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Physics · Electromagnetic Induction

JEE Advanced 2020 — Paper 1 — Question 4

A circular coil of radius R and N turns has negligible resistance. As shown in the schematic figure, its two ends are connected to two wires and it is hanging by those wires with its plane being vertical. The wires are connected to a capacitor with charge Q through a switch. The coil is in a horizontal uniform magnetic field Bo parallel to the plane of the coil. When the switch is closed, the capacitor gets discharged through the coil in a very short time. By the time the capacitor is discharged fully, magnitude of the angular momentum gained by the coil will be (assume that the discharge time is so short that the coil has hardly rotated during this time)

Question figure
  1. Option A:

    π2NQB0R2\frac{\pi}{2} \mathrm{NQB}_{0} \mathrm{R}^{2}

  2. Option B:

    πNQB0R2\pi \mathrm{NQB}_{0} \mathrm{R}^{2}

    Correct
  3. Option C:

    2πNQB0R22 \pi \mathrm{NQB}_{0} \mathrm{R}^{2}

  4. Option D:

    4πNQB0R24 \pi \mathrm{NQB}_{0} \mathrm{R}^{2}

Answer: B

Step-by-step solution

A.M.=∫τdt=∫NlπR2 Bdt=NπR2 B0Q\mathrm{A} . \mathrm{M} .=\int \tau \mathrm{dt}=\int \mathrm{Nl} \pi \mathrm{R}^{2} \mathrm{~B} \mathrm{dt}=\mathrm{N} \pi \mathrm{R}^{2} \mathrm{~B}_{0} \mathrm{Q}

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2020
Paper
Paper 1
Subject
Physics
Chapter
Electromagnetic Induction
Topic
Magnetic Flux, Faraday's Law and Lenz's Law

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