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JEE Main20202 Sep 2020Evening ShiftPhysicsElectromagnetic InductionActual

An inductance coil has a reactance of 100 Ω . When an AC signal of frequency 1000 Hz is applied to the coil, the applied voltage leads the current by 45 ° . The self-inductance of the coil is

Options

  1. A1 . 1 × 10 - 2   H
  2. B1 . 1 × 10 - 1   H
  3. C5 . 5 × 10 - 5   H
  4. D6 . 7 × 10 - 7   H

Correct answer

A. 1 . 1 × 10 - 2   H

Step-by-step solution

tanθ = x L R = tan 45 ° x L = R = 100 = x L 2 + R 2 100 = R 2 + R 2 2 R = 100 R = 50 2 ∴ x L = 50 2 Lω = 50 2 L = 50 2 2 π × 1000 = 25 2 π mH = 1 . 1 × 10 - 2 H

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