A long solenoid of radius R carries a time (t) - dependent current I(t)=I0t(1 - t). A ring of radius 2R is placed coaxially near its middle. During the time interval 0 ≤ t ≤ 1, the induced current (IR) and the induced EMF(VR) in the ring change as: (a) Direction of IR remains unchanged and VR is zero at t = 0.25 (b) Direction of IR remains unchanged and VR is maximum at t = 0.5 (c) At t = 0.25 direction of IR reverses and VR is maximum (d) At t = 0.5 direction of IR reverses and VR is zero

Answer: (d). Flux through the ring comes only from inside the solenoid: φ = μ₀ n I π R² ∝ I. VR ∝ dI/dt = I₀(1-2t). At t = 0.5: dI/dt = 0, so VR = 0; for t 0.5 negative, so IR reverses. So the answer is (d) at t = 0.5 the direction of IR reverses and VR is zero.

A long solenoid of radius R carries a time (t) - dependent current… | JEE Main 2020 (7 Jan, Shift 1)

Q26JEE Main 2020 · 7 JAN · SHIFT 1Difficult

Lenz's Law, Eddy Currents & Induced Electric Field

A long solenoid of radius R carries a time (t) - dependent current

I(t)=I0t(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 (IR) and the induced EMF(VR) in the ring change as:

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Answer: (d)

Flux through the ring comes only from inside the solenoid: $\phi = \mu_0 n I\,\pi R^2 \propto I$ .
$V_R \propto \frac{dI}{dt} = I_0(1-2t)$ .
At $t = 0.5$ : $\frac{dI}{dt} = 0$ , so $V_R = 0$ ; for $t<0.5$ it is positive, for $t>0.5$ negative, so $I_R$ reverses.
So the answer is (d) at $t = 0.5$ the direction of $I_R$ reverses and $V_R$ is zero.

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