A wire of length 1\ m is moving at a speed of 2\ m/s perpendicular to its length in a homogeneous magnetic field of 0.5\ T . If the ends of the wire are joined to a circuit of resistance 6\ , then the rate at which work

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Correct answer: C. (1)/(6)\ W

The motional emf induced in the wire is given by

e = Bvl

Substituting the given values, we get

e = 0.5 \times 2 \times 1 = 1\ \text{V}

The rate at which work is done to keep the wire moving at a constant speed is equal to the electrical power dissipated in the circuit.

P = \dfrac{e^2}{R}

P = \dfrac{1^2}{6} = \dfrac{1}{6}\ \text{W}

Answer:

\dfrac{1}{6}\ \text{W}

A wire of length 1\ m is moving at a speed of 2\ m/s perpendicular...

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