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JEE Advanced Physics Electromagnetic Induction 2020 JEE Advanced 2020 (Paper 2)

JEE Advanced Physics Question (2020) — Solution

Question

The inductors of two L R circuits are placed next to each other, as shown in the figure. The values of the self-inductance of the inductors, resistances, mutual-inductance and applied voltages are specified in the given circuit. After both the switches are closed simultaneously, the total work done by the batteries against the induced E M F in the inductors by the time the currents reach their steady-state values is_______ mJ .

Options

  1. A. A
  2. B. B
  3. C. C
  4. D. D

Answer

A. A

Step-by-step solution

ε 1 = L 1 d i 1 d t + M d i 2 d t d W 1 = ε 1 i 1 d t = L 1 d i 1 i 1 + M d i 2 i 1 d W 2 = L 2 d i 2 i 2 + M d i 1 i 2 d W 1 + d W 2 = L 1 i 1 d i 1 + L 2 i 2 d i 2 + M d i 1 · i 2 W = 1 2 L 1 i 1 2 + 1 2 L 2 i 2 2 + M 1 i 2 = 55 mJ , Note : If mutual induction is taken in opposite direction then W = 35 mJ

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Related: Physics — Electromagnetic Induction · All PYQ Banks