A 0.1 m long conductor carrying a current of 50A is perpendicular to a magnetic field of 1.25mT. The mechanical power to move the conductor with a speed of 1 m/s in a direction perpendicular to its length and perpendicular to its length and perpendicular to the magnetic field is,

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answer is 2.

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Detailed Solution

Magnetic force FB=ilB=50×0.1×1.25×10−3N=6.25×10−3N∴Power FB×v=6.25×10−3×1J=6.25mW

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