A coil of 200 turns and area 0.20 m² is rotated at half a revolution per second and is placed in uniform magnetic field of 0.01 T perpendicular to axis of rotation of the coil. The maximum voltage generated in the coil is (2 π)/β volt. The value of β is _______.

Solution

Physics sir · 4 steps

A rotating coil gives the peak emf NBAω, so first turn half a revolution per second into ω.

1. Flux through the coil. As the coil turns, the angle between B⃗ and the coil's normal is ω t. So the flux is Φ = NBA cos ω t, and Faraday's law gives ε = -dΦ/dt = NBAω sin ω t.

   εₘₐₓ = NBAω

2. Angular speed. Half a revolution per second means f = 0.5 Hz. So ω = 2π f = 2π × 0.5 = π rad s⁻¹.
3. Put in numbers. εₘₐₓ = 200 × 0.01 × 0.20 × π = 0.4π V = 2π/5 V.
4. So. Matching 2π/5 with 2π/β gives β = 5.

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

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