A copper rod of length $l$ is rotated about one end perpendicular to the uniform magnetic field B with constant angular velocity $ω$ .
The induced e.m.f. between the ends is:

\frac{1}{2} B ω l^{2}

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\frac{3}{2} B ω l^{2}

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B ω l^{2}

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2 B ω l^{2}

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Solution

The correct option is A $\frac{1}{2} B ω l^{2}$
$A t a d i s t a n c e x f r o m t h e c e n t e r, l e t^{'} s t a k e a n e l e m e n t o f t h i c k n e s s d x . E l e c t r i c f i e l d i n t h i s e l e m e n t i s v B, w h e r e v i s t h e v e l o c i t y o f t h e e l e m e n t w h i c h i s ω x . \Rightarrow E = B ω x P o t e n t i a l d i f f e r e n c e b e t w e e n t h e t w o e n d s o f t h i s e l e m e n t i s d V = B ω x d x, t h u s t h e P D b e t w e e n t h e e n d s o f t h e r o d i s V = \int_{0}^{l} B ω x d x = \frac{1}{2} B ω l^{2}$

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