A long solenoid of radius R carries a time $$(t)$$ dependent current $$I(t) = I_0 t(1 - t)$$. A ring of radius 2R is placed coaxially near its middle. During the time interval $$0 \le t \le 1$$, the induced current $$(I_R)$$ and the induced EMF $$(V_R)$$ in the ring change as:

We have a very long solenoid of radius $$R$$ carrying a current that varies with time as

$$I(t)=I_0\,t(1-t), \qquad 0 \le t \le 1.$$

First recall the formula for the magnetic field at the centre of a long solenoid. For $$n$$ turns per unit length, the field is

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

This field exists almost uniformly throughout the circular cross-section of the solenoid (radius $$R$$) and is negligible outside it. A coaxial conducting ring of radius $$2R$$ surrounds the solenoid near its middle. Though the ring is larger, only the flux linked with the area of the solenoid itself (radius $$R$$) matters because the field outside the solenoid is practically zero. Hence the magnetic flux through the ring is

$$\Phi(t)=B(t)\,\times\,(\text{area of solenoid cross-section})$$

$$\phantom{\Phi(t)}=\mu_0 n I(t)\;(\pi R^2).$$

Substituting the given current we get

$$\Phi(t)=\mu_0 n \pi R^2\,I_0\,t(1-t) =\bigl(\mu_0 n \pi R^2 I_0\bigr)\,\bigl(t-t^2\bigr).$$

Faraday’s law states that the induced emf is the negative time derivative of the flux,

$$V_R(t)=-\dfrac{d\Phi}{dt}.$$

Differentiating explicitly,

$$\dfrac{d\Phi}{dt}=\bigl(\mu_0 n \pi R^2 I_0\bigr)\,\dfrac{d}{dt}(t-t^2) =\bigl(\mu_0 n \pi R^2 I_0\bigr)\,(1-2t).$$

Therefore

$$V_R(t)=-\bigl(\mu_0 n \pi R^2 I_0\bigr)\,(1-2t),$$

and the magnitude of the emf is

$$|V_R(t)|=\bigl(\mu_0 n \pi R^2 I_0\bigr)\,|1-2t|.$$

The induced current $$I_R(t)$$ in the ring depends on the emf and its direction follows Lenz’s law. The sign of $$V_R(t)$$ (or equally the sign of $$1-2t$$) tells us the direction:

• For $$0 \le t < 0.5$$, $$1-2t>0$$, so $$V_R(t)<0$$ and the induced current has one fixed direction.
• For $$t > 0.5$$, $$1-2t<0$$, so $$V_R(t)>0$$ and the induced current reverses direction.

Exactly at $$t=0.5$$ we have $$1-2t=0$$. Thus

$$V_R(0.5)=0,$$

and at this instant the induced current changes its sense (because the emf passes through zero and changes sign).

For completeness, observe the magnitude behaviour:

$$|V_R(t)|=\bigl(\mu_0 n \pi R^2 I_0\bigr)\,|1-2t|$$

is maximum at the end points $$t=0$$ and $$t=1$$ (value = the coefficient itself) and falls linearly to zero at $$t=0.5$$. So the emf is certainly not maximum at $$t=0.25$$ or $$t=0.5$$; it is actually zero at $$t=0.5$$.

Putting all observations together:

• The induced current $$I_R$$ keeps the same direction from $$t=0$$ up to just before $$t=0.5$$.
• At $$t=0.5$$ the emf becomes zero and immediately after that the current reverses direction.

Hence, the correct answer is Option D.