Thinking Process

The magnetic flux linked with uniform surface of area A in uniform magnetic field is given by

$$ \phi=B . A $$

Answer

(c) Here, $A=L^{2} \hat{\mathbf{k}}$ and $\mathbf{B}=B_{0}(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}) T$

$$ \phi=\mathbf{B} \cdot \mathbf{A}=B_{0}(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}) \cdot L^{2} \hat{\mathbf{k}}=4 B_{0} L^{2} \mathrm{~Wb} $$

Thinking Process

Here, loop $A B C D$ A lies in $x$-y plane whose area vector $A_{1}=L^{2} \hat{k}$ whereas loop ADEFA lies in $y-z$ plane whose area vector $A_{2}=L^{2} \hat{i}$.

Answer

(b) Also, the magnetic flux linked with uniform surface of area $A$ in uniform magnetic field is given by

and

$$ \begin{aligned} \phi & =B \cdot A \\ A & =A_{1}+A_{2}=\left(L^{2} \hat{k}+L^{2} \hat{i}\right) \\ B & =B_{0}(\hat{i}+\hat{k}) T \\ \phi & =B \cdot A=B_{0}(\hat{i}+\hat{k}) \cdot\left(L^{2} \hat{k}+L^{2} \hat{i}\right) \\ & =2 B_{0} L^{2} \mathrm{~Wb} \end{aligned} $$

Now,

Thinking Process

The problem is associated with the phenomenon of electromagnetic induction.

Answer

(b) When cylindrical bar magnet is rotated about its axis, no change in flux linked with the circuit takes place, consequently no emf induces and hence, no current flows through the ammeter $A$.

Thinking Process

The induced emf in $B$ is due to the variation of magnetic flux in it.

Answer

(d) When the $A$ stops moving the current in $B$ become zero, it possible only if the current in $A$ is constant. If the current in $A$ would be variable, there must be an induced emf (current) in $B$ even if the $A$ stops moving.

Thinking Process

Here, the application of Lenz’s law is tested through this problem.

Answer

(a) When the current in $B(a t=0)$ is counter-clockwise and the coil $A$ is considered above to it. The counterclockwise flow of the current in $B$ is equivalent to north pole of magnet and magnetic field lines are emanating upward to coil $A$. When coil $A$ start rotating at $t=0$, the current in $A$ is constant along clockwise direction by Lenz’s rule.

Thinking Process

The self inductance Lof a solenoid depends on its geometry (i.e., length, cross-sectional area, number of turns etc.) and on the permeability of the medium.

Answer

(b) The self-inductance of a long solenoid of cross-sectional area $A$ and length $l$, having $n$ turns per unit length, filled the inside of the solenoid with a material of relative permeability (e.g., soft iron, which has a high value of relative permeability) is given by

$$ L=\mu_{r} \mu_{0} n^{2} A l $$

$$ \text { where, } \quad n=N / l $$

Note The capacitance, resistance, self and mutual inductance depends on the geometry of the devices as well as permittivity/permeability of the medium.

Thinking Process

This problem is associated with the heating effect of current as well as the phenomenon of electromagnetic induction and eddy currents.

Answer

$(a, b, d)$

A metal plate is getting heated when a DC or AC current is passed through the plate, known as heating effect of current. Also, when metal plate is subjected to time varying magnetic field, the magnetic flux linked with the plate changes and eddy currents comes into existence which make the plate hot.

Thinking Process

This problem is associated with the phenomenon of electromagnetic induction.

Answer

$(a, b, c)$

Here, magnetic flux linked with the isolated coil change when the coil being in a time varying magnetic field, the coil moving in a constant magnetic field or in time varying magnetic field.

Note When magnetic flux linked with the coil change, an emf is used in the coil. This is known as electromagnetic induction.

Answer

$(a, d)$

The mutual inductance $M_{12}$ of coil increases when they are brought nearer and is the same as $M_{21}$ of coil 2 with respect to coil 1 . $M_{12}$ i.e., mutual inductance of solenoid $S_{1}$ with respect to solenoid $S_{2}$ is given by

$$ M_{21}=\mu_{0} n_{1} n_{2} \pi r_{1}^{2} l $$

where signs are as usual.

Also, $M_{12}$ i.e., mutual inductance of solenoid $S_{2}$ with respect to solenoid $S_{1}$ is given by

So, we have

$$ \begin{aligned} & M_{21}=\mu_{0} n_{1} n_{2} \pi r_{1}^{2} l \ & M_{12}=M_{21}=M \end{aligned} $$

Thinking Process

The various arrangement are to be thought of in such a way that the magnetic flux linked with the coil do not change even if coil is placed and expanding in magnetic field.

Answer

$(b, c)$

When circular coil expands radially in a region of magnetic field such that the magnetic field is in the same plane as the circular coil or the magnetic field has a perpendicular (to the plane of the coil) component whose magnitude is decreasing suitably in such a way that the cross product of magnetic field and surface area of plane of coil remain constant at every instant.

Thinking Process

The magnetic flux linked with uniform surface of area $A$ in uniform magnetic field is given by

$$ \phi=\mathbf{B} \cdot \mathbf{A}=B A \cos \theta $$

So, flux linked will change only when either B, or A or the angle between B and A change.

Answer

When the switch is thrown from the off position (open circuit) to the on position (closed circuit), then neither $B$, nor $A$ nor the angle between $B$ and $A$ change. Thus, no change in magnetic flux linked with coil occur, hence no electromotive force is produced and consequently no current will flow in the circuit.

Thinking Process

Here, the application of Lenz’s law is tested through this problem.

Answer

When the coil is stretched so that there are gaps between successive elements of the spiral coil i.e., the wires are pulled apart which lead to the flux leak through the gaps. According to Lenz’s law, the emf produced must oppose this decrease, which can be done by an increase in current. So, the current will increase.

Thinking Process

Here, the application of Lenz’s law is tested through this problem.

Answer

When the iron core is inserted in the current carrying solenoid, the magnetic field increase due to the magnetisation of iron core and consequently the flux increases.

According to Lenz’s law, the emf produced must oppose this increase in flux, which can be done by making decrease in current. So, the current will decrease.

Thinking Process

Here, the application of Lenz’s law is tested through this problem.

Answer

When the current is switched on, magnetic flux is linked through the ring. Thus, increase in flux takes place. According to Lenz’s law, this increase in flux will be opposed and it can happen if the ring moves away from the solenoid.

This happen because the flux increases will cause a counter clockwise current (as seen from the top in the ring in figure.) i.e., opposite direction to that in the solenoid.

This makes the same sense of flow of current in the ring (when viewed from the bottom of the ring) and solenoid forming same magnetic pole infront of each other. Hence, they will repel each other and the ring will move upward.

Thinking Process

This problem is based on the application of Lenz’s law.

Answer

(b) When the current is switched off, magnetic flux linked through the ring decreases. According to Lenz’s law, this decrease in flux will be opposed and the ring experience downward force towards the solenoid.

This happen because the flux $i$ decrease will cause a clockwise current (as seen from the top in the ring in figure) i.e., the same direction to that in the solenoid. This makes the opposite sense of flow of current in the ring (when viewed from the bottom of the ring) and solenoid forming opposite magnetic pole infront of each other.

Hence, they will attract each other but as ring is placed at the cardboard it could not be able to move downward.

Thinking Process

This problem is based on the concept of eddy current and application of Lenz’s law.

Answer

When cylindrical bar magnet of radius $0.8 \mathrm{~cm}$ is dropped through the metallic pipe with an inner radius of $1 \mathrm{~cm}$, flux linked with the cylinder changes and consequently eddy currents are produced in the metallic pipe. According to Lenz’s law, these currents will oppose the (cause) motion of the magnet.

Therefore, magnet’s downward acceleration will be less than the acceleration due to gravity $g$. On the other hand, an unmagnetised iron bar will not produce eddy currents and will fall with an acceleration due to gravity $g$.

Thus, the magnet will take more time to come down than it takes for a similar unmagnetised cylindrical iron bar dropped through the metallic pipe.

Thinking Process

This problem requires application of Faraday’s law of EMI and finding mathematical values of emf at different instants.

Answer

At any instant, flux passes through the ring is given by

$$ \begin{aligned} \phi & =\text { B.A }=B A \cos \theta=B A \\ \text { or } \quad & =B_{0}\left(\pi a^{2}\right) \cos \omega t \end{aligned} \quad(\because \theta=0) $$

By Faraday’s law of electromagnetic induction.

Magnitude of induced emf is given by

$$ \varepsilon=B_{0}\left(\pi a^{2}\right) \omega \sin \omega t $$

This causes flow of induced current, which is given by

$$ I=B_{0}\left(\pi a^{2}\right) \omega \sin \omega t / R $$

Now, finding the value of current at different instants, so we have current at

Because

$$ \begin{aligned} t & =\frac{\pi}{2 \omega} \\ I & =\frac{B_{0}\left(\pi a^{2}\right) \omega}{R} \text { along } \hat{\mathbf{j}} \\ \sin \omega t & =\sin \omega \frac{\pi}{2 \omega}=\sin \frac{\pi}{2}=1 \\ t & =\frac{\pi}{\omega}, I=\frac{B\left(\pi a^{2}\right) \omega}{R} \\ \sin \omega t & =\sin \omega \frac{\pi}{\omega}=\sin \pi=0 \\ t & =\frac{3}{2} \frac{\pi}{\omega} \\ I & =\frac{B\left(\pi a^{2}\right) \omega}{R} \text { along }-\hat{\mathbf{j}} \\ \sin \omega t & =\sin \omega \frac{3 \pi}{2 \omega}=\sin \frac{3 \pi}{2}=-1 \end{aligned} $$

Here,

Thinking Process

This problem underline the concept of continuity of magnetic field lines. They can neither be originated nor be destroyed in space.

Answer

The magnetic flux linked with the surface can considered as the number of magnetic field lines passing through the surface. So, let $d \phi=\mathrm{BA}$ represents magnetic lines in an area $A$ to $B$.

By the concept of continuity of lines $B$ cannot end or start in space, therefore the number of lines passing through surface $S_{1}$ must be the same as the number of lines passing through the surface $S_{2}$. Therefore, in both the cases we gets the same answer for flux.

Answer

The motional electric field $E$ along the dotted line $C D$ ( $\Lambda$ to both $\mathbf{v}$ and $\mathbf{B}$ and along $\mathbf{V} \times \mathbf{B})$ $=v B$

Therefore, the motional emf along $P Q=($ length $P Q) \times($ field along $P Q)$

$$ \begin{aligned} & =(\text { length } P Q) \times(v B \sin \theta) \ & =\frac{d}{\sin \theta} \times(v B \sin \theta)=v B d \end{aligned} $$

This induced emf make flow of current in closed circuit of resistance $R$.

$$ I=\frac{d v B}{R} \text { and is independent of } q $$

Thinking Process

When the current is varied, the flux linked with the coil also changes and an emf is induced in the coil. The induced emf is given by

$$ \begin{aligned} & \varepsilon=-\frac{d\left(N \phi_{B}\right)}{d t} \ & \varepsilon=-L \frac{d l}{d t} \end{aligned} $$

Thus, the self-induced emf always opposes any change (increase or decrease) of current in the coil.

Answer

The back electromotive force in solenoid is $(u)$ a maximum when there is maximum rate of change of current. This occurs is in $A B$ part of the graph. So maximum back emf will be obtained between $5 \mathrm{~s}<t<10 \mathrm{~s}$.

Since, the back emf at $t=3$ s is e,

Also,

the rate of change of current at $t=3, \mathrm{~s}=$ slope of $O A$ from $t=0 \mathrm{~s}$ to $t=5 \mathrm{~s}=1 / 5 \mathrm{~A} / \mathrm{s}$.

So, we have

If $u=L 1 / 5$ for $t=3 \mathrm{~s}, \frac{d I}{d t}=1 / 5$ ( $L$ is a constant). Applying $\varepsilon=-L \frac{d I}{d t}$

Similarly, we have for other values

For 5 s $<t<10$ s

Thus,

For 10 s $<t<30$ s

$$ u_{1}=-L \frac{3}{5}=-\frac{3}{5} L=-3 e $$

$$ \text { at } t=7 \mathrm{~s}, u_{1}=-3 \mathrm{e} $$

$$ u_{2}=L \frac{2}{20}=\frac{L}{10}=\frac{1}{2} e $$

For $t>30 \mathrm{~s}, u_{2}=0$

Thus, the back emf at $t=7 \mathrm{~s}, 15 \mathrm{~s}$ and $40 \mathrm{~s}$ are $-3 e, e / 2$ and 0 respectively.

Thinking Process

A current $I_{1}$ is passed through the coil A and the flux linkage with coil B is,

$$ N_{2} \phi_{2}=M_{21} I_{1} $$

where, $M_{21}$ is called the mutual inductance of coil $A$ with respect to coil $B$ and $M_{21}=M_{12}$

And $M_{12}$ is called the mutual inductance of coil $B$ with respect to coil $A$.

Answer

Applying the mutual inductance of coil $A$ with respect to coil $B$

$$ M_{21}=\frac{N_{2} \phi_{2}}{I_{1}} $$

Therefore, we have

$$ \text { Mutual inductance }=\frac{10^{-2}}{2}=5 \mathrm{mH} $$

Again applying this formula for other case

$$ N_{1} \phi_{1}=M_{12} I_{2}=5 \mathrm{mH} \times 1 \mathrm{~A}=5 \mathrm{mWb} . $$

Thinking Process

The emf induced across $A B$ due to its motion and change in magnetic flux linked with the loop change due to change of magnetic field.

Answer

Let us assume that the parallel wires at are $y=0$ i.e., along $x$-axis and $y=d$. At $t=0, A B$ has $x=0$, i.e., along $y$-axis and moves with a velocity $v$. Let at time $t$, wire is at $x(t)=v t$. Now, the motional emf across $A B$ is

$$ =\left(B_{0} \sin \omega t\right) v d(-\hat{\mathbf{j}}) $$

emf due to change in field (along $O B A C$ )

$$ =-B_{0} \omega \cos \omega t x(t) d $$

Total emf in the circuit $=$ emf due to change in field $($ along $O B A C)+$ the motional emf across $A B$

Electric current in clockwise direction is given by

$$ =-B_{0} d[\omega x \cos (\omega t)+v \sin (\omega t)] $$

$$ =\frac{B_{0} d}{R}(\omega x \cos \omega t+v \sin \omega t) $$

The force acting on the conductor is given by $F=i l B \sin 90^{\circ}=i l B$

Substituting the values, we have

$$ \text { Force needed along } \begin{aligned} \mathbf{i} & =\frac{B_{0} d}{R}(\omega x \cos \omega t+v \sin \omega t) \times d \times B_{0} \sin \omega t \\ & =\frac{B_{0}^{2} d^{2}}{R}(\omega x \cos \omega t+v \sin \omega t) \sin \omega t \end{aligned} $$

This is the required expression for force.

Thinking Process

This problem relates EMI, magnetic force, power consumption and mechanics.

Answer

Let us assume that the parallel wires at are $y=0$, i.e., along $x$-axis and $y=l$. At $t=0, X Y$ has $x=0$ i.e., along $y$-axis.

(i) Let the wire be at $x=x(t)$ at time $t$.

The magnetic flux linked with the loop is given by

$$ \phi=\mathrm{B} \cdot \mathrm{A}=\mathrm{BA} \cos 0=\mathrm{BA} $$

at any instant $t$

$$ \text { Magnetic flux }=B(t)(l \times x(t)) $$

Total emf in the circuit $=$ emf due to change in field $($ along $X Y A C)+$ the motional emf across $X Y$

$$ \left.E=-\frac{d \phi}{d t}=-\frac{d B(t)}{d t} l x(t)-B(t) l v(t) \quad \text { [second term due to motional emf }\right] $$

Electric current in clockwise direction is given by

$$ I=\frac{1}{R} E $$

The force acting on the conductor is given by $F=i l B \sin 90^{\circ}=i l B$

Substituting the values, we have

$$ \text { Force }=\frac{I B(t)}{R}-\frac{d B(t)}{d t} I x(t)-B(t) I v(t) \hat{\mathbf{i}} $$

Applying Newton’s second law of motion,

$$ m \frac{d^{2} x}{d t^{2}}=-\frac{I^{2} B(t)}{R} \frac{d B}{d t} x(t)-\frac{I^{2} B^{2}(t)}{R} \frac{d x}{d t} $$

which is the required equation.

(ii) If $\mathbf{B}$ is independent of time i.e., $B=$ Constant

Or

$$ \frac{d B}{d t}=0 $$

Substituting the above value in $\mathrm{Eq}$ (i), we have

res

$$ \begin{aligned} \frac{d^{2} x}{d t^{2}}+\frac{I^{2} B^{2}}{m R} \frac{d x}{d t} & =0 \ \frac{d v}{d t}+\frac{I^{2} B^{2}}{m R} v & =0 \end{aligned} $$

Integrating using variable separable form of differential equation, we have

$$ v=A \exp \frac{-I^{2} B^{2} t}{m R} $$

Applying given conditions,

$$ \begin{aligned} & \text { at } t=0, v=u_{0} \ & v(t)=u_{0} \exp \left(-I^{2} B^{2} t / m R\right) \end{aligned} $$

This is the required equation.

(iii) Since the power consumption is given by $P=I^{2} R$

Here,

$$ \begin{aligned} I^{2} R & =\frac{B^{2} I^{2} v^{2}(t)}{R^{2}} \times R \ & =\frac{B^{2} I^{2}}{R} u_{0}^{2} \exp \left(-2 I^{2} B^{2} t I m R\right) \end{aligned} $$

Now, energy consumed in time interval $d t$ is given by energy consumed $=P d t=I^{2} R d t$ Therefore, total energy consumed in time $t$

$$ \begin{aligned} & \left.=\int_{0}^{t} I^{2} R d t=\frac{B^{2} I^{2}}{R} u_{0}^{2} \frac{m R}{2 I^{2} B^{2}} 1-e^{-\left(l^{2} B^{2} t / m r\right.}\right) \\ & =\frac{m}{2} u_{0}^{2}-\frac{m}{2} v^{2}(t) \\ & =\text { decrease in kinetic energy. } \end{aligned} $$

This proves that the decrease in kinetic energy of $X Y$ equals the heat lost in $R$.

Thinking Process

The pattern of rate of change of area (hence flux) can be considered uniform from $0<\theta<\frac{\pi}{4} ; \frac{\pi}{4}<\theta<\frac{3 \pi}{4}$ and $\frac{3 \pi}{4}<\theta<\frac{\pi}{2}$. Hence, forth finding emf and current.

Answer

Let us consider the position of rotating conductor at time interval

$$ t=0 \text { to } t=\frac{\pi}{4 \omega}(\text { or } T / 8) $$

the rod $O P$ will make contact with the side $B D$. Let the length $O Q$ of the contact at sometime $t$ such that $0<t<\frac{\pi}{4 \omega}$ or $0<t<\frac{T}{8}$ be $x$. The flux through the area $O D Q$ is

$$ \begin{aligned} \phi & =B \frac{1}{2} Q D \times O D=B \frac{1}{2} l \tan \theta \times l \\ & =\frac{1}{2} B i^{2} \tan \theta, \text { where } \theta=\omega t \end{aligned} $$

Applying Faraday’s law of EMI,

Thus, the magnitude of the emf generated is $\varepsilon=\frac{d \phi}{d t}=\frac{1}{2} B l^{2} \omega \sec ^{2} \omega t$

The current is $I=\frac{\varepsilon}{R}$ where $R$ is the resistance of the rod in contact.

where, $R \propto \lambda$

$$ \begin{aligned} & R=\lambda x=\frac{\lambda l}{\cos \omega t} \\ \therefore \quad & I=\frac{1}{2} \frac{B l^{2} \omega}{\lambda l} \sec ^{2} \omega t \cos \omega t=\frac{B l \omega}{2 \lambda \cos \omega t} \end{aligned} $$

Let the length $O Q$ of the contact at some time $t$ such that $\frac{\pi}{4 \omega}<t<\frac{3 \pi}{4 \omega}$ or $\frac{T}{8}<t<\frac{3 T}{8}$ be $x$. The rod is in contact with the side $A B$. The flux through the area $O Q B D$ is

Where,

$$ \phi=l^{2}+\frac{1}{2} \frac{l^{2}}{\tan \theta} B $$

Thus, the magnitude of emf generated in the loop is

$$ \varepsilon=\frac{d \phi}{d t}=\frac{1}{2} B l^{2} \omega \frac{\sec ^{2} \omega t}{\tan ^{2} \omega t} $$

The current is $I=\frac{\varepsilon}{R}=\frac{\varepsilon}{\lambda x}=\frac{\varepsilon \sin \omega t}{\lambda l}=\frac{1}{2} \frac{B l \omega}{\lambda \sin \omega t}$

Similarly for $\frac{3 \pi}{4 \omega}<t<\frac{\pi}{\omega}$ or $\frac{3 T}{8}<t<\frac{T}{2}$, the rod will be in touch with $A C$.

The flux through $O Q A B D$ is given by

$$ \phi=2 l^{2}-\frac{l^{2}}{2 \tan \omega t} B $$

And the magnitude of emf generated in loop is given by

$$ \begin{aligned} & \varepsilon=\frac{d \phi}{d t}=\frac{B \omega l^{2} \sec ^{2} \omega t}{2 \tan ^{2} \omega t} \\ & l=\frac{\varepsilon}{R}=\frac{\varepsilon}{\lambda x}=\frac{1}{2} \frac{B l \omega}{\lambda \sin \omega t} \end{aligned} $$

These are the required expressions.

Thinking Process

This question need the use of integration in order to find the total magnetic flux linked with the loop.

Answer

Let us consider a strip of length $l$ and width $d r$ at a distance $r$ from infinite long current carrying wire. The magnetic field at strip due to current carrying wire is given by

Total flux through the loop is

$$ \text { Field } B(r)=\frac{\mu_{0} I}{2 \pi r} \text { (out of paper) } $$

$$ \text { Flux }=\frac{\mu_{0} I}{2 \pi} l \int_{x_{0}}^{x} \frac{d r}{r}=\frac{\mu_{0} I}{2 \pi} \ln \frac{x}{x_{0}} $$

The emf induced can be obtained by differentiating the e(i. ) wrt $t$ and then applying Ohm’s law

$$ \frac{\varepsilon}{R}=I $$

We have, induced current $=\frac{1}{R} \frac{d \phi}{d t}=\frac{\varepsilon}{R}=\frac{\mu_{0} I}{2 \pi} \frac{\lambda}{R} \ln \frac{x}{x_{0}}$

$$ \because \frac{d \mathrm{I}}{d t}=\lambda $$

Thinking Process

The charge passes through the circuit can be obtained by finding the relation between instantaneous current and instantaneous magnetic flux linked with it.

Answer

The emf induced can be obtained by differentiating the expression of magnetic flux linked wrt $t$ and then applying Ohm’s law

$$ I=\frac{E}{R}=\frac{1}{R} \frac{d \phi}{d t} $$

We know that electric current

$$ I(t)=\frac{d Q}{d t} \text { or } \frac{d Q}{d t}=\frac{1}{R} \frac{d \phi}{d t} $$

Integrating the variable separable form of differential equation for finding the charge $Q$ that passed in time $t$, we have

$$ \begin{aligned} Q\left(t_{1}\right)-Q\left(t_{2}\right) & =\frac{1}{R}\left[\phi\left(t_{1}\right)-\phi\left(t_{2}\right)\right] \\ \phi\left(t_{1}\right) & =L_{1} \frac{\mu_{0}}{2 \pi} \int_{x}^{L_{2}+x} \frac{d x^{\prime}}{x^{\prime}} I\left(t_{1}\right) \quad \text { [Refer to the E(i. ) of answer no.25] } \\ & =\frac{\mu_{0} L_{1}}{2 \pi} I\left(t_{1}\right) \ln \frac{L_{2}+x}{x} \end{aligned} $$

The magnitude of charge is given by,

$$ \begin{aligned} & =\frac{\mu_{0} L_{1}}{2 \pi} \ln \frac{L_{2}+x}{x}\left[I_{0}+0\right] \\ & =\frac{\mu_{0} L_{1}}{2 \pi} I_{1} \ln \frac{L_{2}+x}{x} \end{aligned} $$

This is the required expression.

Thinking Process

The decrease in magnetic field causes induced emf and hence, electric field around ring. The torque experienced by the ring produces change in angular momentum.

Answer

Since, the magnetic field is brought to zero in time $\Delta t$, the magnetic flux linked with the ring also reduces from maximum to zero. This, in turn, induces an emf in ring by the phenomenon of EMI. The induces emf causes the electric field $E$ generation around the ring.

The induced emf $=$ electric field $E \times(2 \pi b)($ Because $V=E \times d)$

By Faraday’s law of EMI

The induced emf =rate of change of magnetic flux

$=$ rate of change of magnetic field $\times$ area

$$ =\frac{B \pi a^{2}}{\Delta t} $$

From Eqs. (i) and (ii), we have

$$ 2 \pi b E=\mathrm{e} m f=\frac{B \pi \mathrm{a}^{2}}{\Delta t} $$

Since, the charged ring experienced a electric force $=Q E$

This force try to rotate the coil, and the torque is given by

$$ \begin{aligned} \text { Torque } & =b \times \text { Force } \\ & =Q E b=Q \frac{B \pi a^{2}}{2 \pi b \Delta t} b \\ & =Q \frac{B a^{2}}{2 \Delta t} \end{aligned} $$

If $\Delta L$ is the change in angular momentum

$$ \Delta \nu=\text { Torque } \times \Delta t=Q \frac{B a^{2}}{2} $$

Since, initial angular momentum $=0$

Now, since $\quad$ Torque $\times \Delta t=$ Change in angular momentum

Final angular momentum $=m b^{2} \omega=\frac{Q B a^{2}}{2}$

$$ \omega=\frac{Q B a^{2}}{2 m b^{2}} $$

On rearranging the terms, we have the required expression of angular speed.

Thinking Process

This problem combines the mechanics, EMI, magnetic force and linear differential equation.

Answer

Here, the component of magnetic field perpendicular the plane $=B \cos \theta$

Now, the conductor moves with speed $v$ perpendicular to $B \cos \theta \operatorname{component}$ of magnetic field. This causes motional emf across two ends of rod, which is given by $=v(B \cos \theta) d$

This makes flow of induced current $i=\frac{v(B \cos \theta) d}{R}$ where, $R$ is the resistance of rod. Now, current carrying rod experience force which is given by $F=i B d$ (horizontally in backward direction). Now, the component of magnetic force parallel to incline plane along upward direction $=F \cos \theta=i B d \cos \theta=\frac{v(B \cos \theta) d}{R} B d \cos \theta$ where, $v=\frac{d x}{d t}$

Also, the component of weight $(\mathrm{mg})$ parallel to incline plane along downward direction $=m g \sin \theta$.

Now, by Newton’s second law of motion

$$ \begin{aligned} m \frac{d^{2} x}{d t^{2}} & =m g \sin \theta-\frac{B \cos \theta d}{R} \frac{d x}{d t} \times(B d) \cos \theta \\ \frac{d v}{d t} & =g \sin \theta-\frac{B^{2} d^{2}}{m R}(\cos \theta)^{2} v \\ \frac{d v}{d t} & +\frac{B^{2} d^{2}}{m R}(\cos \theta)^{2} v=g \sin \theta \end{aligned} $$

But, this is the linear differential equation.

On solving, we get

$$ v=\frac{g \sin \theta}{\frac{B^{2} d^{2} \cos ^{2} \theta}{m R}}+A \exp -\frac{B^{2} d^{2}}{m R}\left(\cos ^{2} \theta\right) t $$

$A$ is a constant to be determine by initial conditions.

The required expression of velocity as a function of time is given by

$$ =\frac{m g R \sin \theta}{B^{2} d^{2} \cos ^{2} \theta} 1-\exp -\frac{B^{2} d^{2}}{m R}\left(\cos ^{2} \theta\right) t $$

Thinking Process

This problem combines the concept of EMI, charging of capacitor and linear differential equation.

Answer

The conductor of length $d$ moves with speed $v$, perpendicular to magnetic field $B$ as shown in figure. This produces motional emf across two ends of rod, which is given by $=v B d$. Since, switch $S$ is closed at time $t=0$. capacitor is charged by this potential difference. Let $Q(t)$ is charge on the capacitor and current flows from $A$ to $B$.

Now, the induced current

On rearranging the terms, we have

$$ I=\frac{v B d}{R}-\frac{Q}{R C} $$

$$ \frac{Q}{R C}+\frac{d Q}{d t}=\frac{v B d}{R} $$

This is the linear differential equation. On solving, we get

$\begin{aligned} & Q=v B d C+A e^{-t / R C} \\ & \Rightarrow \quad Q=v B d C\left[1-e^{-t / R C}\right] \quad \text { (At time } t=0, Q=0=A=-v B d c \text { ). } \\ & \text { Differentiating, we get } I-\frac{V B d}{R} e^{-t / R C} \ & \end{aligned}$

This is the required expression of current.

Thinking Process

This problem combines the concept of EMI, growth of current in inductor and linear differential equation.

Answer

The conductor of length $d$ moves with speed $v$, perpendicular to magnetic field $\mathbf{B}$ as shown in figure. This produces motional emf across two ends of rod, which is given by $=v B d$.

Since, switch $S$ is closed at time $t=0$. current start growing in inductor by the potential difference due to motional emf.

Applying Kirchhoff’s voltage rule, we have

$$ -L \frac{d I}{d t}+v B d=I R \text { or } L \frac{d I}{d t}+I R=v B d $$

This is the linear differential equation. On solving, we get

$$ \begin{aligned} & I=\frac{V B d}{R}+A e^{-R t / 2} \\ & \text { At } t=0 \quad I=0 \\ & \Rightarrow \quad A=-\frac{v B d}{R} \Rightarrow I=\frac{v B d}{R}\left(1-e^{-R t / L}\right) . \end{aligned} $$

This is the required expression of current.

Thinking Process

This problem establishes a relationship between induced current, power lost and velocity acquired by freely falling ring.

Answer

The magnetic flux linked with the metallic ring of mass $m$ and radius $l$ falling under gravity in a region having a magnetic field whose $z$-component of magnetic field is $B_{z}=B_{0}(1+\lambda z)$ is

$$ \phi=B_{z}\left(\pi l^{2}\right)=B_{0}(1+\lambda z)\left(\pi l^{2}\right) $$

Applying Faraday’s law of EMI, we have emf induced given by $\frac{d \phi}{d t}=$ rate of change of flux Also, by Ohm’s law

$$ B_{0}\left(\pi l^{2}\right) \lambda \frac{d z}{d t}=I R $$

On rearranging the terms, we have

$$ I=\frac{\pi l^{2} B_{0} \lambda}{R} V $$

$$ \text { Energy lost/second }=I^{2} R=\frac{\left(\pi l^{2} \lambda\right)^{2} B_{0}^{2} v^{2}}{R} $$

This must come from rate of change in $\mathrm{PE}=m g \frac{d z}{d t}=m g \mathrm{v}$

$$ \text { [as kinetic energy is constant for } v=\text { constant] } $$

Thus,

$$ m g v=\frac{\left(\pi l^{2} \lambda B_{0}\right)^{2} v^{2}}{R} \text { or } v=\frac{m g R}{\left(\pi l^{2} \lambda B_{0}\right)^{2}} $$

This is the required expression of velocity.

Thinking Process

This problem require an insight to magnetic field due to current carrying solenoid having varying current which induces emf in coil of radius $B$.

Answer

Magnetic field due to a solenoid $S, B=\mu_{0} n l$ where signs are as usual.

Magnetic flux in smaller coil $\phi=N B A$, where

$$ A=\pi b^{2} $$

Applying Faraday’s law of EMI, we have So,

$$ \begin{aligned} e & =\frac{-d \phi}{d t}=\frac{-d}{d t}(N B A) \\ & =-N \pi b^{2} \frac{d(B)}{d t} \end{aligned} $$

where,

$$ \begin{aligned} B & =\mu_{0} N i \\ & =-N \pi b^{2} \mu_{0} n \frac{d l}{d t} \\ & =-N n \pi \mu_{0} b^{2} \frac{d}{d t}\left(m t^{2}+C\right)=-\mu_{0} N n \pi b^{2} 2 m t \end{aligned} $$

Since, current varies as a function of $m t^{2}+C$.

$$ e=-\mu_{0} N n \pi b^{2} 2 m t $$