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. `(B pi r^(2) omega)/(2R)`
B. `((B pi r^(2) omega)^(2))/(8R)`
C. `((B pi r omega)^(2))/(2R)`
D. `((B pi r omega H_(2))^(2))/(8R)`

Answer

B

Explanation

`phi = BA cos theta`
`= B((pi r^(2))/(2)) cos omega t`
`E = -(d)/(dt) phi = (B pi r^(2))/(2) omega sin omega t`
`P = (E^(2))/(R ) = (B^(2) pi^(2)r^(4)omega^(2))/(4R) sin^(2) omega t`
But `lt sin^(2) omega t gt (1)/(2)`
`:. lt P gt =((B pi r^(2) omega)^(2))/(8 R)`