In a uniform magnetic field of induction B a wire in the form of a 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 ......

Options

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

Answer

B

Explanation

Magnetic flux `phi=vecA . vecB`
`=AB cos omegat`
`=(pir^2 B cos omegat)/2` ( `because` Area of semicircle `A=(pir^2)/2`)
Induced emf `epsilon=-(dphi)/(dt)`
`=-d/(dt)[(pir^2 B cos omegat)/2]`
`=+1/2 pi Br^2omega sin omegat`
Power `P=epsilon^2/R=(pi^2B^2r^4 omega^2 sin^2 omegat)/(4R)`
`therefore lt sin^2 omegat gt =1/2`
`therefore` Average power = `(piBr^2omega)^2/(4R)xx1/2`
`=(Bpir^2omega)^2/(8R)`