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) (BNAω)/(2R) (d) (N²A²B²ω²)/(R)

Answer: (a). Induced emf: e = NBAωsin ω t, peak e₀ = NBAω. Average power: P = erms²R = (e₀²)/(2R) P = (N²A²B²ω²)/(2R) So the answer is (a).

A circular coil of resistance ' R ', area ' A ', number of turns ' N… | MHT CET 2024 (16 May, Shift 2)

Q15MHT CET 2024 · 16 MAY · SHIFT 2Medium

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: $P = \frac{e_{rms}^2}{R} = \frac{e_0^2}{2R}$
$P = \frac{N^2A^2B^2\omega^2}{2R}$
So the answer is (a).

Report a mistake

Rotating Conductors & AC Generator question from MHT CET 2024 (16 May, Shift 2)

More questions on Rotating Conductors & AC Generator