An elliptical loop having resistance R, of semi major axis a, and semi minor axis b is placed in a magnetic field as shown in the figure. If the loop is rotated about the x-axis with angular frequency `omega`, the average power loss in the loop due to Joule heating is :

Options

A. `(pi ab B omega )/(R )`
B. `(pi^2 a^2 b^2 B^2 omega^2)/(R )`
C. `(pi^2a^2 b^2 B^2 omega^2 )/( 2R )`
D. zero

Answer

C

Explanation

`phi =BA cos omega t`
` in =- ( d phi ) /(dt) = BA sin omega t`
` in = B pi ab omega sin omega t `
`in =B pi ab omega`
`P=(in ^2 )/(R ) =(B^2 pi^2 a^2 b^2 omega^2)/( R ) ( sin ^2 omega t)`
` P_(av ) =(B^2 pi ^2 a^2 b^2 omega^2)/(2R)`