A circular coil of resistance R, area A, number of turns ' N ' is rotated about its vertical diameter with angular speed ' ω ' in a uniform magnetic field of magnitude ' B '. The average power dissipated in a complete cycle is (a) (N²A²B²ω²)/(2R) (b) (BNAω)/(R) (c) (N²AB)/(2Rω²) (d) (BAω)/(2NR)

Answer: (a). Induced emf: e = NBAωsin ω t, peak e₀ = NBAω. Average power: P = ( e²)/(R) = (e₀²)/(2R) since sin ²ω t = 1/2. P = (N²A²B²ω²)/(2R) So the answer is (a).

A circular coil of resistance R, area A, number of turns ' N ' is… | MHT CET 2024 (4 May, Shift 1)

Q16MHT CET 2024 · 4 MAY · SHIFT 1Medium

Rotating Conductors & AC Generator

A circular coil of resistance $R$ , area $A$ , number of turns ' $N$ ' is rotated about its vertical diameter with angular speed ' $\omega$ ' in a uniform magnetic field of magnitude ' $B$ '. The average power dissipated in a complete cycle is

Show answer and solution

Answer: (a)

Induced emf: $e = NBA\omega\sin\omega t$ , peak $e_0 = NBA\omega$ .
Average power: $P = \frac{\langle e^2 \rangle}{R} = \frac{e_0^2}{2R}$ since $\langle\sin^2\omega t\rangle = \frac{1}{2}$ .
$P = \frac{N^2A^2B^2\omega^2}{2R}$
So the answer is (a).

Report a mistake

Rotating Conductors & AC Generator question from MHT CET 2024 (4 May, Shift 1)

More questions on Rotating Conductors & AC Generator