NCERT Exemplar Solutions Class 12 Physics Electromagnetic Induction — Very Short, Short and Long Answer Questions Q11–32

NCERT Exemplar Problems, Class 12 Physics — Electromagnetic Induction, Very Short Answer Questions 11 to 16, Short Answer Questions 17 to 21 and Long Answer Questions 22 to 32. All 22 questions solved. The book numbers them 6.11 to 6.32.

Three ideas carry this page. Faraday’s law: an emf appears whenever the flux $\phi=\int\mathbf B\cdot d\mathbf A$ through a circuit changes, whether because $B$ changes, the area changes or the orientation changes. Lenz’s law fixes the direction: the induced current opposes the change in flux that causes it. Inductance is the same law applied to a circuit’s own current ($e=-L\,dI/dt$) or its neighbour’s ($e=-M\,dI/dt$).

Key insight. Write the flux through the circuit as a function of time, differentiate, and divide by the resistance. Lenz’s law then supplies the sign in words, and the force on a moving conductor always opposes its motion, which is what turns the equations of motion into exponential decays.

Very Short Answer Questions

Question 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.

Solution. No. The magnet is at rest and its field does not change with time, and the loop is fixed. So the flux through the loop never changes, and Faraday’s law gives zero emf. Closing the switch only completes the path; it does not supply an emf.

Question 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.

Solution. The current increases while the coil is being stretched. Pulling the turns apart lets flux leak out through the gaps, so the flux linked with the coil decreases. By Lenz’s law the induced emf opposes this decrease, so it acts in the direction of the existing current and adds to the battery’s emf. Once stretching stops, the current settles back to the steady value $V/R$, since resistance alone limits a steady DC current.

Question 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.

Solution. The current decreases, transiently. Inserting iron raises the flux through the solenoid because the iron is magnetised along the field. The flux is increasing, so the induced emf opposes the rise, which means it opposes the current and the current falls while the core moves in. When the core is at rest again the current returns to $V/R$.

Question 14

Consider a metal ring kept on top of a fixed solenoid (say on a cardboard) (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.

Solution. When the current is switched on, the flux through the ring rises from zero. By Lenz’s law the ring carries an induced current that opposes this rise, so it flows in the sense opposite to the current in the solenoid. Two nearby coaxial loops with opposite currents repel each other, so the ring is pushed up and jumps.

Question 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?

Solution. Switching off makes the flux through the ring fall. The induced current now opposes the decrease, so it flows in the same sense as the old solenoid current. Parallel currents in coaxial loops attract, so the ring is pulled downwards towards the solenoid. The cardboard prevents it moving, so the ring stays where it is while the normal reaction from the cardboard increases for that short time.

Question 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.

Solution. The falling magnet changes the flux through every ring of the pipe it passes, so eddy currents are induced in the metal. By Lenz’s law these currents oppose the cause, which is the magnet’s motion, so they exert an upward drag and the magnet’s acceleration is less than $g$. An unmagnetised iron bar produces no changing flux of its own, so no eddy currents are induced and it falls freely with acceleration $g$. Hence the magnet takes longer.

Short Answer Questions

Question 17

A magnetic field in a certain region is given by $\mathbf B = B_0\cos(\omega t)\,\hat{\mathbf k}$ 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 $t=3\pi/2\omega$.

Solution. The field is perpendicular to the coil, so the flux is $$\phi = B_0\pi a^2\cos\omega t,\qquad \varepsilon=-\frac{d\phi}{dt}=B_0\pi a^2\omega\sin\omega t,\qquad I=\frac{B_0\pi a^2\omega\sin\omega t}{R}.$$

• At $t=\pi/2\omega$: $\sin\omega t=1$, so $I=\dfrac{B_0\pi a^2\omega}{R}$. The positive sign means the induced field points along $+\hat k$, which is a counterclockwise current seen from $+z$, so at $(a,0,0)$ the current is along $+\hat j$. This agrees with Lenz’s law: $B$ is falling from its maximum, and the current tries to maintain the $+\hat k$ flux.
• At $t=\pi/\omega$: $\sin\omega t=0$, so $I=0$.
• At $t=3\pi/2\omega$: $\sin\omega t=-1$, so $I=\dfrac{B_0\pi a^2\omega}{R}$ in magnitude, now along $-\hat j$ at $(a,0,0)$ (clockwise seen from $+z$).

Question 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 $\phi=\mathbf B_1\cdot d\mathbf A_1+\mathbf B_2\cdot d\mathbf A_2+\dots$. 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? Justify your answer.

Solution. Yes. Magnetic field lines have no beginning or end; they form closed loops, which is the statement $\nabla\cdot\mathbf B=0$, or that the net flux through any closed surface is zero. Take $S_1$ and $S_2$ together: they form a closed surface, so the flux leaving through one equals the flux entering through the other. Every field line that passes through $S_1$ must therefore also pass through $S_2$, and the two fluxes are equal. That is why the flux through a loop is well defined without saying which surface is used.

Question 19

Find the current in the wire for the configuration shown in Fig 6.8. Wire PQ has negligible resistance. $\mathbf B$, the magnetic field is coming out of the paper. $\theta$ is a fixed angle made by PQ travelling smoothly over two conducting parallel wires separated by a distance $d$.

Solution. The rod PQ is inclined at $\theta$ and slides with velocity $\mathbf v$ along the rails, which are $d$ apart. Only the component of PQ perpendicular to $\mathbf v$ cuts field lines, and that perpendicular extent is the rail separation $d$, since the slanted rod still joins one rail to the other. Equivalently, the rod’s length is $d/\cos(\text{angle to the normal})$ and the cutting component of its velocity carries the same factor, so the two cancel. The emf is therefore $$\varepsilon = vBd,\qquad I=\frac{\varepsilon}{R}=\frac{Bvd}{R},$$ independent of $\theta$. Direction: the area of the circuit is increasing and the flux out of the page is rising, so the induced current produces flux into the page, which is a clockwise current in the circuit.

Question 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.

Solution. The back emf is $L\left|dI/dt\right|$, so it is proportional to the slope of the graph. Reading the graph: $I=0$ at $t=0$, $+1$ A at $t=5$ s (A), $-2$ A at $t=10$ s (B), and $0$ at $t=30$ s (C), with $I=0$ afterwards.

• OA: slope $=\dfrac{1-0}{5}=+\dfrac15$ A/s.
• AB: slope $=\dfrac{-2-1}{5}=-\dfrac35$ A/s.
• BC: slope $=\dfrac{0-(-2)}{20}=+\dfrac1{10}$ A/s.
• After C: slope $=0$.

The steepest segment is AB, so the back emf is greatest for $5\ \text{s}<t<10\ \text{s}$.

At $t=3$ s the slope is $+\tfrac15$ A/s and the back emf is $e$, so $e\leftrightarrow\tfrac15$. Scaling by the slope:

• $t=7$ s (on AB): $\dfrac{-3/5}{1/5}e=-3e$, so the magnitude is $3e$ and the sense is opposite to that at $t=3$ s.
• $t=15$ s (on BC): $\dfrac{1/10}{1/5}e=\dfrac e2$.
• $t=40$ s: the current is constant at zero, so the back emf is $0$.

Question 21

There are two coils A and B separated 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?

Solution. The flux in B due to the current in A is $\phi_B=M_{21}I_A$, so $$M=\frac{10^{-2}}{2}=5\times10^{-3}\ \text{H}=5\ \text{mH}.$$ Mutual inductance is symmetric, $M_{12}=M_{21}=M$, so a current of 1 A in B puts a flux $\phi_A=M\times1=5\times10^{-3}$ Wb through A.

Long Answer Questions

Question 22

A magnetic field $\mathbf B=B_0\sin(\omega t)\,\hat{\mathbf k}$ covers a large region where a wire AB slides smoothly over two parallel wires 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 AB is moving with velocity $v$, what is the current in the circuit? What is the force needed to keep the wire moving at constant velocity?

Solution. Place the rails along $y=0$ and $y=d$ with the rod at $x=vt$, and the closed loop is the rectangle between the rod and the end OC. Both $B$ and the area vary with time, so $$\phi=B_0\sin\omega t\cdot d\,x=B_0\,d\,x\sin\omega t.$$ Differentiating with $dx/dt=v$, $$\varepsilon=-\frac{d\phi}{dt}=-B_0d\left(\omega x\cos\omega t+v\sin\omega t\right).$$ Hence the current has magnitude $$I=\frac{B_0d}{R}\left(\omega x\cos\omega t+v\sin\omega t\right),$$ with the sign of the emf showing it is clockwise along OBAC when the bracket is positive.

The magnetic force on the rod is $IdB$ and, by Lenz’s law, it opposes the motion. To hold the velocity constant, the external force must be equal and opposite, directed along $\hat i$: $$F=IdB=\frac{B_0^2d^2}{R}\left(\omega x\cos\omega t+v\sin\omega t\right)\sin\omega t,\qquad x=vt.$$

Question 23

A conducting wire XY of mass $m$ and negligible 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 $\mathbf B=B(t)\,\hat{\mathbf k}$. (i) Write down the 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 (ii), show that the decrease in kinetic energy of XY equals the heat lost in $R$.

Solution. (i) With the wire at distance $x(t)$ from AC and rails a distance $l$ apart, the flux is $\phi=B(t)\,l\,x(t)$. Then $$\varepsilon=-\frac{d\phi}{dt}=-\left(\frac{dB}{dt}\,l\,x+B\,l\,v\right),\qquad I=\frac{|\varepsilon|}{R}.$$ The magnetic force on XY is $BIl$ and it opposes the change of flux, so $$m\frac{d^2x}{dt^2}=-\frac{l^2B}{R}\frac{dB}{dt}\,x-\frac{l^2B^2}{R}\frac{dx}{dt}.$$

(ii) If $B$ is constant, $dB/dt=0$ and the equation becomes $m\,dv/dt=-\dfrac{l^2B^2}{R}v$. Separating variables, $$\frac{dv}{v}=-\frac{l^2B^2}{mR}\,dt\ \Rightarrow\ v(t)=u_0\,e^{-l^2B^2t/mR}.$$

(iii) The current is $I=Blv/R$, so the heat produced in $R$ up to time $t$ is $$\int_0^tI^2R\,dt=\frac{B^2l^2}{R}\int_0^tu_0^2e^{-2l^2B^2t/mR}\,dt=\frac{B^2l^2u_0^2}{R}\cdot\frac{mR}{2l^2B^2}\left(1-e^{-2l^2B^2t/mR}\right)=\tfrac12mu_0^2-\tfrac12mv^2(t),$$ which is exactly the loss of kinetic energy of XY. The mechanical energy lost has all become heat.

Question 24

ODBAC is a fixed rectangular conductor of negligible resistance (CO is not connected) 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 $180^\circ$.

Solution. The rectangle has width $2l$ and height $l$, and O is the midpoint of the base CD, so $OD=OC=l$. The rod pivots at O and sweeps from OD through the rectangle to OC, an angle of $180^\circ$. The text says “clockwise”, but the arrow in the figure shows a counterclockwise sweep, and that is the only reading in which the rod stays inside the conductor at all (a clockwise rod would leave it at once), so we take $\theta=\omega t$ measured from OD.

The part of the rod in contact has length $x$, resistance $\lambda x$ and sweeps area at rate $\tfrac12x^2\omega$, so $$\varepsilon=\tfrac12B\omega x^2,\qquad I=\frac{\varepsilon}{\lambda x}=\frac{B\omega x}{2\lambda}.$$ Only the contact length $x$ changes with the side that the rod’s tip touches.

• $0<t<\dfrac{\pi}{4\omega}$: the tip is on the right side DB, $x=\dfrac{l}{\cos\omega t}$. The flux is the triangle $\phi=\tfrac12Bl^2\tan\omega t$, so $\left|d\phi/dt\right|=\dfrac{B\omega l^2}{2\cos^2\omega t}$, and dividing by $\lambda x$, $$I=\frac{B\omega l}{2\lambda\cos\omega t}.$$
• $\dfrac{\pi}{4\omega}<t<\dfrac{3\pi}{4\omega}$: the tip is on the top side AB, $x=\dfrac{l}{\sin\omega t}$. The flux is $\phi=B\left(l^2+\dfrac{l^2}{2\tan\omega t}\right)$, and the same step gives $$I=\frac{B\omega l}{2\lambda\sin\omega t}.$$
• $\dfrac{3\pi}{4\omega}<t<\dfrac{\pi}{\omega}$: the tip is on the left side CA, which is vertical, so $x=\dfrac{l}{\lvert\cos\omega t\rvert}$. The flux is $\phi=B\left(2l^2-\tfrac12l^2\tan(\pi-\omega t)\right)$, and $$I=\frac{B\omega l}{2\lambda\lvert\cos\omega t\rvert}.$$

Printed key error. In the third interval the key prints $I=\dfrac{1}{2}\dfrac{Bl\omega}{\lambda\sin\omega t}$, which is wrong: the rod’s tip is then on the vertical side CA, where the contact length is $l/\lvert\cos\omega t\rvert$, not $l/\sin\omega t$. The key also misprints the limit of that interval as $3\pi/\omega$ instead of $3\pi/4\omega$. The two intervals before it are correct.

Question 25

Consider an infinitely long wire carrying a current $I(t)$, with $\dfrac{dI}{dt}=\lambda=$ constant. Find the current produced in the rectangular loop of wire ABCD if its resistance is $R$ (Fig. 6.13).

Solution. The field of the long wire at distance $r$ is $B=\mu_0I/2\pi r$. Take a strip of width $dr$ and length $l$ at distance $r$ from the wire. Its flux is $\dfrac{\mu_0I}{2\pi r}\,l\,dr$, and the loop extends from $r=x_0$ to $r=x$, so $$\phi=\frac{\mu_0Il}{2\pi}\int_{x_0}^{x}\frac{dr}{r}=\frac{\mu_0Il}{2\pi}\ln\frac{x}{x_0}.$$ Then, since $dI/dt=\lambda$, $$\varepsilon=-\frac{d\phi}{dt}=-\frac{\mu_0l\lambda}{2\pi}\ln\frac{x}{x_0},\qquad I_{\text{loop}}=\frac{\mu_0l\lambda}{2\pi R}\ln\frac{x}{x_0}.$$ This is constant in time because $\lambda$ is constant, and by Lenz’s law it flows so as to oppose the growing flux.

Question 26

A rectangular loop of wire ABCD is kept close to an infinitely long wire carrying a current $I(t)=I_0\left(1-t/T\right)$ for $0\le t\le T$ 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 $R$.

Solution. The loop has width $L_1$ and its sides are at distances $x$ and $L_2+x$ from the wire. The flux is $$\phi=\int_x^{x+L_2}\frac{\mu_0I}{2\pi r}\,L_1\,dr=\frac{\mu_0L_1I}{2\pi}\ln\frac{L_2+x}{x}.$$ The charge through any point of a loop is $Q=\int I\,dt=\dfrac{1}{R}\int\left|d\phi\right|=\dfrac{\Delta\phi}{R}$, so only the change in flux matters. Over time $T$ the current falls from $I_0$ to $0$, so $\Delta\phi$ is the flux at $t=0$: $$Q=\frac{\mu_0L_1I_0}{2\pi R}\ln\frac{L_2+x}{x}.$$

Printed key error. Minor: the key’s last line drops $R$ from the denominator and writes $I_1$ where $I_0$ is meant. The method and the logarithm are right; this is a typographical slip, not a different answer.

Question 27

A magnetic field $\mathbf B$ is confined to a region $r\le a$ and points out of the paper (the $z$-axis), $r=0$ being the centre of the circular region. A charged ring (charge $=Q$) 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.

Solution. The changing flux induces a circulating electric field along the ring. By Faraday’s law, with flux only in $r\le a$, $$\oint\mathbf E\cdot d\mathbf l=2\pi bE=\frac{\pi a^2B}{\Delta t}.$$ The force on the charge is $QE$ and the torque about the centre is $QEb$, so the angular momentum gained in time $\Delta t$ is $$\Delta L=QEb\,\Delta t=\frac{Qb}{\Delta t}\cdot\frac{a^2B}{2b}\,\Delta t=\frac{QBa^2}{2}.$$ Setting this equal to $mb^2\omega$ gives $$\omega=\frac{QBa^2}{2mb^2}.$$ The result does not depend on $\Delta t$, because a slower change gives a smaller field acting for longer. For positive $Q$ the ring turns counterclockwise seen from $+z$, since the induced field circulates so as to keep the outward flux from falling.

Question 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. 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.

Solution. The field is vertical but the rod moves along the incline, so only the component of $\mathbf v$ perpendicular to $\mathbf B$ cuts flux. The velocity makes angle $\theta$ with the horizontal, so that component is $v\cos\theta$ and $$\varepsilon=Bd\,v\cos\theta,\qquad I=\frac{Bdv\cos\theta}{R}.$$ The magnetic force $IdB$ on the rod is horizontal. Its component up the incline is $IdB\cos\theta=\dfrac{B^2d^2v\cos^2\theta}{R}$. Along the incline, $$m\frac{dv}{dt}=mg\sin\theta-\frac{B^2d^2\cos^2\theta}{R}\,v\ \Rightarrow\ \frac{dv}{dt}=g\sin\theta-\frac{B^2d^2\cos^2\theta}{mR}\,v.$$ Writing $k=\dfrac{B^2d^2\cos^2\theta}{mR}$, the solution with $v(0)=0$ is $v=\dfrac{g\sin\theta}{k}\left(1-e^{-kt}\right)$, that is $$v=\frac{mgR\sin\theta}{B^2d^2\cos^2\theta}\left(1-e^{-B^2d^2\cos^2\theta\,t/mR}\right).$$ The rod approaches the terminal speed $\dfrac{mgR\sin\theta}{B^2d^2\cos^2\theta}$, where the magnetic drag balances the pull of gravity along the incline.

Question 29

Find the current in the sliding rod AB (resistance $=R$) for the arrangement shown in Fig 6.16. $\mathbf B$ is constant and is out of the paper. Parallel wires have no resistance. $\mathbf v$ is constant. Switch S is closed at time $t=0$.

Solution. The rod’s motional emf is $vBd$. It drives a current through $R$ and charges the capacitor, so at any instant, with charge $Q$ on the capacitor and current $I=dQ/dt$, $$vBd=IR+\frac QC\ \Rightarrow\ R\frac{dQ}{dt}=vBd-\frac QC.$$ Solving with $Q(0)=0$, $$Q=vBdC\left(1-e^{-t/RC}\right),\qquad I=\frac{dQ}{dt}=\frac{vBd}{R}\,e^{-t/RC}.$$ The current starts at $vBd/R$, as if the capacitor were a wire, and dies away as the capacitor charges up to $vBdC$. Its direction in the rod is from A to B.

Question 30

Find the current in the sliding rod AB (resistance $=R$) for the arrangement shown in Fig 6.17. $\mathbf B$ is constant and is out of the paper. Parallel wires have no resistance. $\mathbf v$ is constant. Switch S is closed at time $t=0$.

Solution. Now the inductor opposes any change of current. The motional emf $vBd$ equals the sum of the voltage drops: $$vBd=L\frac{dI}{dt}+IR.$$ This is a standard $LR$ growth equation. With $I(0)=0$, $$I=\frac{vBd}{R}\left(1-e^{-Rt/L}\right).$$ The current starts at zero because the inductor resists a sudden change, and grows to the steady value $vBd/R$ with time constant $L/R$.

Printed key error. Minor: an intermediate line of the key has $e^{-Rt/2}$ where $e^{-Rt/L}$ is meant. The final expression is the same as ours.

Question 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 $B_z=B_0(1+\lambda z)$. 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 $g$.

Solution. The flux through the horizontal ring is $\phi=B_0(1+\lambda z)\pi l^2$. As it falls, $dz/dt=-v$ in magnitude $v$, so $$\varepsilon=\left|\frac{d\phi}{dt}\right|=\pi l^2B_0\lambda v.$$ The power lost in the resistance is $$P=\frac{\varepsilon^2}{R}=\frac{\left(\pi l^2\lambda B_0\right)^2v^2}{R}.$$ At constant (terminal) velocity the ring’s kinetic energy is not changing, so all the work done by gravity goes into heat: $mgv=P$. Therefore $$mgv=\frac{\left(\pi l^2\lambda B_0\right)^2v^2}{R}\ \Rightarrow\ v=\frac{mgR}{\left(\pi l^2\lambda B_0\right)^2}.$$

Question 32

A long solenoid ‘S’ has ‘$n$’ turns per metre, with diameter ‘$a$’. At the centre of this long solenoid 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 a graph showing the nature of variation in emf, if current varies as a function of $mt^2+C$.

Solution. Inside a long solenoid the field is uniform, $B=\mu_0nI$. The small coil has diameter $b$, so its radius is $b/2$ and its area is $\pi(b/2)^2=\pi b^2/4$. With $N$ turns, $$e=-N\frac{d\phi}{dt}=-N\mu_0n\frac{\pi b^2}{4}\frac{dI}{dt}.$$

Linear rise. If $I=kt$, then $dI/dt=k$ is constant, so the emf is constant: $$e=-\frac{\mu_0nN\pi b^2k}{4}.$$

Quadratic variation. If $I=mt^2+C$, then $dI/dt=2mt$ and $$e=-\frac{\mu_0nN\pi b^2m\,t}{2}.$$ The emf is proportional to $t$, so the graph of $\left|e\right|$ against $t$ is a straight line through the origin whose slope is $\dfrac{\mu_0nN\pi b^2m}{2}$ (negative in sign if plotted as $e$ itself, because the induced emf opposes the increase). The constant $C$ drops out, as it should, because it does not change.

Printed key error. The key takes the coil area as $\pi b^2$, although $b$ is the diameter, so its result $e=-\mu_0Nn\pi b^2\cdot2mt$ is four times too large; it also never states the constant emf for the linear case. The correct area is $\pi b^2/4$.