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COMEDK · Physics · 22. Electromagnetic Induction

An AC generator having 400 turns and an area of cross section of \(2 \times 10^{-3}\ m^2\) rotates with an angular speed of \(200\pi\ rad\ s^{-1}\) in a uniform magnetic field of strength 0.4 T. The generator is connected to the primary of an ideal transformer having 500 turns in the primary and 2000 turns in the secondary. The secondary is connected to a \(400\ \Omega\) resistive load. What is the rms current in the secondary of the transformer? Assume, the transformer is ideal and the resistance of the coil is negligible

  1. A \(2.84\ \text{A}\)
  2. B \(28.4\ \text{A}\)
  3. C \(1.41\ \text{A}\)
  4. D \(14.2\ \text{A}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(1.41\ \text{A}\)

Step-by-step Solution

Detailed explanation

The peak emf generated by the AC generator is given by \(E_0 = NBA\omega\). Substituting the given values, \(E_0 = 400 \times 0.4 \times (2 \times 10^{-3}) \times 200\pi = 64\pi \text{ V}\). The rms voltage across the primary coil of the transformer is…