In an ac generator, a rectangular coil of 100 turns each having area 14 × 10⁻² m² is rotated at 360 rev / min about an axis perpendicular to a uniform magnetic field of magnitude 3.0 T. The maximum value of the emf produced will be ________ V.
( Take π=22/7)

Solution

Physics sir · 4 steps

Peak emf is NBAω, so the whole job is turning rev/min into rad/s correctly.

1. Peak emf formula. A coil of N turns spinning at angular speed ω in a field B has flux NBA cos ω t. Differentiating gives an emf whose largest value is:

   ε₀ = NBAω

2. Find omega. 360 rev/min = 6 rev/s. Each revolution is 2π radians, so ω = 2π × 6 = 12π = 12 × 22/7 = 264/7 rad s⁻¹.
3. Put in numbers. NBA = 100 × 3.0 × 0.14 = 42. Then ε₀ = 42 × 264/7 = 6 × 264 = 1584 V.
4. So. The maximum emf produced is 1584 V.

The common slip: 252. Using ω = 6 (rev/s) without the factor 2π gives 42 × 6 = 252 V, not 1584.

The idea to keep. For any rotating coil, peak emf is NBAω. Always convert rev/min to rad/s with ω = 2π × rpm/60 before you multiply.

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