In a uniform magneitc field of induced B a wire in the form of a semicircle of radius r rotates about the diameter of the circle with an 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. `((pi B r^(2) omega)^(2))/(8 R ) `
B. `(pi B r omega^(2))/(8 R )`
C. `((pi B r omega)^(2))/(8 R )`
D. 0

Answer

B

Explanation

The flux linked with coil (made of wire of radius r) of area A and magnetic field B, is given by
`phi= BA cos theta = B ((pi r^(2))/(2)) cos omegat`
`phi = (piB r^(2))/(2) cos omega t " " [because A = (pir^(2))/(2)" " "and" theta = omegat]`
By Faraday’s law of electromagnetic induction, induced emf, ` e =- (dphi)/(dt )`
Hence after differentiating
`e =- (d)/(dt ) ((piBr^(2))/(2) " cos " omega t ) = (pi B r^(2) omega)/(2) sin omega t `
(where, R = resistance in the circuit)
`:. ` Power P `= (e^(2))/(R ) = (pi^(2)B^(2)r^(4) omega^(2) sin^(2) omegat)/(4 R)`
`:.` Mean power `bar(P)= (pi^(2) B^(2) r^(4) omega^(2))/(4 R) . bar(sin^(2)omegat)`
`=(pi^(2)B^(2)r^(4)omega^(2))/(4R) .(1)/(2) " [because bar(sin^(2)omegat)=(1)/(2)]`
`=((pi Br^(2) omega)^(2))/(8R)`