1. Magnetic field of a long solenoid

For a very long solenoid, the magnetic field inside is $B(t)=\mu_0 n I(t)=\mu_0 n k t e^{-at}$ and outside it is approximately zero.

Here, $n$ is the number of turns per unit length.

2. Flux through the outer circular coil

The outer conducting coil has radius $2R$, but the magnetic field exists only inside the solenoid of radius $R$.

So the magnetic flux linked with the outer coil is only through the area of the solenoid: $\Phi(t)=B(t)\cdot \pi R^2=\mu_0 n k \pi R^2 \, t e^{-at}$

3. Induced emf in the outer coil

By Faraday's law, $\mathcal{E}=-\frac{d\Phi}{dt}$

Thus, $\mathcal{E}=-\mu_0 n k \pi R^2 \frac{d}{dt}(t e^{-at})$

Now, $\frac{d}{dt}(t e^{-at})=e^{-at}-at e^{-at}=e^{-at}(1-at)$

Hence, $\mathcal{E}=-C e^{-at}(1-at)$ where $C=\mu_0 n k \pi R^2 >0$

So, $\mathcal{E}=C e^{-at}(at-1)$

4. Direction of induced current

Counterclockwise current is taken as positive.

The solenoid current is counterclockwise (positive), so the magnetic field through the equatorial coil is in a fixed direction. Since $I_s(t)=k t e^{-at}$ first increases from $0$ to a maximum at $t=\frac{1}{a}$ and then decreases, the induced current must oppose the change:

• For $0<t<\frac{1}{a}$: flux is increasing, so induced current is clockwise to oppose increase. Thus induced current is negative.
• At $t=\frac{1}{a}$: flux is maximum, so $\frac{d\Phi}{dt}=0$ hence induced current is zero.
• For $t>\frac{1}{a}$: flux is decreasing, so induced current is counterclockwise to oppose decrease. Thus induced current is positive.

Therefore the induced current:

• starts negative,
• becomes zero at $t=\frac{1}{a}$,
• then becomes positive,
• and finally decays to zero as $t\to\infty$ because of the factor $e^{-at}$.

Also at $t=0$, $\mathcal{E}(0)=-C$ so it starts from a finite negative value.

5. Shape of the graph

The graph of induced current is proportional to $i_{\text{ind}}(t) \propto -e^{-at}(1-at)=e^{-at}(at-1)$

This means:

• finite negative value at $t=0$,
• crosses zero at $t=\frac{1}{a}$,
• positive hump afterwards,
• approaches zero for large $t$.

This corresponds to Option A.

6. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

So they agree.