A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad s⁻¹ in a uniform horizontal magnetic field of 3.0 × 10⁻² T. The maximum emf induced the coil will be ................. × 10⁻² volt (rounded off to the nearest integer)

Solution

Physics sir · 4 steps

Turning the coil keeps changing the flux through it, so the peak emf is just NBAω.

1. Why emf appears. The coil spins about a vertical diameter while B is horizontal, so the angle between the coil's normal and B keeps changing. The flux is Φ = NBA cos ω t, and the emf is -dΦ/dt = NBAω sin ω t.

   εₘₐₓ = NBAω

2. Area in metres. With r = 8.0 cm = 0.08 m, the area is A = π r² = π × 0.0064 = 2.01 × 10⁻² m².
3. Put in numbers. εₘₐₓ = 20 × (3.0 × 10⁻²) × (2.01 × 10⁻²) × 50 = 0.603 V.
4. So. εₘₐₓ = 60.3 × 10⁻² V, which rounds to 60.

The common slip: 3. Leaving out the 20 turns gives 0.0302 V, that is 3, not 60. Every turn adds its own emf, so multiply by N.

The idea to keep. For any coil rotating in a uniform field with its axis perpendicular to B, the emf is sinusoidal and peaks at NBAω. Always convert the radius to metres first.

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