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JEE Main 2019 (9 Apr, Shift 2) · Lenz's Law, Eddy Currents & Induced Electric Field

Lenz's Law, Eddy Currents & Induced Electric Field question from JEE Main 2019 (9 Apr, Shift 2)

Q28JEE Main 2019 · 9 APR · SHIFT 2Difficult

Lenz's Law, Eddy Currents & Induced Electric Field

A very long solenoid of radius R is carrying current I(t) = kte–at(k > 0), as a function of time (t ≥\ge 0). counter clockwise 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 :-

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Answer: (a)

  1. Induced current in the coil ∝−dIdt\propto -\frac{dI}{dt} (Lenz's law), since the flux ∝I\propto I.
  2. dIdt=ke−at(1−at)\frac{dI}{dt} = k e^{-at}(1 - at): equal to kk (non-zero) at t=0t=0, zero at t=1at = \frac{1}{a}, negative after that and tending to zero.
  3. So the induced current starts negative (non-zero), crosses zero at t=1/at = 1/a, becomes positive, peaks and decays to zero.
  4. So the answer is (a).
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