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

Answer

B

Explanation

`psi = vec(B).vec(A), psi = BA cos omega t`
`epsilon=-(d psi)/(dt)=omega A sin omega t , i=(omega BA)/(R )sin omega t`
`P_("inst")=i^(2)R=((omega BA)/(R ))^(2)xx R sin^(2)omega t`
`P_(avg)=(int_(0)^(T)P_("inst")xx dt)/(int_(0)^(T)dt)=((omega BA)^(2))/(R )(int_(0)^(T)sin^(2)omega tdt)/(int_(0)^(T)dt)=(1)/(2)((omega BA)^(2))/(R )`
`therefore P_(avg)=((omega B pi r^(2))^(2))/(8R) " " [A=(pi r^(2))/(2)]`