A circular coil of 1000 turns each with area 1m² is rotated about its vertical diameter at the rate of one revolution per second in a uniform horizontal magnetic field of 0.07T. The maximum voltage generation will be ___________ V.

Solution

Physics sir · 4 steps

Peak emf of a rotating coil is NBAω, and one revolution per second means ω = 2π, not 1.

1. Why emf appears. As the coil turns, the angle between its normal and B is ω t, so the flux is Φ = NBA cos ω t. By Faraday's law the emf is -dΦ/dt = NBAω sin ω t, and it is largest when sin ω t = 1.

   εₘₐₓ = NBAω

2. Angular speed. One revolution per second is a frequency f = 1 Hz. So ω = 2π f = 2π rad s⁻¹.
3. Put in numbers. εₘₐₓ = 1000 × 0.07 × 1 × 2π = 140π ≈ 439.8 V.
4. So. The maximum voltage is about 440 V.

The common slip: 70. Taking ω = 1 rad s⁻¹ instead of 2π gives 70, not 440.

The idea to keep. For any coil spinning in a uniform field, peak emf is NBAω. Always turn revolutions per second into ω = 2π f before you multiply.

Answer from the NTA final answer key. Solution by PrepWiser.