In a uniform magneitc field of induced B a wire in the form of a semicircle of radius r rotates about the diameter of hte circle with an angular frequency `omega`. The axis of rotation is perpendicular to hte field. If the total resistance of hte circuit is R, the mean power generated per period of rotation is

Options

A. `(Bpiromega)^(2)/(2R)`
B. `(Bpir^(2)omega)^(2)/(8R)`
C. `(Bpir^(2)omega)/(2R)`
D. `(Bpiromega^(2))^(2)/(8R)`

Answer

B

Explanation

`phi = vec B.vec A, phi = BA cos omegat`
`epsilon = -(dphi)/(dt) = omegaBA sin omegat, i = (omegaBA)/(R) sin omega t`
`p_(inst) = i^(2)R = ((omegaBA)/(R))^(2) xx R sin^(2)omegat`
`P_(avg) = (int_(0)^(T) P_(inst)xxdt)/(int_(0)^(T) dt) = (omegaBA)^(2)/(R) (int_(0)^(T) sin^(2)omegatdt)/(int_(0)^(T) dt ) = (1)/(2)(omegaBA)^(2)/(R)`
`therefore P_(avg) = ((omegaBpir^(2))^(2))/(8R)`
`[A = (pir^(2))/(2)]`