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A cionducing rod moves with constant velocity `v` perpendicular to the straight wire carrying a current I as shown compute that the emf generated between the ends of the rod A. `(mu_(0)vIl)/(pir)`B. `(mu_(0)vIl)/(2pir)`C. `(2mu_(0)vIl)/(pir)`D. `(mu_(0)vIl)/(4pir)` |
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Answer» Correct Answer - B `e=Blv=mu_(0)/(4pi)(2I)/rlv` `e=(mu_(0)Ilv)/(2pir)` |
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