A very long solenoid of radius R is carrying current $$I(t) = kte^{-\alpha t}$$ (k > 0), as a function of time (t $$\ge$$ 0). Counterclockwise current is taken to be positive. A circular conducting coil of radius 2R is placed in the equatorial plane of the solenoid and concentric with the solenoid. The current induced in the outer coil is correctly depicted, as a function of time, by:

We have a very long solenoid of radius $$R$$. Its current varies with time as $$I(t)=k\,t\,e^{-\alpha t}$$, where $$k\gt 0$$ and $$t\ge 0$$. Counter-clockwise (ccw) current is defined as positive.

The magnetic field at the centre of a long solenoid is given by the standard formula

$$B=\mu_0\,n\,I(t),$$

where $$n$$ is the number of turns per unit length and $$\mu_0$$ is the permeability of free space. This field is directed along the axis of the solenoid. For a ccw current, the right-hand rule shows that the magnetic field points out of the page (towards the observer).

A circular conducting loop of radius $$2R$$ is placed coaxially in the same plane (the “equatorial” plane). Because the solenoid is long, the field outside it is negligible, so magnetic flux through the outer loop is contributed only by the field inside the solenoid’s cross-section of area $$\pi R^2$$.

Hence the magnetic flux linked with the outer loop is

$$\Phi(t)=B\;(\text{area inside solenoid})=\mu_0\,n\,I(t)\;\pi R^2.$$

Faraday’s law of electromagnetic induction states

$$\mathcal E = -\dfrac{d\Phi}{dt},$$

where $$\mathcal E$$ is the induced emf. Substituting the expression for $$\Phi(t)$$ gives

$$\mathcal E = -\mu_0\,n\,\pi R^2\,\dfrac{dI}{dt}.$$

Because $$\mu_0\,n\,\pi R^2$$ is a positive constant, the sign (direction) of the induced emf, and therefore of the induced current in the outer loop, is completely determined by the sign of $$\dfrac{dI}{dt}$$. A negative emf corresponds to a clockwise current (negative, by problem convention), and a positive emf corresponds to a counter-clockwise current (positive).

So we next compute $$\dfrac{dI}{dt}$$ for

$$I(t)=k\,t\,e^{-\alpha t}.$$

Using the product rule of differentiation,

$$\dfrac{dI}{dt} = k\,e^{-\alpha t} + k\,t\,\dfrac{d}{dt}\!\big(e^{-\alpha t}\big) = k\,e^{-\alpha t} - k\,\alpha t\,e^{-\alpha t} = k\,e^{-\alpha t}\,\bigl(1-\alpha t\bigr).$$

Now we analyse the sign of this derivative:

• For $$0\le t \lt \dfrac{1}{\alpha}$$, we have $$1-\alpha t \gt 0$$, so $$\dfrac{dI}{dt}\gt 0.$$br • At $$t=\dfrac{1}{\alpha}$$, we get $$1-\alpha t=0$$, so $$\dfrac{dI}{dt}=0.$$br • For $$t \gt \dfrac{1}{\alpha}$$, we have $$1-\alpha t \lt 0$$, so $$\dfrac{dI}{dt}\lt 0.$$

Because $$\mathcal E = -\text{(constant)}\dfrac{dI}{dt}$$, the induced emf (and thus the induced current) has the opposite sign:

• When $$\dfrac{dI}{dt}\gt 0$$ (i.e., $$t\lt \dfrac{1}{\alpha}$$), $$\mathcal E\lt 0$$. The current in the outer loop is clockwise, hence negative.

• At $$t=\dfrac{1}{\alpha}$$, $$\dfrac{dI}{dt}=0$$, so $$\mathcal E=0$$. The induced current is zero.

• When $$\dfrac{dI}{dt}\lt 0$$ (i.e., $$t\gt \dfrac{1}{\alpha}$$), $$\mathcal E\gt 0$$. The current in the outer loop is counter-clockwise, hence positive.

Therefore, the induced current is negative for early times, crosses zero at $$t=\dfrac{1}{\alpha}$$, and becomes positive thereafter. The qualitative plot is a curve that starts below the time axis, touches the axis once, and then rises above it.

This behaviour corresponds exactly to the graph in Option 2, which shows the current starting negative, becoming zero once, and then becoming positive.

Hence, the correct answer is Option 2.