In a uniform magnetic field of induction $B$ a wire in the form of semicircle of radius $r$ rotates about the diameter of the circle with angular frequency $\omega$. The axis of rotation is perpendicular to the field. If the total resistance of the circuit is $R$ the mean power generated per period of rotation is

(B) $\frac{(B\pi r^2\omega)^2}{8R}$

(A) $\frac{B\pi r^2\omega}{2R}$

(C) $\frac{(B\pi r\omega)^2}{2R}$

(D) $\frac{(B\pi r\omega^2)^2}{8R}$

JEE Main · 2004 · 24 Apr 2004 · Offline · Electromagnetic Induction

In a uniform magnetic field of induction

B

a wire in the form of semicircle of radius

r

rotates about the diameter of the circle with angular frequency

ω

. The axis of rotation is perpendicular to the field. If the total resistance of the circuit is

R

the mean power generated per period of rotation is

Answer: (B)

The semicircular loop encloses area

A = \frac{π r ^{2}}{2}

. Rotation gives flux

Φ = B A cos ω t

, emf

ε = B A ω sin ω t

with peak

ε_{0} = \frac{B π r ^{2} ω}{2}

. Mean power

\overset{ˉ}{P} = \frac{ε _{r m s}^{2}}{R} = \frac{ε _{0}^{2}}{2 R} = \frac{( B π r ^{2} ω ) ^{2}}{8 R}

JEE Main · 2004 · 24 Apr 2004 · Offline · Single correct option