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An elliptical loop having resistance R, ...

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 :

A

`(pi^2a^2b^2B^2omega^2)/(2R)`

B

Zero

C

`(piabBomega)/R`

D

`(pi^2a^2b^2B^2omega^2)/(R)`

Text Solution

Verified by Experts

The correct Answer is:
A
`e = NABomegacos omegat`
`P= (e^2)/R = (A^2 B^2 omega^2)/(R)`
`implies P = (A^2 B^2omega^2)/R(1/2)=(pi^2a^2b^2B^2)/(2R)(omega^2)`
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