A square loop of side L, resistance R placed in a uniform magnetic field B acting normally to the plane of the loop. If we attempt to pull it out of the field with a constant velocity v, then the power needed is
A. BRlv
B. `(B^(2)l^(2)v^(2))/(R)`
C. `(Bl^(2)v^(2))/(R)`
D. `(Bvl)/(R)`

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1 Answers

Correct Answer - B
Power, P = Fv = ilBv
`=((Bvl)/(R))lBv=(B^(2)l^(2)v^(2))/(R)`

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