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#### Page No 22:

#### Question 1:

Two charged particles traverse *identical *helical paths in a completely opposite sense in a uniform magnetic field * B* =

*B*

_{0}$\hat{\mathit{k}}$.

#### Answer:

Since path of the charge particle is helix, it means it is thrown in the field $\overrightarrow{B}$ at some angle *θ *and its component of velocity along the field be *v* cos*θ *and perpendicular to the field be *v* sin*θ*.

So, the pitch of the helix will be,*p* = Time period × *v* cos*θ*

$=\frac{2\mathrm{\pi}m}{qB}\times v\mathrm{cos}\theta \left(\because T=\frac{2\mathrm{\pi}m}{qB}\right)$

$\Rightarrow \frac{q}{m}=\frac{2\pi}{pB}\times v\mathrm{cos}\theta $

For a given pitch, $\frac{q}{m}$ must be constant since, the two charged particles traverse identical helical paths (same pitch) in completely opposite sense (i.e. nature of charge must be opposite) in the uniform magnetic field *B*, so, $\frac{q}{m}$ for the two particles must be same but of opposite sign.

So, $\frac{{q}_{1}}{{m}_{1}}+\frac{{q}_{2}}{{m}_{2}}=0$

If

${q}_{1}={e}_{1}\mathrm{and}{q}_{2}={e}_{2}\phantom{\rule{0ex}{0ex}}\mathrm{Then}\mathrm{the}\mathrm{given}\mathrm{expression}\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{as}:\phantom{\rule{0ex}{0ex}}$

${\left(\frac{e}{m}\right)}_{1}+{\left(\frac{e}{m}\right)}_{2}=0$

Hence, the correct answer is option (d).

#### Page No 22:

#### Question 2:

**v**) produce a magnetic field

**B**such that

**B**⊥

**v**.

**B**||

**v**.

#### Answer:

According to Biot-Savart law:

$\overrightarrow{B}=\frac{{\mu}_{\circ}}{4\mathrm{\pi}}\times \frac{q\overrightarrow{v}\times \overrightarrow{r}}{{r}^{3}}$

Here, *q* is charge, *v* is velocity, *r* is position vector.

$\left|\overrightarrow{B}\right|=\frac{{\mu}_{\circ}}{4\mathrm{\pi}}\times \frac{qvr\mathrm{sin}\theta}{{r}^{3}}=\frac{{\mu}_{\circ}}{4\mathrm{\pi}}\times \frac{qv\mathrm{sin}\theta}{{r}^{2}}$

(Here, *θ* is angle between $\overrightarrow{v}\mathrm{and}\overrightarrow{r}$)

So, in the given options, the moving charge will produce magnetic field if:

$\theta =90\xb0,i.e.,\overrightarrow{B}\perp \overrightarrow{v}.$

But if $\overrightarrow{B}\left|\right|\overrightarrow{v},\mathrm{then}\theta \mathit{=}0\Rightarrow \overrightarrow{\mathrm{B}}=0$

Hence, the correct answer is option (a).

#### Page No 23:

#### Question 3:

*R*is placed in the

*x*-

*y*plane with centre at the origin. Half of the loop with

*x*> 0 is now bent so that it now lies in the

*y*-

*z*plane.

*at (0.0.*

**B***z*),

*z*>>

*R*increases.

*at (0.0.*

**B***z*),

*z*>>

*R*is unchanged.

#### Answer:

When the loop of radius *R* is in *x*-*y* plane then magnetic moment,

${\overrightarrow{m}}_{1}=I\times {A}_{1}=I\mathrm{\pi}{R}^{\mathit{2}}\mathit{}(\mathrm{along}z\mathit{-}\mathrm{direction})$

Now, when the loop is bent that one half in y-z plane, then magnetic moment,

${\overrightarrow{m}}_{2}=I\times \left(\frac{\mathrm{\pi}{R}^{2}}{2}\right)$ along *z*-axis + $I\times \left(\frac{\mathrm{\pi}{R}^{2}}{2}\right)$ along *x*-axis

So, the effective magnetic moment due to entire loop when bent,

$\left|{\overrightarrow{m}}_{2}\right|=\sqrt{{\left(\frac{I\mathrm{\pi}{R}^{2}}{2}\right)}^{2}+{\left(\frac{I\mathrm{\pi}{R}^{2}}{2}\right)}^{2}}=\frac{I\mathrm{\pi}{R}^{\mathit{2}}}{\sqrt{2}}$

$\Rightarrow \left|{\overrightarrow{m}}_{2}\right|<\left|{\overrightarrow{m}}_{1}\right|$

Hence, the correct answer is option (a).

#### Page No 23:

#### Question 4:

#### Answer:

If an electron (e) is projected with velocity $\overrightarrow{v}$ along the axis of a solenoid (having uniform magnetic field $\overrightarrow{B}$ along its axis), the magnetic force of the electron,

$\left|\overrightarrow{F}\right|=\left|e\left(\overrightarrow{v}\times \overrightarrow{B}\right)\right|=\left|evB\mathrm{sin}\theta \right|$

But, *θ* = 0° (as $\overrightarrow{v}$ is along $\overrightarrow{B}$)

So, $\left|\overrightarrow{F}\right|$ = *e* *v* *B* × 0 = 0

Hence, the electron will continue to move with uniform velocity along the axis of the solenoid.

Hence, the correct answer is option (d).

#### Page No 23:

#### Question 5:

#### Answer:

Cyclotron is a device, which is used to accelerate the charged particles. In cyclotron, charged particles accelerates all the time.

Hence, the correct answer is option (a).

#### Page No 23:

#### Question 6:

*M*is in an arbitrary orientation in an external magnetic field

*B*. The work done to rotate the loop by 30° about an axis perpendicular to its plane is

(b) $\sqrt{3}\frac{MB}{2}.$

(c) $\frac{MB}{2}.$

(d) zero.

#### Answer:

According to the question, the loop should be rotated about an axis perpendicular to its plane. It means, the angle between area vector and the field will always zero. So, the work done to rotate the loop is zero.

Hence, the correct answer is option (d).

#### Page No 23:

#### Question 7:

#### Answer:

For, an electron in *n*^{th} orbit of hydrogen atom,

Angular momentum, *L* = $\frac{nh}{2\mathrm{\pi}}$

Magnetic moment, *M* = $\frac{neh}{4\mathrm{\pi}m}$

Now,

Gyromagnetic ratio = $\frac{M}{L}=\frac{\left({\displaystyle \frac{neh}{4\mathrm{\pi}m}}\right)}{\left({\displaystyle \frac{nh}{2\mathrm{\pi}}}\right)}$

$=\frac{e}{2m}$

This expression is independent of *n*, i.e., gyromagnetic ratio is independent of the orbit in which the electron is revolving.

Now, charge on an electron is negative so, this ratio is negative.

Hence, the correct answers are options (a) and (b).

#### Page No 24:

#### Question 8:

*perpendicular to its length. Consider the charges inside the wire. It is known that magnetic forces do no work. This implies that,*

**B****since they do not absorb energy.**

*B***.**

*B***, no work is done by the force.**

*B**, no work is done by the magnetic force on the ions, assumed fixed within the wire.*

**B**#### Answer:

According to question, the wire is perpendicular to the field. Magnetic force on a moving charge in a uniform magnetic field:

$\overrightarrow{F}=q\overrightarrow{v}\times \overrightarrow{B}\phantom{\rule{0ex}{0ex}}\left|\overrightarrow{F}\right|=qvB\mathrm{sin}\theta $

This force may cause the charge particles to move to the surface of the wire. Now, if the wire is moving, but the charge particles inside the wire are fixed, then the magnetic force will be perpendicular to the displacement of the ion, and hence, there will be no work done.

Hence, the correct answers are options (b) and (d).

#### Page No 24:

#### Question 9:

*I*in an opposite sense.

*A*simple amperian loop passes through both of them once. Calling the loop as

*C*,

*C*.

*C*where

**B**and

**dl**are perpendicular.

*vanishes everywhere on*

**B***C*.

#### Answer:

Both the current loops having current flowing in opposite directions. Now if we draw an amperian loop (*C*) including both the loops once then, total current within the loop

$\underset{}{\sum i={i}_{1}+{i}_{2}=i-i=0}$

(âˆµ currents are in opposite direction.)

Now, according to Ampere's law

$\oint \overrightarrow{B}\xb7d\overrightarrow{l}={\mu}_{\circ}\underset{}{\sum i=0}$

Which is independent of sense of *C*.

It also indicates that, there may be a point on *C* where, $\overrightarrow{B}$ and $d\overrightarrow{l}$ are perpendicular for which angle between them be 90° and hence, $\overrightarrow{B}\xb7d\overrightarrow{l}=0$

Hence, the correct answers are options (b) and (c).

#### Page No 24:

#### Question 10:

*v*and a positron enters via opposite face with velocity -

*v*. At this instant,

#### Answer:

Here,

$\overrightarrow{E}=\mathrm{electric}\mathrm{field}\phantom{\rule{0ex}{0ex}}{\overrightarrow{F}}_{E}=\mathrm{force}\mathrm{due}\mathrm{to}\mathrm{electric}\mathrm{field}\phantom{\rule{0ex}{0ex}}{\overrightarrow{F}}_{m}=\mathrm{magnetic}\mathrm{force}\phantom{\rule{0ex}{0ex}}\overrightarrow{B}=\mathrm{magnetic}\mathrm{field}$

Since, magnitude of charge and mass on an electron (e^{–}) and positron (e^{+}) is same, then the magnitude of electrostatic force (F_{E}) and magnetic force (F_{m}) are,

$\left|{F}_{E}\right|$ = *e E* (are same for both electron and positron)

$\left|{F}_{m}\right|$ = *e v B* (are also same for both electron and same)

But the nature of charge on electron is negative (–e) and on positron is positive (+e), so the acceleration caused due to electrostatic force are opposite to each-other as shown in the diagram.

As, shown in the diagram magnetic force on both the charged particles are equal and in the same direction. So, the acceleration caused due to magnetic field on both the particles are equal, i.e., $\frac{evB}{m}$. Also, both the particles will lose energy at same rate.

Now, the net force on the positron and electron pair will be:

2 *F _{m}* in downward direction

So, the motion of centre of mass (COM) will be determined by magnetic field $\left(\overrightarrow{B}\right)$.

Hence, the correct answers are options b, c, and d.

#### Page No 24:

#### Question 11:

**E**= 0,

**B**≠ 0.

**E**≠ 0,

**B**≠ 0.

**E**≠ 0,

**B**= 0.

**E**= 0,

**B**= 0.

#### Answer:

Let the charge on the particle be *q* and it's velocity be *$\overrightarrow{v}$*, which is constant. Now, if $\overrightarrow{E}=0\mathrm{and}\overrightarrow{\mathrm{B}}\ne 0$ then, electrostatic force will not act and only magnetic force will act. Due to which the particle will move in a circular path and centripetal force will be governed by the magnetic force, F_{m} = *qvB *which acts towards the centre of the circular path. The magnitude of the velocity will remain constant.

If $\overrightarrow{E}\ne 0$ and $\overrightarrow{B}\ne 0,$ and the direction of electric and magnetic field is such that the net force on the charged particle is balanced, then also the magnitude of velocity remains constant.

If $\overrightarrow{E}\ne 0,\overrightarrow{B}=0,$then electrostatic force will cause the change in velocity of the charged particle.

If $\overrightarrow{E}=0\mathrm{and}\overrightarrow{\mathrm{B}}=0,$ then neither electrostatic nor magnetic force will act on the charged particle. So, velocity will remain constant.

Hence, the correct answers are options a, b and d.

#### Page No 25:

#### Question 12:

*eB*/

*m*has the correct dimensions of [T]

^{–1}.

#### Answer:

If a charged particle (*e*) is moving perpendicular to the magnetic field (*B*) with velocity (*v*), then centripetal force,

${F}_{c}=\frac{m{v}^{2}}{R}=qvB\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{qB}{m}=\frac{v}{R}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{qB}{m}=\omega (\because v=\omega \times R)$

So, dimensionally

$\Rightarrow \left[\omega \right]=\left[\frac{qB}{m}\right]=\left[\frac{v}{R}\right]=\frac{\left[L{T}^{-1}\right]}{\left[L\right]}\phantom{\rule{0ex}{0ex}}\Rightarrow \left[\omega \right]=\left[{T}^{-1}\right]$

#### Page No 25:

#### Question 13:

#### Answer:

Work done,

$W=\overrightarrow{F}\xb7d\overrightarrow{x}\phantom{\rule{0ex}{0ex}}\mathrm{If},W\mathit{}\mathit{=}\overrightarrow{\mathit{F}}\mathit{\xb7}d\overrightarrow{\mathit{x}}\mathit{=}\mathit{0}\phantom{\rule{0ex}{0ex}}\mathit{\Rightarrow}\overrightarrow{\mathit{F}}\mathit{\xb7}\overrightarrow{\mathit{v}}dt\mathit{=}\mathit{0}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left(\because \mathrm{velocity}v=\frac{dx}{dt}\right)\phantom{\rule{0ex}{0ex}}\mathit{\Rightarrow}\overrightarrow{\mathit{F}}\mathit{\xb7}\overrightarrow{\mathit{v}}\mathit{=}\mathit{0}$

⇒ Angle between force and velocity is 90°.

Hence, force F must be velocity dependent. If *v* changes (direction) then (directions) F should also change such that above condition is satisfied.

#### Page No 25:

#### Question 14:

#### Answer:

Magnetic force on a charged particle (*q*) moving with velocity, $\overrightarrow{v}$ in an inertial frame in magnetic field region $\left(\overrightarrow{B}\right),$

${\overrightarrow{F}}_{m}=q\overrightarrow{v}\times \overrightarrow{B}=qvB\mathrm{sin}\theta $

(*θ* = angle between the velocity and $\overrightarrow{B}$)

So, the magnetic force on a charged particle will differ from one inertial frame to another frame of reference i.e., magnetic force is frame dependent. But, net acceleration arising from this force is frame independent for inertial frames.

#### Page No 25:

#### Question 15:

*rf*) field were doubled.

#### Answer:

Time period of the charged particle (*q*)

T = $\frac{2\mathrm{\pi}m}{qB}$ (*q* = charge, *B* = magnetic field, *m* = mass of the particle)

Frequency, *f* = $\frac{1}{T}=\frac{qB}{2\mathrm{\pi m}}$

Now, if frequency is increased double,

i.e., $f\text{'}=2f$

Then, its time period, $T\text{'}=\frac{1}{f\text{'}}=\frac{1}{2f}=\frac{T}{2}=\frac{\mathrm{\pi}m}{qB}$

So, it can be said that the particle will accelerate but the radius of the path will remain unchanged.

#### Page No 25:

#### Question 16:

*I*

_{1}and

*I*

_{2}are arranged as shown in the given figure. The one carrying current

*I*

_{1}is along is the

*x*-axis. The other carrying current

*I*

_{2}is along a line parallel to the

*y*-axis given by

*x*= 0 and

*z = d*. Find the force exerted at O

_{2}because of the wire along the

*x*-axis.

#### Answer:

Magnetic field due to current carrying wire (*I*_{1}) at O_{2} be,

${\overrightarrow{B}}_{12}=\frac{{\mu}_{\circ}{I}_{1}}{2\mathrm{\pi}d}(\mathrm{along}y-\mathrm{axis})$

But the second wire carrying current (*I*_{2}) is along y-axis. So, magnetic force on the second wire carrying current (I_{2}) at O_{2 }will be,

$F={I}_{2}d\overrightarrow{l}\times {\overrightarrow{B}}_{12}\phantom{\rule{0ex}{0ex}}=0$ (âˆµ angle between $d\overrightarrow{l}$ and ${\overrightarrow{B}}_{12}$ is zero, i.e. both are in *y*-direction)

#### Page No 25:

#### Question 17:

*R*, lying in the positive quadrants of the

*x-y, y-z*and

*z-x*planes with their centres at the origin, joined together. Find the direction and magnitude of

**B**at the origin.

#### Answer:

As shown in the diagram three quarter circles in *x *– *y*, *y *– *z* and *z *– *x* planes.

The magnetic field due the quarter current carrying loop in *x *– *y *plane is:

$\overrightarrow{{B}_{1}}=\frac{{\mu}_{\circ}i}{2R}\times \left(\frac{90\xb0}{360\xb0}\right)=\frac{{\mu}_{\circ}i}{8R}\left(\hat{k}\right)$

Similarly, for the loop in *y* – *z *plane:

${\overrightarrow{B}}_{2}=\frac{{\mu}_{\circ}i}{8R}\hat{i}$

For the loop in *z*-*x* plane:

${\overrightarrow{B}}_{3}=\frac{{\mu}_{\circ}i}{8R}\hat{j}$

So, the resultant magnetic field on the completed loop will be,

$\overrightarrow{B}={\overrightarrow{B}}_{1}+{\overrightarrow{B}}_{2}+{\overrightarrow{B}}_{3}=\frac{{\mu}_{\circ}i}{8R}\hat{i}+\frac{{\mu}_{\circ}i}{8R}\hat{j}+\frac{{\mu}_{\circ}i}{8R}\hat{k}\phantom{\rule{0ex}{0ex}}=\frac{1}{4}\left(\hat{i}+\hat{j}+\hat{k}\right)\frac{{\mu}_{\circ}i}{2R}$

â€‹

#### Page No 25:

#### Question 18:

*e*and mass

*m*is moving in an electric field

**E**and magnetic field

**B**. Construct dimensionless quantities and quantities of dimension [T ]

^{}

^{–1}.

#### Answer:

When a charged particle enters in a region where electric field and magnetic field are perpendicular, then the charged particle will follow the helical path and then, its time period,

T = $\frac{2\mathrm{\pi}r}{v}=\frac{2\mathrm{\pi}}{\omega}\left(\because \omega =\frac{v}{r}\right)$

Here, *r* is the radius and ω is angular velocity.

$\Rightarrow \frac{1}{T}=\frac{\omega}{2\mathrm{\pi}}\phantom{\rule{0ex}{0ex}}\Rightarrow \omega =\frac{2\mathrm{\pi}}{T}\phantom{\rule{0ex}{0ex}}\Rightarrow \left[\omega \right]=\left[{T}^{-1}\right]$â€‹

#### Page No 25:

#### Question 19:

**v**=

*v*

_{0}

**i**into a cubical region (faces parallel to coordinate planes) in which there are uniform electric and magnetic fields. The orbit of the electron is found to spiral down inside the cube in plane parallel to the

*x-y*plane. Suggest a configuration of fields

**E**and

**B**that can lead to it.

#### Answer:

Since, the orbit of the electron spiral down inside the cube in the plane parallel to *x* – *y* plane, so

Magnetic field $\left(\overrightarrow{B}\right)$ must be along +*z*-axis and electric field $\left(\overrightarrow{E}\right)$ must be along +*x*-axis.

Because, ${\overrightarrow{F}}_{m}=q\overrightarrow{v}\times \overrightarrow{B}$ will try to move the particle in circular path but, electrostatic force,

${\overrightarrow{F}}_{e}=q\overrightarrow{E}$ will act in *x*-direction.

So,

$\overrightarrow{B}={B}_{\circ}\hat{k}\phantom{\rule{0ex}{0ex}}\overrightarrow{E}={E}_{\circ}\hat{i}\phantom{\rule{0ex}{0ex}}$

Such that, ${E}_{\circ},{B}_{\circ}0$

This condition will ensure the movement of the charged particle in spiral path parallel to the *x*-*y* plane.

#### Page No 25:

#### Question 20:

**dl**_{1}=

*dl $\hat{\mathbf{i}}$*located at the origin and

*dl*_{2}=

*dl $\hat{\mathbf{j}}$*located at (0,

*R*, 0). Both carry current

*I*.

#### Answer:

Given, $d\overrightarrow{{l}_{1}}=dl\hat{i}$

and, $d\overrightarrow{{l}_{2}}=dl\hat{j}$

Now, magnetic field at current element *dl*_{2} due to current element *dl*_{1},

${\overrightarrow{B}}_{12}=\frac{{\mu}_{\circ}Idl}{4\mathrm{\pi}{R}^{2}}\hat{k}$

So, force on current element *dl*_{2} be,

$\overrightarrow{{F}_{12}}=Idl\hat{j}\times {\overrightarrow{B}}_{12}=\left(Idl\right)\hat{j}\times \left(\frac{{\mu}_{\circ}Idl}{4\mathrm{\pi}{R}^{2}}\right)\hat{k}\phantom{\rule{0ex}{0ex}}=\left[\frac{{\mu}_{\circ}{I}^{2}{\left(dl\right)}^{2}}{4\mathrm{\pi}{R}^{2}}\right]\hat{l}\left(\because \hat{j}\times \hat{k}=\hat{i}\right)$

Which is non zero,

But, there is no effect of magnetic field of current element *dl*_{2} on current element *dl*_{1}

So, $\overrightarrow{{F}_{21}}=0$

it means, ${F}_{12}\ne {F}_{21}$

Which violates the Newton's third law.

#### Page No 26:

#### Question 21:

*R*

_{1},

*R*

_{2}and

*R*

_{3}that have to be used.

#### Answer:

Resistance of the galvanometer, *R* = 10 Ω

It will produce maximum deflection, when current,*I *= 1 mA = 1 × 10^{–3} A

for, 2 V range,

I (*R* + *R*_{1}) = 2

$\Rightarrow R+{R}_{1}=\frac{2}{I}\phantom{\rule{0ex}{0ex}}\Rightarrow {R}_{1}=\frac{2}{I}-R\phantom{\rule{0ex}{0ex}}\Rightarrow {R}_{1}=\frac{2}{1\times {10}^{-3}}-10=1990\mathrm{\Omega}$

For, 20 V range,*I* (*R* + *R*_{1} + *R*_{2}) = 20

$\Rightarrow {R}_{2}=\frac{20}{I}-R-{R}_{1}\phantom{\rule{0ex}{0ex}}=\frac{20}{1\times {10}^{-3}}-10-1990=18000\mathrm{\Omega}$

For, 200 V range,*I* (*R* + *R*_{1} + *R*_{2 }+ *R*_{3}) = 200

â€‹$\Rightarrow {R}_{3}=\frac{200}{I}-R-{R}_{1}-{R}_{2}\phantom{\rule{0ex}{0ex}}=\frac{200}{1\times {10}^{-3}}-10-1990-18000\phantom{\rule{0ex}{0ex}}=180000\mathrm{\Omega}$

#### Page No 26:

#### Question 22:

#### Answer:

Current through the straight long wire,*I*_{1} = 25 A (right to left)

Current through *PQ*,*I*_{2} = 25 A (left to right)

i.e., *I*_{1} = *I*_{2} = *I*

Length of *PQ*,*l* = 1 m

Mass of the wire*m* = 2.5 g = 2.5 × 10^{–3} kg

Let, the wire *PQ* rises to the height h.

Gravitational force on the wire be,*F _{G}* = mg

Magnetic force on the wire

*PQ*due the long straight wire,

*F*= $\frac{{\mu}_{\circ}I\times I\times l}{2\mathrm{\pi}h}$

_{m}At, height

*h*,

*F*=

_{G}*F*

_{m}$\Rightarrow mg=\frac{{\mu}_{\circ}{I}^{2}l}{2\mathrm{\pi}h}$

$\Rightarrow h=\frac{{\mu}_{\circ}{I}^{2}l}{2\mathrm{\pi}mg}=\frac{4\mathrm{\pi}\times {10}^{-7}\times {\left(25\right)}^{2}\times 1}{2\mathrm{\pi}\times 2.5\times {10}^{-3}\times 9.8}\left(\because {\mu}_{\circ}=4\mathrm{\pi}\times {10}^{-7},g=9.8\mathrm{m}/{\mathrm{s}}^{2}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow h=51.02\times {10}^{-4}\mathrm{m}\phantom{\rule{0ex}{0ex}}=0.51\mathrm{cm}$

â€‹

Hence, the required height will be 0.51 cm.

#### Page No 26:

#### Question 23:

*xz*plane) is switched on such that only arm CD of length 1 cm lies in the field. How much additional mass ‘

*m*’ must be added to regain the balance?

#### Answer:

Mass, *m*_{1 }= 500 *g* = 500 × 10^{–3 }kg,

Magnetic field,* B* = 0.2 T,

Current*, I *= 4.9 A and

CD = 1 cm

Initially, when there is no magnetic field, then net torque about the pivot must be zero, i.e.,

So, *m*_{1}*g* × *l *= *W _{coil }*×

*l*

⇒

*W*

_{coil }=

*m*

_{1}

*g*

Weight of the coil

*= m*

_{1}

*g*.....(1)

Now, when the magnetic field is switched on and additional mass

*m*is added, then magnetic force (

*IlB*) starts acting on the coil in downward direction:

So, (

*m*

_{1 }+

*m*)

*gl*=

*W*×

_{coil }*l +*(

*IlB*) ×

*l*

⇒

*mgl =*

*W*

_{coil }×

*l – m*

_{1}

*gl + (IlB*)

*l*

= 0 + (

*IlB*)

*l*(âˆµ

*W*=

_{coil }*m*

_{1}

*g*)

$\Rightarrow m=\frac{\left(IlB\right)l}{gl}\phantom{\rule{0ex}{0ex}}=\frac{IBl}{g}\phantom{\rule{0ex}{0ex}}=\frac{4.9\times 0.2\times \left(1\times {10}^{-2}\right)}{9.8}\phantom{\rule{0ex}{0ex}}\Rightarrow m=1\times {10}^{-3}\mathrm{kg}$

Hence, the required additional mass will be 10

^{–3}kg.

#### Page No 26:

#### Question 24:

*l*joined together by rods of length

*d*. The wires are each of the same material but with cross-sections differing by a factor of 2. The thicker wire has a resistance

*R*and the rods are of low resistance, which in turn are connected to a constant voltage source

*V*

_{0}. The loop is placed in uniform a magnetic field

**B**at 45° to its plane. Find

**τ**, the torque exerted by the magnetic field on the loop about an axis through the centres of rods.

#### Answer:

Given length of the parallel wires is *l* and distance between them is* d*.

Let the current through the thicker and thinner wire be *I*_{1} and *I*_{2} respectively.

Let the resistance of the thicker wire be *R*_{1} = *R*

But, ${R}_{2}=\rho \frac{l}{{A}_{2}}$

So, resistance of the thinner wire will be:

${R}_{2}=\rho \frac{l}{{A}_{2}}=\rho \frac{l}{\left({\displaystyle \frac{{A}_{1}}{2}}\right)}=2R\left(\because \mathrm{area},\frac{{A}_{1}}{{A}_{2}}=2\right)$

Now, magnetic force on the thicker wire be, *F*_{1} = *I*_{1}*lB *(âˆµ loop is connected to the voltage source of voltage ${V}_{\mathrm{o}}$)* $=\frac{{V}_{\mathrm{o}}}{{R}_{1}}lB\left({V}_{\mathrm{o}}={I}_{1}{R}_{1}\right)\phantom{\rule{0ex}{0ex}}=\frac{{V}_{\mathrm{o}}}{R}lB$*

But the magnetic field is in the direction of 45º to the plane of the loop.

So, torque about the centre of the rods

${\tau}_{1}=\frac{d}{2\sqrt{2}}{F}_{1}=\frac{d}{2\sqrt{2}}\times \frac{{V}_{\mathrm{o}}}{R}lB$

Similarly,

${\tau}_{\mathit{2}}=\frac{d}{2\sqrt{2}}\times \frac{{V}_{\mathrm{o}}}{2R}lB$

Net torque,

${\tau}_{1}-{\tau}_{2}=\frac{d}{2\sqrt{2}}\times \frac{{V}_{\mathrm{o}}}{R}lB-\frac{d}{2\sqrt{2}}\times \frac{{V}_{\mathrm{o}}}{2R}lB\phantom{\rule{0ex}{0ex}}=\frac{1}{4\sqrt{2}}\times \frac{{V}_{\mathrm{o}}lB}{R}$

#### Page No 26:

#### Question 25:

*R*) respectively, in a uniform magnetic field

**B**= B

_{0}$\hat{\mathbf{i}}$, each with an equal momentum of magnitude

*p = e*

*BR*. Under what conditions on the direction of momentum will the orbits be non-intersecting circles?

#### Answer:

Here, *p*_{1} and *p*_{2} are the momentum of electron(*e*^{–}) and positron (*e*^{+}).*p*_{1 }and *p*_{2} makes an angle *θ* with *y-*axis.

Let, both electron and positron traverse a circle of radius *R*.

Let *C*_{e }and *C*_{p }be the centre line of their respective circular path.

So, the co-ordinates of *C*_{e }and *C*_{p} can be given as,*C*_{e $\equiv $}(0, – *R* sin *θ, R *cos *θ*)*C*_{p}$\equiv $(0, + *R* sin *θ, *1.5* R – R*cos *θ*)

Now, if the distance between the centres of the two circular paths is more than 2*R* i.e., more than the diameter of either circle, then the two circles will never overlap.

Distance between the centres,

$D=\sqrt{{\left(0-0\right)}^{2}+{\left(-R\mathrm{sin}\theta -R\mathrm{sin}\theta \right)}^{2}+{\left(R\mathrm{cos}\theta -1.5R+R\mathrm{cos}\theta \right)}^{2}}$

⇒ *D*^{2} = 4*R*^{2} + 2.25 *R*^{2} – 6*R*^{2 }cos *θ *

But for not overlapping the circles *D *> 2*R*

i.e. *D*^{2 }> 4*R*^{2}

So, $4{R}^{2}+\frac{9}{4}{R}^{2}-6{R}^{2}\mathrm{cos}\theta \text{4}{R}^{2}$

$\begin{array}{l}\Rightarrow \frac{9}{4}{R}^{2}>6{R}^{2}\text{cos}\theta \\ \Rightarrow \text{cos}\theta \frac{3}{8}\end{array}$

This is the required condition.

#### Page No 26:

#### Question 26:

A uniform conducting wire of length 12a and resistance R is wound up as a current carrying coil in the shape of (i) an equilateral triangle of side a; (ii) a square of sides a and, (iii) a regular hexagon of sides *a*. The coil is connected to a voltage source V_{0}. Find the magnetic moment of the coils in each case.

#### Answer:

Total length of the conducting wire, *l* = 12*a*

Let the current flowing in each loop be *I*.

(i) So, number of loops, if the wire is wound in the form of an equilateral triangle of side *a*,

n_{1} = $\frac{12a}{a+a+a}=4$

So, area of the triangle

${A}_{1}=\frac{\sqrt{3}}{4}{a}^{2}$

Magnetic moment, *m*_{1} = *n*_{1}(*IA*_{1}) $=4\times I\times \frac{\sqrt{3}}{4}{a}^{2}=\sqrt{3}I{a}^{2}.....\left(1\right)$

(ii) Now for square of side *a*,

Number of loops,

${n}_{2}=\frac{12a}{a+a+a+a}=3$

Area of the square loops,*A*_{2} = *a*_{2}

Magnetic moments,

m_{2} = n_{2}(IA_{2})

= 3 × I × a_{2}

= 3Ia_{2}

${a}_{2}={a}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {m}_{2}=3I{a}^{2}....\left(2\right)$

(iii) Now, for regular hexagon loop:

Number of loops,

${n}_{3}=\frac{12a}{a+a+a+a+a}=2$

â€‹

Area of hexagonal loops,

${A}_{3}=\frac{3\sqrt{3}}{2}{a}^{2}$

Magnetic moments,*m*_{3} = *n*_{3}(*IA*_{3})

$=2\times I\times \frac{3\sqrt{3}}{2}\times {a}^{2}\phantom{\rule{0ex}{0ex}}=3\sqrt{3}I{a}^{2}$

#### Page No 27:

#### Question 27:

*R*in the

*x-y*plane with centre at origin. Consider the line intergral

$\mathfrak{I}\left(L\right)=\left|\underset{-L}{\overset{L}{\int}}\mathbf{B}\mathbf{.}\mathbf{dl}\right|\mathrm{taken}\mathrm{along}z\u2010\mathrm{axis}.$

*L*.

*I*, where

*I*is the current in the wire.

*R*carrying the same current

*I*. What can you say about ℑ(

*L*) andℑ(∞) ?

#### Answer:

Consider the circular loop of radius *R *in *x-y *plane, such that its axis passes through its centre 0 along *z*-axis.

Consider two points at a distance +*L* and –*L* along the *z*-axis from the centre of the loop 0.

(a) As shown in the diagram, the sense of magnetic field due to the current carrying loop is along *z*-axis.

$\mathfrak{I}\left(L\right)=\left|\underset{-L}{\overset{L}{\int}}\overrightarrow{B}.d\overrightarrow{l}\right|$

The angle between $\overrightarrow{B}$ and $d\overrightarrow{l}$ is 0º.

So, â€‹$\mathfrak{I}\left(L\right)=\underset{-L}{\overset{L}{\int}}\overrightarrow{B}.d\overrightarrow{l}=\underset{-L}{\overset{L}{\int}}Bdl\mathrm{cos}0\xba$

$=B{\left[l\right]}_{-L}^{L}=2BL$

It means, its value varies monotonically with *L*.

(b) Now assume an amperian loop MPNM* *as shown in the diagram.

$\underset{\mathrm{MPNM}}{\oint}\overrightarrow{B}.d\overrightarrow{l}={\mu}_{\mathrm{o}}I$

So, for $\mathfrak{I}\left(L\right)atL\to \infty $

$\mathfrak{I}\left(\infty \right)+0={\mu}_{o}I$

Because contribution from large distance on contour → 0 $\left(\mathrm{as},B\propto \frac{1}{{r}^{3}}\right)$

So, $\mathfrak{I}\left(\infty \right)={\mu}_{o}I$

(c) Now, magnetic field due to the circular loops along the *z*-axis, at a distance *r *from the origin

${B}_{z}=\frac{{\mu}_{o}I{R}^{2}}{2{\left({r}^{2}+{R}^{2}\right)}^{{\displaystyle \frac{3}{2}}}}$

For, large distance

$\mathfrak{I}\left(\infty \right)=\underset{-\infty}{\overset{\infty}{\int}}\frac{{\mu}_{\mathrm{o}}I{R}^{2}}{2{\left({r}^{2}+{R}^{2}\right)}^{{\displaystyle \frac{3}{2}}}}\times dr\phantom{\rule{0ex}{0ex}}\Rightarrow \mathfrak{I}\left(\infty \right)=\frac{{\mu}_{o}I}{2}\underset{-\infty}{\overset{\infty}{\int}}\frac{{R}^{2}}{{\left({r}^{2}+{R}^{2}\right)}^{{\displaystyle \frac{3}{2}}}}dr$

Let, *r = R *tan *θ*, then *dr = R*(*sec*^{2 }*θ*)*d **θ*

$\mathrm{So},\mathfrak{I}\left(\infty \right)=\frac{{\mu}_{\mathrm{o}}I}{2}\underset{-\frac{\mathrm{\pi}}{2}}{\overset{\frac{\mathrm{\pi}}{2}}{\int}}\mathrm{cos}\theta d\theta \phantom{\rule{0ex}{0ex}}=\frac{{\mu}_{\mathrm{o}}I}{2}{\left[\mathrm{sin}\theta \right]}_{-\frac{\mathrm{\pi}}{2}}^{\frac{\mathrm{\pi}}{2}}\phantom{\rule{0ex}{0ex}}=\frac{{\mu}_{\mathrm{o}}I}{2}\left[\mathrm{sin}\frac{\mathrm{\pi}}{2}-\mathrm{sin}\left(-\frac{\mathrm{\pi}}{2}\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{{\mu}_{\mathrm{o}}I}{2}\times \left(1+1\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \mathfrak{I}\left(\infty \right)={\mu}_{\mathrm{o}}I$

(d) Now, if we replace the circular coil with the square coil then magnetic field,*B _{z}â€‹*(square) <

*B*(circular coil)

_{z}$\Rightarrow \mathfrak{I}{\left(L\right)}_{square}<\mathfrak{I}{\left(L\right)}_{\left(\mathrm{circular}\mathrm{coil}\right)}$

But for large distance,

$\Rightarrow \mathfrak{I}{(\infty )}_{square}<\mathfrak{I}{(\infty )}_{\left(\mathrm{circular}\mathrm{coil}\right)}$

[The same is proved in part (b)]

#### Page No 27:

#### Question 28:

A multirange current meter can be constructed by using a galvanometer circuit as shown in the given figure. We want a current meter that can measure 10mA, 100mA and 1A using a galvanometer of resistance 10â„¦ and that prduces maximum deflection for current of 1mA. Find S_{1}, S_{2} and S_{3} that have to be used.

#### Answer:

Resistance of the galvanometer, *R *= 10 Ω

The maximum deflection for current *I *= 1 mA = 1 × 10^{–3} A

Now, for *I*_{1 }= 10 mA current,*I* × *R *= (*I*_{1} – *I*)(*S*_{1} + *S*_{2} + *S*_{3}) .....(1)

For, *I*_{2 }= 100 mA current,*I* × (*R+S*_{1}) = (*I*_{2} – *I*)(*S*_{2} + *S*_{3}) .....(2)

and for,* I*_{3} = 1 A current*I* × (*R+S*_{1}*+S*_{2}) = (*I*_{3} –* I*) (*S*_{3}) .....(3)

Now, by putting the values of *I*, *I*_{1}, *I*_{2}, *I*_{3} and *R *and solving the equations (1), (2) and (3) we get:*S*_{1} = 1 Ω*S*_{2} = 0.1 Ω*S*_{3 }= 0.01 Ωâ€‹

#### Page No 27:

#### Question 29:

*I*are arranged to form edges of a pentagonal prism as shown in the given figure. Each carries current out of the plane of paper.

*R*from each wire.

#### Answer:

(a) Since, each wire carries equal current (*I*), and point O is at equal distance *R* from each wire.

Hence, due to symmetry of the shape, the net field at point O will be zero.

$\overrightarrow{{B}_{A}}+\overrightarrow{{B}_{B}}+\overrightarrow{{B}_{c}}+\overrightarrow{{B}_{D}}+\overrightarrow{{B}_{\mathit{E}}}=0.....\left(1\right)$

i.e. $\overrightarrow{{B}_{net}}=0$

(b) Now if the current in one of the wire (let for A) is switched off,

then, resultant of remaining fields will be.

$\overrightarrow{{B}_{A}}+\overrightarrow{{B}_{B}}+\overrightarrow{{B}_{c}}+\overrightarrow{{B}_{D}}+\overrightarrow{{B}_{E}}=0\left[\mathrm{from}\left(1\right)\right]\phantom{\rule{0ex}{0ex}}\Rightarrow \left|\overrightarrow{{B}_{B}}+\overrightarrow{{B}_{c}}+\overrightarrow{{B}_{D}}+\overrightarrow{{B}_{E}}\right|=\left|-\overrightarrow{{B}_{A}}\right|\phantom{\rule{0ex}{0ex}}\Rightarrow \left|\overrightarrow{{B}_{B}}+\overrightarrow{{B}_{c}}+\overrightarrow{{B}_{D}}+\overrightarrow{{B}_{E}}\right|=\left|-\frac{{\mu}_{\mathrm{o}}I}{2\pi R}\right|=\frac{{\mu}_{\mathrm{o}}I}{2\pi R}$

But the direction must be opposite.

(c)

Now, if the current in a wire (let for A) is reversed then magnetic field due to that wire will be in opposite direction but of same magnitude as shown in figure. Now using vector sum of magnetic fields due to currents at point A, B, C, D and E we, get

â€‹$\left|{\overrightarrow{B}}_{net}\right|=\frac{{\mu}_{\mathrm{o}}I}{\pi R}\mathrm{perpendicular}\mathrm{to}AO\mathrm{towards}\mathrm{left}.$

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