A copper rod of length lis 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:
A
12Bωl2
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B
32Bωl2
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C
Bωl2
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D
2Bωl2
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Solution
The correct option is A12Bωl2 Atadistancexfromthecenter,let′stakeanelementofthicknessdx.ElectricfieldinthiselementisvB,wherevisthevelocityoftheelementwhichisωx.⇒E=BωxPotentialdifferencebetweenthetwoendsofthiselementisdV=Bωxdx,thusthePDbetweentheendsoftherodisV=∫l0Bωxdx=12Bωl2