1.

A copper rod AB of length L, pivoted at one end A, rotates at constant angular velocity omega, at right angles to a uniform magnetic field of induction B. The e.m.f developed between the mid point C of the rod and B is

Answer»

`(Bomegal^(2))/4`
`(Bomegal^(2))/2`
`(3Bomegal^(2))/4`
`(3Bomegal^(2))/8`

SOLUTION :`de=Bdxv`
`de=Bomegaxdx`

`e=Bomegaoverset(l)underset(l//2)intxdx = (Bomega)/2[X^(2)]_(l//2)^(l)`
`=(Bomega)/2xx(3l^(2))/4impliese=(3Bomegal^(2))/8`


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