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.

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

$$ B \perp v $$

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.

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.

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.

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.

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.

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.

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,

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

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.

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 $B \neq 0$

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.