A conducting rod of length 2l is rotating with constant angular speed ω about its perpendicular bisector. A uniform magnetic field exists parallel to the axis of rotation. The e.m.f. induced between two ends of the rod is
A
Bωl2
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B
12Bωl2
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C
18Bωl2
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D
Zero
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Solution
The correct option is D Zero Potential difference between O and A is V0−VA=12Bl2ω O and B is V0−VB=12Bl2ω so VA−VB=0