A conducting rod of length 2l is rotating with constant angular speed $omega$ about its perpendicular bisector. A uniform magnetic field $overrightarrow B$ exists parallel to the axis of rotation. The e.m.f. induced between two ends of the rod is

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B $omega$ l²

$frac{1}{2}Bomega {l^2}$

$frac{1}{8}Bomega {l^2}$

Zero

answer is D.

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Detailed Solution

Potential difference between
O and A is ${V_0} - {V_A} = frac{1}{2}B{l^2}omega$
O and B is ${V_0} - {V_B} = frac{1}{2}B{l^2}omega$
so ${V_A} - {V_B} = 0$
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