2a}{l}}\right)}^{2}}}$

(b) When *a* > > *l*,

$\mathrm{B}=\frac{{\mu}_{0}ni}{2a}$

#### Page No 253:

#### Answer:

Given:

Magnitude of current in the solenoid,* i* = 2 A

Frequency of the electron, $f=1\times {10}^{8}\mathrm{rev}/\mathrm{s}$

Mass of the electron, $m=9.1\times {10}^{-31}\mathrm{kg}$

Charge of the electron, $q=1.6\times {10}^{-19}\mathrm{C}$

We know that the magnetic field inside a solenoid is given by

*B = **µ*_{0}*ni*

If a particle executes uniform circular motion inside a magnetic field, the frequency of the particle is given by

$f=\frac{q\mathrm{B}}{2\mathrm{\pi}m}\phantom{\rule{0ex}{0ex}}\Rightarrow B=\frac{2\mathrm{\pi}mf}{q}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{\mu}}_{0}ni=\frac{2\mathrm{\pi}mf}{q}[\mathrm{Using}(1\left)\right]\phantom{\rule{0ex}{0ex}}\Rightarrow n=\frac{2\mathrm{\pi}mf}{{\mathrm{\mu}}_{0}qi}\phantom{\rule{0ex}{0ex}}=\frac{2\mathrm{\pi}\times 9.1\times {10}^{-31}\times 1\times {10}^{8}}{4\mathrm{\pi}\times {10}^{-7}\times 1.6\times {10}^{-19}\times 2}\phantom{\rule{0ex}{0ex}}=1420\mathrm{turns}/\mathrm{m}$

#### Page No 253:

#### Answer:

Given:

Magnitude of current in the solenoid = *i*

Number of turns per unit length = *n*

When a particle is projected perpendicular to the magnetic field, it describes a circular path.

And for the particle (projected from a point on the axis in a direction perpendicular to the axis) to not strike the solenoid, the maximum radius of that circular path should be *r*/2.

∴ Radius of the circle = $\frac{r}{2}$

We know,

Centripetal force = Magnetic force

$\frac{m{V}^{2}}{r}=qVB\phantom{\rule{0ex}{0ex}}\Rightarrow V=\frac{qBr}{m}=\frac{q{\mu}_{0}nir}{2m}$

#### Page No 253:

#### Answer:

Given:

Number of turns per unit length of the solenoid = *n*

(a) Since the net magnetic field near the centre of the solenoid is 0,

$\therefore {\overrightarrow{B}}_{\mathrm{plate}}={\overrightarrow{B}}_{\mathrm{solenoid}}\phantom{\rule{0ex}{0ex}}\mathrm{Using}\mathrm{Guass}\text{'}\mathrm{s}\mathrm{law}\mathrm{for}\mathrm{the}\mathrm{plate},\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}{B}_{\mathrm{plate}}\times 2l={\mathrm{\mu}}_{0}i={\mathrm{\mu}}_{0}kl\phantom{\rule{0ex}{0ex}}{B}_{\mathrm{plate}}=\frac{{\mathrm{\mu}}_{0}k}{2}\dots \left(1\right)$

For the solenoid,

${B}_{\mathrm{solenoid}}={\mathrm{\mu}}_{0}ni$ ...(2)

From (1) and (2), we get

$i=\frac{k}{2n}$

(b) On putting the value of *n *in (2), we get

${\overrightarrow{B}}_{\mathrm{solenoid}}=\frac{{\mathrm{\mu}}_{0}k}{2}$

Now, ${\overrightarrow{B}}_{\mathrm{plate}}\mathrm{and}{\overrightarrow{B}}_{\mathrm{solenoid}}$ are perpendicular to each other.

Thus, the net magnetic field near the centre of the solenoid is given by

${B}_{\mathrm{net}}=\sqrt{{\left(\frac{{\mathrm{\mu}}_{0}k}{2}\right)}^{2}+{\left(\frac{{\mathrm{\mu}}_{0}k}{2}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{\mu}}_{0}k}{\sqrt{2}}$

#### Page No 253:

#### Answer:

Given:

Capacitance, C = 100 microfarad

Voltage, V = 20 V

Charge stored in the capacitor, Q = CV

$=100\times {10}^{-6}\times 20\phantom{\rule{0ex}{0ex}}=2\times {10}^{-3}\mathrm{C}$

It is given that the potential difference across the capacitor drops to 90% of its maximum value.

Thus,

$V\text{'}=\frac{90}{100}\times 20=18\mathrm{V}\phantom{\rule{0ex}{0ex}}\mathrm{New}\mathrm{charge},Q\text{'}=CV\text{'}=1.8\times {10}^{-3}\mathrm{C}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}\mathrm{Current},i=\frac{Q-Q\text{'}}{t}\phantom{\rule{0ex}{0ex}}\Rightarrow i=\frac{(2-1.8)\times {10}^{-3}}{2}=\frac{2\times {10}^{-4}}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow i=1\times {10}^{-4}\mathrm{A}$

No. of turns per metre, *n* = 4000

Thus, the average magnetic field at the centre of the solenoid is given by

$B={\mathrm{\mu}}_{0}ni\phantom{\rule{0ex}{0ex}}=4\mathrm{\pi}\times {10}^{-7}\times 4000\times {10}^{-4}\phantom{\rule{0ex}{0ex}}=16\mathrm{\pi}\times {10}^{-8}\mathrm{T}$

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