Thinking Process

The uniqueness of helical path is determined by its pitch which is given by

$$ \text { Pitch }=\frac{2 \pi m v \cos \theta}{q B} $$

Answer

(d) For given pitch d correspond to charge particle, we have

$$ \frac{q}{m}=\frac{2 \pi v \cos \theta}{q B}=\text { constant } $$

Since, charged particles traverse identical helical paths in a completely opposite sense in a uniform magnetic field B, LHS for two particles should be same and of opposite sign. Therefore,

$$ \frac{e}{m_{1}}+\frac{e}{m_{2}}=0 $$

Note Consider e in place of $q$ in solution.

• (a) They have equal z-components of momenta: This is incorrect because the z-component of momentum depends on both the charge and the velocity of the particles. Since the particles traverse identical helical paths in opposite senses, their velocities in the z-direction could be different, leading to different z-components of momenta.

• (b) They must have equal charges: This is incorrect because the particles can have different charges as long as their charge-to-mass ratios are such that they produce identical helical paths in opposite senses. The condition for identical helical paths in opposite senses is that the charge-to-mass ratios are equal in magnitude but opposite in sign, not that the charges themselves are equal.

• (c) They necessarily represent a particle, anti-particle pair: This is incorrect because having opposite helical paths in a magnetic field does not necessarily imply that the particles are a particle and its anti-particle. They could simply be two different particles with charge-to-mass ratios that are equal in magnitude but opposite in sign.

Thinking Process

Here use of Biot-Savart law play vital role.

Answer

(a) In Biot-Savart’s law, magnetic field $\mathbf{B} | \mathrm{idll} \times \mathbf{r}$ and idl due to flow of electron is in opposite direction of $\mathbf{v}$ and by direction of cross product of two vectors

$$ B \perp v $$

• (b) $B$ is parallel to $v$: This is incorrect because, according to the Biot-Savart law, the magnetic field $\mathbf{B}$ is given by the cross product of the current element $\mathbf{idll}$ and the position vector $\mathbf{r}$. The cross product of two vectors is always perpendicular to the plane containing the vectors, hence $\mathbf{B}$ cannot be parallel to $\mathbf{v}$.

• (c) it obeys inverse cube law: This is incorrect because the Biot-Savart law states that the magnetic field $\mathbf{B}$ produced by a current element is inversely proportional to the square of the distance from the element, not the cube. Mathematically, $\mathbf{B} \propto \frac{1}{r^2}$, where $r$ is the distance from the current element to the point of observation.

• (d) it is along the line joining the electron and point of observation: This is incorrect because the magnetic field $\mathbf{B}$ produced by a moving charge or current element is not along the line joining the electron and the point of observation. Instead, it is perpendicular to both the direction of the current (or velocity of the electron) and the line joining the electron to the point of observation, as determined by the right-hand rule for the cross product.

Thinking Process

The magnetic moment of circular loop and the net magnitudes of magnetic moment of each semicircular loop of radius R lie in the $x-y$ plane and the $y-z$ plane are compared.

Answer

(a) The direction of magnetic moment of circular loop of radius $R$ is placed in the $x-y$ plane is along $z$-direction and given by $M=I\left(\pi^{2}\right)$, when half of the loop with $x>0$ is now bent so that it now lies in the $y-z$ plane, the magnitudes of magnetic moment of each semicircular loop of radius $R$ lie in the $x-y$ plane and the $y-z$ plane is $M^{\prime}=I\left(\pi^{2}\right) / 4$ and the direction of magnetic moments are along $z$-direction and $x$-direction respectively.

Their resultant

$$ M_{\text {net }}=\sqrt{M^{\prime 2}+M^{\prime 2}}=\sqrt{2} M^{\prime}=\sqrt{2} I\left(\pi^{2}\right) / 4 $$

So, $M_{\text {net }}<M$ or $M$ diminishes.

• (b) The magnetic moment does not change: This option is incorrect because the magnetic moment does change when the loop is bent. Initially, the magnetic moment is along the z-direction. After bending, the magnetic moments of the two semicircular loops are in different planes (x-y and y-z), resulting in a net magnetic moment that is less than the original magnetic moment.

• (c) The magnitude of $B$ at $(0,0, z), z>R$ increases: This option is incorrect because the magnetic field at a point on the z-axis (for $z > R$) is primarily determined by the net magnetic moment of the loop. Since the net magnetic moment diminishes after bending, the magnetic field at $(0,0,z)$ for $z > R$ would also decrease, not increase.

• (d) The magnitude of $B$ at $(0,0, z), z \gg R$ is unchanged: This option is incorrect because the magnetic field at a point far along the z-axis (for $z \gg R$) is proportional to the net magnetic moment of the loop. Since the net magnetic moment diminishes after bending, the magnetic field at $(0,0,z)$ for $z \gg R$ would also decrease, not remain unchanged.

Thinking Process

Here, magnetic lorentz force comes into existence when a charge moves in uniform magnetic field produces by current carrying long solenoid.

Answer

(d) Magnetic Lorentz force electron is projected with uniform velocity along the axis of a current carrying long solenoid $F=-\operatorname{ev} B \sin 180^{\circ}=0\left(\theta=0^{\circ}\right)$ as magnetic field and velocity are parallel. The electron will continue to move with uniform velocity along the axis of the solenoid.

• (a) The electron will not be accelerated along the axis because the magnetic Lorentz force acts perpendicular to both the velocity of the electron and the magnetic field. Since the electron’s velocity is parallel to the magnetic field inside the solenoid, the force is zero, and there is no acceleration along the axis.

• (b) The electron path will not be circular about the axis because a circular path requires a perpendicular component of velocity relative to the magnetic field. In this case, the electron’s velocity is parallel to the magnetic field, resulting in no perpendicular force to cause circular motion.

• (c) The electron will not experience a force at $45^{\circ}$ to the axis and hence execute a helical path because the force on a charged particle in a magnetic field is given by the Lorentz force, which depends on the sine of the angle between the velocity and the magnetic field. Here, the angle is $0^{\circ}$, making the force zero, so the electron will not follow a helical path.

Thinking Process

Here, understanding of working of cyclotron is needed.

Answer

(a) The charged particle undergoes acceleration as

(i) speeds up between the dees because of the oscillating electric field and

(ii) speed remain the same inside the dees because of the magnetic field but direction undergoes change continuously.

• (b) The charged particle does not speed up between the dees because of the magnetic field; it speeds up due to the oscillating electric field present in the gap between the dees. The magnetic field only changes the direction of the particle within the dees.

• (c) The charged particle does not speed up in a dee; it maintains a constant speed within the dees because the magnetic field only changes its direction, not its speed.

• (d) The charged particle does not slow down within a dee; it maintains a constant speed within the dees due to the magnetic field. It speeds up only in the gap between the dees due to the oscillating electric field.

Thinking Process

The rotation of the loop by $30^{\circ}$ about an axis perpendicular to its plane imply that the axis of the loop still continues to inclined with the same angle with the direction of magnetic field.

Answer

(a) The rotation of the loop by $30^{\circ}$ about an axis perpendicular to its plane make no change in the angle made by axis of the loop with the direction of magnetic field, therefore, the work done to rotate the loop is zero.

Note The work done to rotate the loop in magnetic field $W=M B\left(\cos \theta_{1}-\cos \theta_{2}\right)$. where signs are as usual.

• Option (b) $\sqrt{3} \frac{M B}{2}$: This option is incorrect because it suggests that there is a non-zero work done to rotate the loop by $30^{\circ}$. However, since the rotation is about an axis perpendicular to the plane of the loop, the angle between the magnetic moment and the magnetic field does not change, resulting in zero work done.

• Option (c) $\frac{M B}{2}$: This option is incorrect for the same reason as option (b). It implies that there is a non-zero work done, but the angle between the magnetic moment and the magnetic field remains unchanged during the rotation, leading to zero work done.

• Option (d) zero: This option is actually correct, as explained in the provided answer. The work done to rotate the loop by $30^{\circ}$ about an axis perpendicular to its plane is indeed zero because the angle between the magnetic moment and the magnetic field does not change.

Thinking Process

The gyro-magnetic ratio of an electron in an $\mathrm{H}$-atom is equal to the ratio of the magnetic moment and the angular momentum of the electron.

Answer

(a) If $I$ is the magnitude of the angular momentum of the electron about the central nucleus (orbital angular momentum). Vectorially,

$$ \mu_{l}=-\frac{e}{2 m_{e}} l $$

The negative sign indicates that the angular momentum of the electron is opposite in direction to the magnetic moment.

• (b) negative: The gyro-magnetic ratio is defined as the ratio of the magnetic moment to the angular momentum. While the magnetic moment itself is negative (due to the negative charge of the electron), the gyro-magnetic ratio is a scalar quantity and is positive because it is the magnitude of the ratio of two vectors (magnetic moment and angular momentum).

• (c) positive: This option is not incorrect. The gyro-magnetic ratio is indeed positive as it is the magnitude of the ratio of the magnetic moment to the angular momentum.

• (d) increases with the quantum number ( n ): According to the Bohr model, the gyro-magnetic ratio is given by (\frac{e}{2m_e}), which is a constant and does not depend on the quantum number ( n ). Therefore, it does not increase with ( n ).

Answer

$(b, d)$

Magnetic forces on a wire carrying a steady current, $I$ placed in a uniform magnetic field $B$, a wire carrying a steady current, $I$ placed in a uniform magnetic field $\mathbf{B}$ perpendicular to its length is given by

$$ F=I l B $$

The direction of force is given by Fleming’s left hand rule and $F$ is perpendicular to the direction of magnetic field $\mathbf{B}$. Therefore, work done by the magnetic force on the ions is zero.

• (a) Motion of charges inside the conductor is affected by $\mathbf{B}$ because the magnetic field exerts a force on the moving charges, causing them to experience a Lorentz force perpendicular to both their velocity and the magnetic field. This can lead to a redistribution of charges within the conductor, even though the magnetic force does no work.

• (c) If the wire moves under the influence of $\mathbf{B}$, work can be done by the force. The magnetic force can cause the wire to move, and if the wire moves in the direction of the force, mechanical work is done. However, this work is not done by the magnetic force directly on the charges but rather through the interaction of the magnetic field with the current in the wire.

Thinking Process

The Ampere’s circuital law is to be applied on given situation.

Answer

( $b, c$ )

Applying the Ampere’s circuital law, we have

$$ \int_{C} \mathrm{~B} \cdot \mathrm{dl}=\mu_{0}(I-I)=0 \quad \text { (because current is in opposite sense.) } $$

Also, there may be a point on $\mathbf{C}$ where $\mathbf{B}$ and $\mathbf{d l}$ are perpendicular and hence,

$$ \int_{c} \mathrm{~B} \cdot \mathrm{dl}=0 $$

• Option (a) is incorrect: The expression $\oint \mathbf{B} \cdot \mathbf{dI} = m \mu_{0} I$ is incorrect because the correct application of Ampere’s circuital law in this scenario, where the currents are in opposite directions, results in the net current being zero. Therefore, $\oint \mathbf{B} \cdot \mathbf{dI} = \mu_{0}(I - I) = 0$, not $m \mu_{0} I$.

• Option (d) is incorrect: The statement that $\mathbf{B}$ vanishes everywhere on $C$ is incorrect. While the net integral $\oint \mathbf{B} \cdot \mathbf{dI}$ is zero due to the opposite currents canceling each other out, this does not imply that the magnetic field $\mathbf{B}$ is zero at every point on $C$. The magnetic field can still exist at various points on $C$, but the contributions from the two loops will cancel out when integrated over the entire loop.

Thinking Process

The Lorentz force is experienced by the single moving charge in space is filled with some uniform electric and magnetic fields is given by $\mathbf{F}=\mathbf{q} \mathbf{E}+q(\mathbf{v} \times \mathbf{B})$.

Answer

$(b, c, d)$

The magnetic forces $\mathbf{F}=q(\mathbf{v} \times \mathbf{B})$, on charge particle is either zero or $\mathbf{F}$ is perpendicular to $\mathbf{v}$ (or component of $\mathbf{v}$ ) which in turn revolves particles on circular path with uniform speed. In both the cases particles have equal accelerations.

Both the particles gain or loss energy at the same rate as both are subjected to the same electric force $(\mathbf{F}=q \mathbf{E})$ in opposite direction.

Since, there is no change of the Centre of Mass (CM) of the particles, therefore the motion of the Centre of Mass (CM) is determined by $\mathbf{B}$ alone.

• (a) the electric forces on both the particles cause identical accelerations

  This option is incorrect because the electric force on a charged particle is given by (\mathbf{F} = q \mathbf{E}). Since the electron and positron have opposite charges ((q) and (-q)), the electric forces on them will be in opposite directions. Therefore, the accelerations will also be in opposite directions, not identical.

• (b) the magnetic forces on both the particles cause equal accelerations

  This option is correct, so no reason is needed here.

• (c) both particles gain or lose energy at the same rate

  This option is correct, so no reason is needed here.

• (d) the motion of the Centre of Mass (CM) is determined by B alone

  This option is correct, so no reason is needed here.

Thinking Process

The Lorentz force is experienced by the single moving charge in space is filled with some uniform electric and magnetic fields is given by $\mathbf{F}=q \mathbf{E}+q(\mathbf{v} \times \mathbf{B})$.

Answer

$(a, b, d)$

Here, force on charged particle due to electric field $\mathrm{F}_{\mathrm{E}}=q \mathrm{E}$.

Force on charged particle due to magnetic field, $\mathbf{F}_{m}=q(\mathbf{v} \times \mathbf{B})$

Now, $F_{E}=0$ if $E=0$ and $F_{m}=0$ if $\sin \theta=0$ or $\theta^{\circ}=0^{\circ}$ or $180^{\circ}$

Hence, $B \neq 0$.

Also, $E=0$ and $B=0$ and the resultant force $q \mathbf{E}+q(\mathbf{v} \times \mathbf{B})=0$. In this case $E \neq 0$ and $B \neq 0$

• For option (c) $\mathbf{E} \neq 0, \mathbf{B}=0$: The presence of a non-zero electric field $\mathbf{E}$ would exert a force $\mathbf{F}_{\mathbf{E}} = q\mathbf{E}$ on the charged particle, causing it to accelerate rather than move with a constant velocity.

Answer

For a charge particle moving perpendicular to the magnetic field, the magnetic Lorentz forces provides necessary centripetal force for revolution.

$$ \frac{m v^{2}}{R}=q v B $$

On simplifying the terms, we have

$$ \therefore \quad \frac{q B}{m}=\frac{v}{R}=\omega $$

Finding the dimensional formula of angular frequency

$$ \begin{array}{ll} \therefore & {[\omega]=\frac{q B}{m}=\frac{v}{R}=[T]^{-1}} \end{array} $$

Answer

Let no work is done by a force, so we have

$$ \begin{array}{rlrl} \mathrm{d} \mathbf{W}=\mathbf{F} \cdot \mathbf{d l}=0 \ \Rightarrow & \text { F.v } d t & =0 \ \Rightarrow & \text { F.v } & =0 & \end{array} $$

Thus, $F$ must be velocity dependent which implies that angle between $F$ and $v$ is $90^{\circ}$. If $v$ changes (direction), then (directions) $F$ should also change so that above condition is satisfied,

Answer

Yes, the magnetic force differ from inertial frame to frame. The magnetic force is frame dependent.

The net acceleration which comes into existing out of this is however, frame independent (non -relativistic physics) for inertial frames.

Thinking Process

The relationship of radio frequency and charge particle frequency must be equal in order to accelerate the charge particle between the dees in cyclotron.

Answer

Here, the condition of magnetic resonance is violated.

When the frequency of the radio frequency (rf) field were doubled, the time period of the radio frequency ( $\mathrm{rf}$ ) field were halved. Therefore, the duration in which particle completes half revolution inside the dees, radio frequency completes the cycle.

Hence, particle will accelerate and decelerate alternatively. So, the radius of path in the dees will remain same.

Thinking Process

Here, the understanding of application of the rule of finding directions of magnetic field and magnetic force on current carrying wire placed in magnetic field is beautifully tested.

Answer

In Biot- Savart law, magnetic field $\mathbf{B}$ is parallel to $i \mathrm{dl} \times \mathbf{r}$ and $i \mathbf{d l}$ have its direction along the direction of flow of current.

$\text {Here, for the direction of magnetic field, At due to wire carrying $I_{1}$ current is}$

$$ \text { B } | \text { parallel } i \text { dl } \times r \text { or } \hat{\mathbf{i}} \times \hat{\mathbf{k}} \text {, but } \hat{\mathbf{i}} \times \hat{\mathbf{k}}=-\hat{\mathbf{j}} $$

So, the direction at $\mathrm{O}_{2}$ is along $Y$ - direction.

The direction of magnetic force exerted at $\mathrm{O}_{2}$ because of the wire along the, $x$-axis.

$$ \mathrm{F}=I l \times \mathrm{B} \approx \hat{\mathbf{j}} \times(-\hat{\mathbf{j}})=0 $$

So, the magnetic field due to $l_{1}$ is along the $y$-axis. The second wire is along the $y$-axis and hence, the force is zero.

Thinking Process

The magnetic field due to arc of current carrying coil which subtends an angle $\theta$ at centre is given by $\mathbf{B}=\frac{\mu_{0}}{4 \pi} \frac{I \theta}{R}$.

Answer

For the current carrying loop quarter circles of radius $R$, lying in the positive quadrants of the $x$-y plane

$$ B_1=\frac{\mu_{0}}{4 \pi} \frac{I(\pi / 2)}{R} \hat{k}=\frac{\mu_{0}}{4} \frac{I}{2 R} \hat{{k}} $$

For the current carrying loop quarter circles of radius $R$, lying in the positive quadrants of the $y-z$ plane

$$ B_2=\frac{0}{4} \frac{I}{2R} \hat{i} $$

For the current carrying loop quarter circles of radius $R$, lying in the positive quadrants of the $z-x$ plane

$$ B_{3}=\frac{\mu_{0}}{4} \frac{I}{2 R} \hat{i} $$

Current carrying loop consists of 3 identical quarter circles of radius $R$, lying in the positive quadrants of the $x-y, y-y$ and $z-z$ planes with their centres at the origin, joined together is equal to the vector sum of magnetic field due to each quarter and given by

$$ \mathbf{B}=\frac{1}{4 \pi}(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}) \frac{\mu_{0} I}{2 R} . $$

Answer

No dimensionless quantity can be constructed using given quantities.

For a charge particle moving perpendicular to the magnetic field, the magnetic Lorentz forces provides necessary centripetal force for revolution.

$$ \frac{m v^{2}}{R}=q v B $$

On simplifying the terms, we have

$$ \therefore \quad \frac{q B}{m}=\frac{v}{R}=\omega $$

Finding the dimensional formula of angular frequency

$$ \begin{array}{ll} \therefore & {[\omega]=\frac{q B}{m}=\frac{v}{R}=[T]^{-1}} \end{array} $$

This is the required expression.

Thinking Process

The magnetic field revolves the charge particle in uniform circular motion in $x-y$ plane and electric field along $x$-direction increases the speed, which in turn increases the radius of circular path and hence, particle traversed on spiral path.

Answer

Considering magnetic field $\mathbf{B}=B_{0} \hat{\mathbf{k}}$, and an electron enters with a velocity $\mathbf{v}=v_{0} \hat{\mathbf{i}}$ into a cubical region (faces parallel to coordinate planes).

The force on electron, using magnetic Lorentz force, is given by

$$ F=-e\left(v_{0} \hat{i} \times B_{0} \hat{k}\right)=e v_{0} B_{0} \hat{i} $$

which revolves the electron in $x$ - $y$ plane.

The electric force $\mathbf{F}=-\mathbf{e} \mathrm{E}_{0} \hat{\mathbf{k}}$ accelerates $e$ along $z$-axis which in turn increases the radius of circular path and hence particle traversed on spiral path.

Thinking Process

Here, the understanding of application of the rules of finding directions of magnetic field and magnetic force on current carrying wire placed in magnetic field is needed.

Answer

In Biot-Savart’s law, magnetic field $\mathbf{B}$ is parallel (II) to $i \mathbf{d l} \times \mathbf{r}$ and $i d$ have its direction along the direction of flow of current.

Here, for the direction of magnetic field, At $dl_2$, located at $(0, R, 0)$ due to wire $d_1$ is given by $B | i dl \times r$ or $ \hat {i} \times \hat{j} $ (because point $(0, R, 0)$ lies on $y$-axis), but $\hat{i} \times \hat{j}=\hat{k}$

So, the direction of magnetic field at $d_{2}$ is along $z$-direction.

The direction of magnetic force exerted at $d_{2}$ because of the first wire along the $x$-axis.

$F=i(I \times B)$ i.e., $F |(i \times k)$ or along $-\hat{\mathbf{j}}$ direction.

Therefore, force due to $dl_1$ on $dl_2$ is non-zero.

Now, for the direction of magnetic field, At $d_{1}$, located at $(0,0,0)$ due to wire $d_{2}$ is given by $B | i \mathbf{d l} \times \mathbf{r}$ or $\hat{\mathbf{j}} \times \hat{\mathbf{j}}$ (because origin lies on $y$-direction w.r.t. point $(0, R, 0)$.), but $\mathbf{j} \times \mathbf{j}=0$.

So, the magnetic field at $d_{1}$ does not exist.

Force due to $dl_2$ on $dl_1$ is zero.

So, magnetic forces do not obey Newton’s third law.

Thinking Process

A galvanometer can be converted into voltmeter by connecting a very high resistance wire connected in series with galvanometer. The relationship is given by $I_{g}(G+R)=V$ where $I_{g}$ is range of galvanometer, $G$ is resistance of galvanometer and $R$ is resistance of wire connected in series with galvanometer.

Answer

Applying expression in different situations

$$ \begin{array}{lll} \text { For } & i_{G}\left(G+R_{1}\right)=2 & \text { for 2V range } \ \text { For } & i_{G}\left(G+R_{1}+R_{2}\right)=20 & \text { for 20V range } \ \text { and For } & i_{G}\left(G+R_{1}+R_{2}+R_{3}\right)=200 & \text { for 200V range } \end{array} $$

On solving, we get $R_{1}=1990 \Omega, R_{2}=18 \mathrm{k} \Omega$ and $R_{3}=180 \mathrm{k} \Omega$.

Thinking Process

The force applied on PQ by long straight wire carrying current of 25A rests on a table must balance the weight of small current carrying wire.

Answer

The magnetic field produced by long straight wire carrying current of $25 \mathrm{~A}$ rests on a table on small wire

$$ B=\frac{\mu_{0} I}{2 \pi h} $$

The magnetic force on small conductor is

$$ F=B I l \sin \theta=B I l $$

Force applied on $P Q$ balance the weight of small current carrying wire.

$$ \begin{aligned} F & =m g=\frac{\mu_{0} I^{2} l}{2 \pi h} \ h & =\frac{\mu_{0} I^{2} l}{2 \pi m g}=\frac{4 \pi \times 10^{-7} \times 25 \times 25 \times 1}{2 \pi \times 2.5 \times 10^{-3} \times 9.8}=51 \times 10^{-4} \ h & =0.51 \mathrm{~cm} \end{aligned} $$

Thinking Process

The magnetic force applied on CD by magnetic field must balance the weight.

Answer

For equilibrium/ balance, net torque should also be equal to zero.

When the field is off $\Sigma t=0$ considering the separation of each hung from mid-point be $I$.

$$ \begin{aligned} M g l & =W_{\text {coil }} l \newline \ 500 g l & =W_{\text {coil }} l \newline \ W_{\text {coil }} & =500 \times 9.8 \mathrm{~N} \end{aligned} $$

Taking moment of force about mid-point, we have the weight of coil When the magnetic field is switched on

$$ \begin{aligned} M g l+m g l & =W_{\text {coil }} l+I B L \sin 90^{\circ} I \newline \ m g l & =B I L l \newline \ m & =\frac{B I L}{g}=\frac{0.2 \times 4.9 \times 1 \times 10^{-2}}{9.8}=10^{-3} \mathrm{~kg}=1 \mathrm{~g} \end{aligned} $$

Thus, $1 \mathrm{~g}$ of additional mass must be added to regain the balance.

Thinking Process

After finding current in both wires, magnetic forces and torques need to be calculated for finding the net torque.

Answer

Front view

Side view

The thicker wire has a resistance $R$, then the other wire has a resistance $2 R$ as the wires are of the same material but with cross-sections differing by a factor 2 .

Now, the force and hence, torque on first wire is given by

$$ F_{1}=i_{1} l B=\frac{V_{0}}{2 R} l B, \tau_{1}=\frac{d}{2 \sqrt{2}} F_{1}=\frac{V_{0} l d B}{2 \sqrt{2} R} $$

Similarly, the force hence torque on other wire is given by

$$ F_{2}=i_{2} l B=\frac{V_{0}}{2 R} l B, \tau_{2}=\frac{d}{2 \sqrt{2}} F_{2}=\frac{V_{0} l d B}{4 \sqrt{2} R} $$

So, net torque, $\quad \tau=\tau_{1}-\tau_{2}$

$$ \tau=\frac{1}{4 \sqrt{2}} \frac{V_{0} l d B}{R} $$

Thinking Process

The circles of the electron and a positron shall not overlap if the distance between the two centers are greater than $2 R$

Answer

Since, $B$ is along the $x$-axis, for a circular orbit the momenta of the two particles are in the $y-z$ plane. Let $p_{1}$ and $p_{2}$ be the momentum of the electron and positron, respectively. Both traverse a circle of radius $R$ of opposite sense. Let $p_{1}$ make an angle $\theta$ with the $y$-axis $p_{2}$ must make the same angle.

The centres of the respective circles must be perpendicular to the momenta and at a distance $R$. Let the centre of the electron be at $C_{e}$ and of the positron at $C_{p}$. The coordinates of $C_{e}$ is

$$ C_{e} \equiv(0,-R \sin \theta, R \cos \theta) $$

The coordinates of $C_{p}$ is

$$ C_{p} \equiv\left(0,-R \sin \theta, \frac{3}{2} R-R \cos \theta\right) $$

The circles of the two shall not overlap if the distance between the two centers are greater than $2 R$.

Let $d$ be the distance between $C_{p}$ and $C_{e}$.

Let $d$ be the distance between $C_{p}$ and $C_{e}$

Then,

$$ \begin{aligned} d^{2} & =(2 R \sin \theta)^{2}+\frac{3}{2} R-2 R \cos \theta \newline \ & =4 R^{2} \sin ^{2} \theta+\frac{9^{2}}{4} R-6 R^{2} \cos \theta+4 R^{2} \cos ^{2} \theta \newline \ & =4 R^{2}+\frac{9}{4} R^{2}-6 R^{2} \cos \theta \end{aligned} $$

Since, $d$ has to be greater than $2 R$

$$ \begin{array}{rlrl} d^{2} >4 R^{2} \newline \ \Rightarrow 4 R^{2}+\frac{9}{4} R^{2}-6 R^{2} \cos \theta >4 R^{2} \newline \ \Rightarrow \frac{9}{4} >6 \cos \theta \newline \ \text { or, } \cos \theta <\frac{3}{8} \end{array} $$

Thinking Process

The different shapes forms figures of different area and hence, there magnetic moments varies.

Answer

We know that magnetic moment of the coils $m=n / A$.

Since, the same wire is used in three cases with same potentials, therefore, same current flows in three cases.

(i) for an equilateral triangle of side $a$,

$n=4$ as the total wire of length $=12 a$

Magnetic moment of the coils $m=n I A=4 I \quad \frac{\sqrt{3}}{4} a^{2}$

$\therefore \quad m=I a^{2} \sqrt{3}$

(ii) For a square of sides a,

$$ A=a^{2} $$

$n=3$ as the total wire of length $=12 a$

Magnetic moment of the coils $m=n I A=3 I\left(a^{2}\right)=3 I a^{2}$

(iii) For a regular hexagon of sides a,

$n=2$ as the total wire of length $=12 a$

Magnetic moment of the coils $m=n \left\lvert, A=21 \frac{6 \sqrt{3}}{4} a^{2}\right.$

$$ m=3 \sqrt{3} a^{2} I $$

$m$ is in a geometric series.

Thinking Process

This question revolves around the application of Ampere circuital law.

Answer

(a) B (z) points in the same direction on $z$-axis and hence, $J(L)$ is a monotonically function of $L$.

Since, $B$ and $\mathbf{d l}$ along the same direction, therefore $\mathbf{B} \cdot \mathbf{d l}=\mathbf{B} \cdot \mathbf{d l}$ as $\cos 0=1$

(b) $J(L)+$ contribution from large distance on contour $C=\mu_{0} I$

$\therefore \quad$ as $L \rightarrow \infty$

Contribution from large distance $\rightarrow 0\left(\right.$ as $\left.B \propto 1 / r^{3}\right)$

$$ J(\infty)-\mu_{0} I $$

(c) The magnetic field due to circular current-carrying loop of radius $R$ in the $x$-y plane with centre at origin at any point lying at a distance of from origin.

$$ \begin{aligned} B_{z} & =\frac{\mu_{0} I R^{2}}{2\left(z^{2}+R^{2}\right)^{3 / 2}} \ \int_{-\infty}^{\infty} B_{z} d z & =\int_{-\infty}^{\infty} \frac{\mu_{0} / R^{2}}{2\left(z^{2}+R^{2}\right)^{3 / 2}} d z \end{aligned} $$

Put $\quad z=R \tan \theta_{1}$

$\Rightarrow \quad d z=R \sec ^{2} \theta d \theta$

$\therefore \quad \int_{-\infty}^{\infty} B_{z} d z=\frac{\mu_{0} I}{2} \int_{-\pi / 2}^{\pi / 2} \cos \theta d \theta=\mu_{0} I$

(d)$Bz_{\text {square}}<Bz_{\text {circular coil }}$

$\therefore \quad IL_{\text {square }}< IL_{\text {circular coil }}$

But by using arguments as in (b)

$$ I\infty_{\text {square }}=I\infty_{\text {circular }} $$

Thinking Process

A galvanometer can be converted into ammeter by connecting a very low resistance wire (shunt S) connected in parallel with galvanometer. The relationship is given by $I_{g} G=\left(I-I_{g}\right)$ S, where $I_{g}$ is range of galvanometer, $G$ is resistance of galvanometer.

Answer

$$ \begin{aligned} I_{G} \cdot G & =\left(I_{1}-I_{G}\right)\left(S_{1}+S_{2}+S_{3}\right) \text { for } I_{1}=10 \mathrm{~mA} \newline \ I_{G}\left(G+S_{1}\right) & =\left(I_{2}-I_{G}\right)\left(S_{2}+S_{3}\right) \text { for } I_{2}=100 \mathrm{~mA} \newline \ I_{G}\left(G+S_{1}+S_{2}\right) & =\left(I_{3}-I_{G}\right)\left(S_{3}\right) \text { for } I_{3}=1 \mathrm{~A} \newline \ S_{1} & =1 \mathrm{~W}, S_{2}=0.1 \mathrm{~W} \newline \ S_{3} & =0.01 \mathrm{~W} \end{aligned} $$

Thinking Process

The vector sum of magnetic field produced by each wire at $O$ is equal to 0.

Answer

(a) Suppose the five wires $A, B, C, D$ and $E$ be perpendicular to the plane of paper at locations as shown in figure.

Thus, magnetic field induction due to five wires will be represented by various sides of a closed pentagon in one order, lying in the plane of paper. So, its value is zero.

(b) Since, the vector sum of magnetic field produced by each wire at $O$ is equal to 0 . Therefore, magnetic induction produced by one current carrying wire is equal in magnitude of resultant of four wires and opposite in direction.

Therefore, the field if current in one of the wires (say $A$ ) is switched off is $\frac{\mu_{0}}{2 \pi} \frac{i}{R}$ perpendicular to $A O$ towards left.

(c) If current in wire $A$ is reversed, then

total magnetic field induction at $O$

$=$ Magnetic field induction due to wire $A+$ magnetic field induction due to wires $B, C, D$ and $E$

$$ =\frac{\mu_{0}}{4 \pi R} \frac{2 I}{R} $$

(acting perpendicular to $A O$ towards left) $+\frac{\mu_{0}}{\pi} \frac{2 I}{R}$ (acting perpendicular $A O$ towards left) $=\frac{\mu_{0} I}{\pi R}$ acting perpendicular $A O$ towards left.