Electromagnetic Induction
Chapter 6
ELECTROMAGNETIC INDUCTION
MCQ 1
6.1 A square of side $L$ meters lies in the $x$ - $y$ plane in a region, where the magnetic field is given by $\mathbf{B}=B_o(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}) T$, where $B_o$ is constant. The magnitude of flux passing through the square is
(a) $2 B_o L^{2} Wb$.
(b) $3 B_o L^{2} Wb$.
(c) $4 B_o L^{2} Wb$.
(d) $\sqrt{29} B_o L^{2} Wb$.
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Answer: (c)6.2 A loop, made of straight edges has six corners at $A(0,0,0), B(L, O, O)$ $C(L, L, 0), D(0, L, O) E(0, L, L)$ and $F(0, O, L)$. A magnetic field $\mathbf{B}=B_o(\hat{\mathbf{i}}+\hat{\mathbf{k}}) T$ is present in the region. The flux passing through the loop ABCDEFA (in that order) is
(a) $B_o L^{2} Wb$.
(b) $2 B_o L^{2} Wb$.
(c) $\sqrt{2} B_o L^{2} Wb$.
(d) $4 B_o L^{2} Wb$.
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Answer: (b)6.3 A cylindrical bar magnet is rotated about its axis (Fig 6.1). A wire is connected from the axis and is made to touch the cylindrical surface through a contact. Then
Fig. 6.1
(a) a direct current flows in the ammeter $A$.
(b) no current flows through the ammeter A.
(c) an alternating sinusoidal current flows through the ammeter A with a time period $T=2 \pi / \omega$.
(d) a time varying non-sinosoidal current flows through the ammeter A.
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Answer: (a)6.4 There are two coils A and B as shown in Fig 6.2. A current starts flowing in $B$ as shown, when $A$ is moved towards $B$ and stops when A stops moving. The current in A is counterclockwise. B is kept stationary when A moves. We can infer that
Fig. 6.2
(a) there is a constant current in the clockwise direction in $A$.
(b) there is a varying cuirent in A.
(c) there is no current in $A$.
(d) there is a constant current in the counterclockwise direction in $A$.
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Answer: (d)6.5 Same as problem 4 except the coil $A$ is made to rotate about a vertical axis (Fig 6.3). No current flows in $B$ if $A$ is at rest. The current in coil $A$, when the current in $B(at t=0)$ is counterclockwise and the coil $A$
Fig. 6.3
is as shown at this instant, $t=0$, is
(a) constant current clockwise.
(b) varying current clockwise.
(c) varying current counterclockwise.
(d) constant current counterclockwise.
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Answer: (a)6.6 The self inductance $L$ of a solenoid of length $l$ and area of crosssection $A$, with a fixed number of turns $N$ increases as
(a) $l$ and A increase.
(b) $l$ decreases and $A$ increases.
(c) $l$ increases and A decreases.
(d) both $l$ and A decrease.
MCQ II
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Answer: (b)6.7 A metal plate is getting heated. It can be because
(a) a direct current is passing through the plate.
(b) it is placed in a time varying magnetic field.
(c) it is placed in a space varying magnetic field, but does not vary with time.
(d) a current (either direct or alternating) is passing through the plate.
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Answer: (a), (b), (d)6.8 An e.m.f is produced in a coil, which is not connected to an external voltage source. This can be due to
(a) the coil being in a time varying magnetic field.
(b) the coil moving in a time varying magnetic field.
(c) the coil moving in a constant magnetic field.
(d) the coil is stationary in external spatially varying magnetic field, which does not change with time.
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Answer: (a), (b), (c)6.9 The mutual inductance $M _{12}$ of coil 1 with respect to coil 2
(a) increases when they are brought nearer.
(b) depends on the current passing through the coils.
(c) increases when one of them is rotated about an axis.
(d) is the same as $M _{21}$ of coil 2 with respect to coil 1.
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Answer: (a), (d)6.10 A circular coil expands radially in a region of magnetic field and no electromotive force is produced in the coil. This can be because
(a) the magnetic field is constant.
(b) the magnetic field is in the same plane as the circular coil and it may or may not vary.
(c) the magnetic field has a perpendicular (to the plane of the coil) component whose magnitude is decreasing suitably.
(d) there is a constant magnetic field in the perpendicular (to the plane of the coil) direction.
VSA
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Answer: (b), (c)6.11 Consider a magnet surrounded by a wire with an on/off switch $S$ (Fig 6.4). If the switch is thrown from the off position (open circuit) to the on position (closed circuit), will a current flow in the circuit? Explain.
Fig. 6.4
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Answer: No part of the wire is moving and so motional e.m.f. is zero. The magnet is stationary and hence the magnetic field does not change with time. This means no electromotive force is produced and hence no current will flow in the circuit.6.12 A wire in the form of a tightly wound solenoid is connected to a DC source, and carries a current. If the coil is stretched so that there are gaps between successive elements of the spiral coil, will the current increase or decrease? Explain.
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Answer: The current will increase. As the wires are pulled apart the flux will leak through the gaps. Lenz’s law demands that induced e.m.f. resist this decrease, which can be done by an increase in current.6.13 A solenoid is connected to a battery so that a steady current flows through it. If an iron core is inserted into the solenoid, will the current increase or decrease? Explain.
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Answer: The current will decrease. As the iron core is inserted in the solenoid, the magnetic field increases and the flux increases. Lent’s law implies that induced e.m.f. should resist this increase, which can be achieved by a decrease in current.6.14 Consider a metal ring kept on top of a fixed solenoid (say on a carboard) (Fig 6.5). The centre of the ring coincides with the axis of the solenoid. If the current is suddenly switched on, the metal ring jumps up. Explain
Fig. 6.5
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Answer: No flux was passing through the metal ring initially. When the current is switched on, flux passes through the ring. According to Lenz’s law this increase will be resisted and this can happen if the ring moves away from the solenoid. One can analyse this in more detail (Fig 6.5). If the current in the solenoid is as shown, the flux (downward) increases and this will cause a counterclockwise current (as seen form the top in the ring). As the flow of current is in the opposite direction to that in the solenoid, they will repel each other and the ring will move upward.6.15 Consider a metal ring kept (supported by a cardboard) on top of a fixed solenoid carrying a current I (see Fig 6.5). The centre of the ring coincides with the axis of the solenoid. If the current in the solenoid is switched off, what will happen to the ring?
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Answer: When the current in the solenoid decreases a current flows in the same direction in the metal ring as in the solenoid. Thus there will be a downward force. This means the ring will remain on the cardboard. The upward reaction of the cardboard on the ring will increase.6.16 Consider a metallic pipe with an inner radius of $1 cm$. If a cylindrical bar magnet of radius $0.8 cm$ is dropped through the pipe, it takes more time to come down than it takes for a similar unmagnetised cylindrical iron bar dropped through the metallic pipe. Explain.
SA
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Answer: For the magnet, eddy currents are produced in the metallic pipe. These currents will oppose the 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 $g$. Thus the magnet will take more time.6.17 A magnetic field in a certain region is given by $\mathbf{B}=B_o \cos (\omega t) \hat{\mathbf{k}}$ and a coil of radius $a$ with resistance $R$ is placed in the $x$ - $y$ plane with its centre at the origin in the magnetic field (see Fig 6.6) . Find the magnitude and the direction of the current at $(a, 0,0)$ at $t=\pi / 2 \omega, t=\pi / \omega$ and $t=3 \pi / 2 \omega$.
Fig. 6.6
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Answer: Flux through the ring
$\phi=B_o(\pi a^{2}) \cos \omega t$
$\varepsilon=B(\pi a^{2}) \omega \sin \omega t$
$I=B(\pi a^{2}) \omega \sin \omega t / R$
Current at
$ \begin{aligned} & t=\frac{\pi}{2 \omega} ; I=\frac{B(\pi a^{2}) \omega}{R} \text{ along } \hat{\mathbf{j}} \\ & t=\frac{\pi}{\omega} ; I=0 \\ & t=\frac{3}{2} \frac{\pi}{\omega} ; I=\frac{B(\pi a^{2}) \omega}{R} \text{ along }-\hat{\mathbf{j}} . \end{aligned} $
6.18 Consider a closed loop $C$ in a magnetic field (Fig 6.7). The flux passing through the loop is defined by choosing a surface whose edge coincides with the loop and using the formula $\phi=\mathbf{B} _1 \cdot \boldsymbol{{}d} \mathbf{A} _1+\boldsymbol{{}B}_2 \cdot \boldsymbol{{}d} \mathbf{A} _2+\ldots$. Now if we chose two different surfaces $S_1$ and $S_2$ having $C$ as their edge, would we get the same answer for flux. Jusity your answer.
Fig. 6.7
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Answer: One gets the same answer for flux. Flux can be throught of as the number of magnetic field lines passing through the surface (we draw $d N=B \Delta A$ lines in an area $\Delta A \perp$ to $\mathbf{B}$ ), As lines of of $\mathbf{B}$ cannot end or start in space (they form closed loops) number of lines passing through surface $S_1$ must be the same as the number of lines passing through the surface $S_2$.6.19 Find the current in the wire for the configuration shown in Fig 6.8. Wire PQ has negligible resistance. B, the magnetic field is coming out of the paper. $\theta$ is a fixed angle made by $PQ$ travelling smoothly over two conducting parallel wires seperated by a distance $d$.
Fig. 6.8
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Answer: Motional electric field $E$ along the dotted line $CD(\perp$ to both $\mathbf{v}$ and $\mathbf{B}$ and along $\mathbf{v} \times \mathbf{B})=v B$
E.M.F. along $PQ=($ length $PQ) \times($ Field along $PQ)$
$ =\frac{d}{\cos \theta} \times v B \cos \theta=d v B . $
Therefore,
$ I=\frac{d v B}{R} \text{ and is independent of } q . $
6.20 A (current vs time) graph of the current passing through a solenoid is shown in Fig 6.9. For which time is the back electromotive force $(u)$ a maximum. If the back emf at $t=3 s$ is $e$, find the back emf at $t=7 s, 15 s$ and $40 s$. $OA, AB$ and $BC$ are straight line segments.
Fig. 6.9
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Answer: Maximum rate of change of current is in $AB$. So maximum back emf will be obtained between $5 s<t<10$ s.
If $u=L 1 / 5(.$ for $.t=3 s, \frac{d I}{d t}=1 / 5) \quad(L$ is a constant $)$
For $5 s<t<10 s \quad u_1=-L \frac{3}{5}=-\frac{3}{5} L=-3 e$
Thus at $t=7 s, u_1=-3 e$.
For $10 s<t<30 s$
$u_2=L \frac{2}{20}=\frac{L}{10}=\frac{1}{2} e$
For $t>30 s \quad u_2=0$
6.21 There are two coils $A$ and $B$ seperated by some distance. If a current of $2 A$ flows through $A$, a magnetic flux of $10^{-2} Wb$ passes through $B$ (no current through $B$ ). If no current passes through $A$ and a current of 1 A passes through $B$, what is the flux through $A$ ?
LA
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Answer: Mutual inductance $=\frac{10^{-2}}{2}=5 mH$
Flux $=5 \times 10^{-3} \times 1=5 \times 10^{-3} Wb$.
6.22 A magnetic field $\mathbf{B}=B_o \sin (\omega t) \hat{\mathbf{k}}$ covers a large region where a wire $A B$ slides smoothly over two parallel conductors separated by a distance $d$ (Fig. 6.10). The wires are in the $x$ - $y$ plane. The wire $AB$ (of length $d$ ) has resistance $R$ and the parallel wires have negligible resistance. If $A B$ is moving with velocity $\boldsymbol{{}v}$, what is the current in the circuit. What is the force needed to keep the wire moving at constant velocity?
Fig. 6.10
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Answer: Let us assume that the parallel wires at are $y=0$ and $y=d$. At $t=0, AB$ has $x=0$ and moves with a velocity $v \hat{\mathbf{i}}$.
At time $t$, wire is at $x(t)=v t$.
Motional e.m.f. $=(B_o \sin \omega t) v d(-\hat{\mathbf{j}})$
E.m.f due to change in field (along OBAC)
$ =-B_o \omega \cos \omega t x(t) d $
Total e.m.f $=-B_o d[\omega x \cos (\omega t)+v \sin (\omega t)]$
Along OBAC, Current (clockwise) $=\frac{B_o d}{R}(\omega x \cos \omega t+v \sin \omega t)$
Force needed along $\hat{\mathbf{i}}=\frac{B_o d}{R}(\omega x \cos \omega t+v \sin \omega t) \times d \times B_o \sin \omega t$
$ =\frac{B_o^{2} d^{2}}{R}(\omega x \cos \omega t+v \sin \omega t) \sin \omega t $
6.23 A conducting wire $XY$ of mass $m$ and neglibile resistance slides smoothly on two parallel conducting wires as shown in Fig 6.11. The closed circuit has a resistance $R$ due to $AC . AB$ and $CD$ are perfect conductors. There is a magnetic field $\mathbf{B}=B(t) \hat{\mathbf{k}}$.
Fig. 6.11
(i) Write down equation for the acceleration of the wire $XY$.
(ii) If $\mathbf{B}$ is independent of time, obtain $v(t)$, assuming $v$ $(0)=u_0$.
(iii) For (b), show that the decrease in kinetic energy of XY equals the heat lost in $R$.
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Answer: (i) Let the wire be at $x=x(t)$ at time $t$.
Flux $=B(t) l x(t)$
$E=-\frac{d \phi}{d t}=-\frac{d B(t)}{d t} l x(t)-B(t) l . v(t)$ (second term due to motional emf)
(iii)
$ \begin{gathered} I^{2} R=\frac{B^{2} l^{2} v^{2}(t)}{R^{2}} \times R=\frac{B^{2} l^{2}}{R} u^{2} \exp (-2 l^{2} B^{2} t / m R) \\ \text{ Power lost }=\int_0^{t} I^{2} R d t=\frac{B^{2} l^{2}}{R} u^{2} \frac{m R}{2 l^{2} B^{2}}[1-e^{-(l^{2} B^{2} t / m R)}] \\ \quad=\frac{m}{2} u^{2}-\frac{m}{2} v^{2}(t) \\ =\text{ decrease in kinetic energy. } \end{gathered} $
6.24 ODBAC is a fixed rectangular conductor of negilible resistance (CO is not connnected) and $OP$ is a conductor which rotates clockwise with an angular velocity $\omega$ (Fig 6.12). The entire system is in a uniform magnetic field $\mathbf{B}$ whose direction is along the normal to the surface of the rectangular conductor $ABDC$. The conductor $OP$ is in electric contact with $ABDC$. The rotating conductor has a resistance of $\lambda$ per unit length. Find the current in the rotating conductor, as it rotates by $180^{\circ}$.
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Answer: Between time $t=0$ and $t=\frac{\pi}{4 \omega}$, the rod OP will make contact with the side $BD$. Let the length $OQ$ of the contact at some time $t$ such that
$0<t<\frac{\pi}{4 \omega} \quad$ be $x$. The flux through the area ODQ is
$ \begin{aligned} & \phi=B \frac{1}{2} QD \times OD=B \frac{1}{2} l \tan \theta \times l \\ & =\frac{1}{2} B l^{2} \tan \theta \text{ where } \theta=\omega t \end{aligned} $
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. $R=\lambda x=\frac{\lambda l}{\cos \omega t}$
$\therefore \quad I=\frac{1}{2} \frac{B l^{2} \omega}{\lambda l} \sec ^{2} \omega t \cos \omega t=\frac{B l \omega}{2 \lambda \cos \omega t}$
For $\frac{\pi}{4 \omega}<t<\frac{3 \pi}{\omega}$ the rod is in contact with the side $AB$. Let the length of the rod in contact (OQ) be $x$. The flux through OQBD is $\phi=(l^{2}+\frac{1}{2} \frac{l^{2}}{\tan \theta}) B$ where $\theta=\omega t$ Thus the magnitude of emf generated is $\varepsilon=\frac{d \phi}{d t}=\frac{1}{2} Bl^{2} \omega \frac{\sec ^{2} \omega t}{\tan ^{2} \omega t}$
The current is $I=\frac{\varepsilon}{R}=\frac{\varepsilon}{\lambda x}=\frac{\varepsilon \sin \omega t}{\lambda l}=\frac{1}{2} \frac{B l \omega}{\lambda \sin \omega t}$
For $\frac{3 \pi}{\omega}<t<\frac{\pi}{\omega}$ the rod will be in touch with OC. The Flux through
OQABD is $\phi=(2 l^{2}-\frac{l^{2}}{2 \tan \omega t}) B$
Thus the magnitude of emf
$\varepsilon=\frac{d \phi}{d t}=\frac{B \omega l^{2} \sec ^{2} \omega t}{2 \tan ^{s} \omega t}$
$I=\frac{\varepsilon}{R}=\frac{\varepsilon}{\lambda x}=\frac{1}{2} \frac{Bl \omega}{\lambda \sin \omega t}$
6.25 Consider an infinitely long wire carrying a current $I(t)$, with $\frac{d I}{d t}=\lambda=$ constant. Find the current produced in the rectangular loop of wire ABCD if its resistance is $R$ (Fig. 6.13).
Fig. 6.13
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Answer: At a distance $r$ from the wire,
Field $B(r)=\frac{\mu_o I}{2 \pi r}$ (out of paper).
Total flux through the loop is
Flux $ =\frac{\mu_o I}{2 \pi} l \int _{x_0 }^ x \frac{d r}{r}=\frac{\mu_o I} {2 \pi} \ln \frac{x} {x_o}$
$ \frac{1}{R} \frac{d I}{d t}=\frac{\varepsilon}{R}=I=\frac{\mu_0 l}{2 \pi} \frac{\lambda}{R} \ln \frac{x}{x_0} $
6.26 A rectangular loop of wire $ABCD$ is kept close to an infinitely long wire carrying a current $I(t)=I_o(1-t / T)$ for $0 \leq t \leq T$ and $I(0)=0$ for $t>T$ (Fig. 6.14). Find the total charge passing through a given point in the loop, in time $T$. The resistance of the loop is $R$.
Fig. 6.14
Q.27 A magnetic field $\mathbf{B}$ is confined to a region $r \leq a$ and points out of the paper (the $z$-axis), $r=0$ being the centre of the circular region. A charged ring (charge $=0$ ) of radius $b, b>a$ and mass $m$ lies in the $x$ - $y$ plane with its centre at the origin. The ring is free to rotate and is at rest. The magnetic field is brought to zero in time $\Delta t$. Find the angular velocity $\omega$ of the ring after the field vanishes.
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Answer: If $I(t)$ is the current in the loop.
$ I(t)=\frac{1}{R} \frac{d \phi}{d t} $
If $Q$ is the charge that passed in time $t$,
$I(t)=\frac{d Q}{d t}$ or $\frac{d Q}{d t}=\frac{1}{R} \frac{d \phi}{d t}$
I (t)
Integrating $\quad Q(t_1)-Q(t_2)=\frac{1}{R}[\phi(t_1)-\phi(t_2)]$
$ \begin{aligned} \phi(t_1)= & L_1 \frac{\mu_o}{2 \pi} \int_x^{L_2+x} \frac{d x^{\prime}}{x^{\prime}} I(t_1) \\ & =\frac{\mu_o L_1}{2 \pi} I(t_1) \ln \frac{L_2+x}{x} \end{aligned} $
The magnitute of charge is
$ \begin{aligned} Q & =\frac{\mu_o L_1}{2 \pi} \ln \frac{L_2+x}{x}[I_o-0] \\ & =\frac{\mu_o L_1 I_1}{2 \pi} \ln (\frac{L_2+x}{x}) . \end{aligned} $
$6.272 \pi b E=E . M . F=\frac{B . \pi a^{2}}{\Lambda t}$ where $E$ is the electric field generated around the ring.
$ \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
$\Delta L=$ Torque $\times \Delta t=Q \frac{B a^{2}}{2}$
Initial angular momentum $=0$
Final angular momentum $=m b^{2} \omega=\frac{Q B a^{2}}{2}$
$\omega=\frac{Q B a^{2}}{2 m b^{2}}$.
6.28 A rod of mass $m$ and resistance $R$ slides smoothly over two parallel perfectly conducting wires kept sloping at an angle $\theta$ with respect to the horizontal (Fig. 6.15). The circuit is closed through a perfect conductor at the top. There is a constant magnetic field $\mathbf{B}$ along the vertical direction. If the rod is initially at rest, find the velocity of the rod as a function of time.
Fig. 6.15
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Answer: $ 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$
$\frac{d v}{d t}=g \sin \theta-\frac{B^{2} d^{2}}{m R}(\cos \theta)^{2} v$
$\frac{d v}{d t}+\frac{B^{2} d^{2}}{m R}(\cos \theta)^{2} v=g \sin \theta$
$v=\frac{g \sin \theta}{(\frac{B^{2} d^{2} \cos ^{2} \theta}{m R})}+A \exp (-\frac{B^{2} d^{2}}{m R}(\cos ^{2} \theta) t) \quad(A$ is a constant to be
determine by initial conditions)
$ =\frac{m g R \sin \theta}{B^{2} d^{2} \cos ^{2} \theta}(1-\exp (-\frac{B^{2} d^{2}}{m R}(\cos ^{2} \theta) t)) $
6.29 Find the current in the sliding rod $AB$ (resistance $=R$ ) for the
arrangement shown in Fig 6.16. B is constant and is out of the paper. Parallel wires have no resistance. $\mathbf{v}$ is constant. Switch $S$ is closed at time $t=0$.
Fig. 6.16
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Answer: If $Q(t)$ is charge on the capacitor (note current flows from $A$ to $B$ )
$ \begin{aligned} I & =\frac{v B d}{R}-\frac{Q}{R C} \\ & \Rightarrow \frac{Q}{R C}+\frac{d Q}{d t}=\frac{v B d}{R} \end{aligned} $
$ Q=v B d C+A e^{-t / R C} $
$\therefore \Rightarrow Q=v B d C[1-e^{-t / R C}]$
(At time $t=0, Q=0=A=-v B d c$ ). Differentiating, we get
$I=\frac{v B d}{R} e^{-t / R C}$
6.30 Find the current in the sliding rod $AB$ (resistance $=R$ ) for the arrangement shown in Fig 6.17. B is constant and is out of the paper. Parallel wires have no resistance. $\boldsymbol{{}v}$ is constant. Switch $S$ is
closed at time $t=0$.
Fig. 6.17
Show Answer
Answer: $ -L \frac{d I}{d t}+v B d=I R$
$L \frac{d I}{d t}+I R=v B d$
$I=\frac{v B d}{R}+A^{-R t / 2}$
At $\quad t=0 \quad I=0 \Rightarrow A=-\frac{v B d}{R}$
$I=\frac{v B d}{R}(1-e^{-R t / L})$.
6.31 A metallic ring of mass $m$ and radius $l$ (ring being horizontal) is falling under gravity in a region having a magnetic field. If $z$ is the vertical direction, the $z$-component of magnetic field is $B_z=B_o$ $(1+\lambda z)$. If $R$ is the resistance of the ring and if the ring falls with a velocity $v$, find the energy lost in the resistance. If the ring has reached a constant velocity, use the conservation of energy to determine $v$ in terms of $m, B, \lambda$ and acceleration due to gravity $g$.
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Answer: $\frac{d \phi}{d t}=$ rate of change in flux $=(\pi l^{2}) B_0 l \frac{d z}{d t}=I R$.
$I=\frac{\pi l^{2} B_o \lambda}{R} v$
Energy lost/second $=I^{2} R=\frac{(\pi l^{2} \lambda)^{2} B_o{ }^{2} v^{2}}{R}$
This must come from rate of change in $PE=m g \frac{d z}{d t}=m g v$
(as kinetic energy is constant for $v=$ constant)
Thus, $m g v=\frac{(\pi l^{2} \lambda B_0)^{2} v^{2}}{R}$
Or, $v=\frac{m g R}{(\pi l^{2} \lambda B_o)^{2}}$.
6.32 A long solenoid ’ $S$ ’ has ’ $n$ ’ turns per meter, with diameter ’ $a$ ‘. At the centre of this coil we place a smaller coil of ’ $N$ ’ turns and diameter ’ $b$ ’ (where $b<a$ ). If the current in the solenoid increases linearly, with time, what is the induced emf appearing in the smaller coil. Plot graph showing nature of variation in emf, if current varies as a function of $m t^{2}+C$.
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Answer: Magnetic field due to a solnoid S, $B=\mu_0 n I$
Magnetic flux in smaller coil $\phi=N B A$ where $A=\pi b^{2}$
So $e=\frac{-d \phi}{d t}=\frac{-d}{d t}$ (NBP
$ \begin{aligned} & =-N \pi b^{2} \frac{d(B)}{d t}=-N \pi b^{2} \frac{d}{d t}(\mu_1. \\ & =-N \pi b^{2} \mu_0 n \frac{d I}{d t} \\ & =-N n \pi \mu_0 b^{2} \frac{d}{d t}(m t^{2}+C)=-\mu_0 N n \pi b^{2} 2 m t \end{aligned} $
$e=-\mu_0 N n \pi b^{2} 2 m t$
Negative sign signifies opposite nature of induced emf. The magnitude of emf varies with time as shown in the Fig.
