JEE Main 2020 · 7 January · morning · Single correct · +4 / −1
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 (I R ) and the induced EMF(V R ) in the ring change as :
- (a) Direction of I R remains unchanged and V R is zero at t = 0.25
- (b) Direction of I R remains unchanged and V R is maximum at t = 0.5
- (c) At t = 0.25 direction of I R reverses and V R is maximum
- (d) At t = 0.5 direction of I R reverses and V R is zero
Answer and solution
Answer (d) At t = 0.5 direction of I R reverses and V R is zero
Solution
Physics sir · 4 steps
The EMF follows the slope of the current, not the current itself. Find where dI/dt changes sign.
- Flux through ring. The field of a long solenoid exists only inside it, so only the area π R² carries flux, even though the ring has radius 2R. The flux is Φ = μ₀ n I π R².
V_R = -dΦ/dt = -μ₀ n π R² dI/dt = -μ₀ n π R² I₀ (1 - 2t)
- Read the factor. V_R is proportional to (1 - 2t). It has its largest size at t = 0 and t = 1, and it is zero at t = 0.5, where the current I(t) peaks and stops rising.
- Sign change. For t < 0.5 the factor (1 - 2t) is positive, and for t > 0.5 it is negative. So V_R, and with it I_R, flips direction at t = 0.5.
- So. At t = 0.5 the induced current reverses and V_R is zero, which is option (d).
The idea to keep. Induced EMF tracks the rate of change of current. It is zero where the current peaks, and that is exactly where its direction reverses.
Answer from the NTA final answer key. Solution by PrepWiser.