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Question

A straight conducting rod of length 2 m is moved with a speed of 2 m/s in a uniform magnetic field of magnitude 10 T. The maximum induced emf across the ends of the rod can be

[0.77 Mark]

A
10 V
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B
20 V
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C
40 V
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D
56 V
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Solution

The correct option is C 40 V
As we know that, emf,

E=ba(v×B)dl

Emax=Bvl=10×2×2=40 V

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