First slide

Motional EMF

Question

24. A wire of length 2 m is moving at a speed of $5 {ms}^{- 1}$ perpendicular to its length and a homogeneous magnetic field of 0.5 T. The ends of the wire are joined to a circuit of resistance 10 $Ω$ . The rate at which work is being done to keep the wire moving at constant speed is

Moderate

A

$\frac{5}{12} W$
B

$\frac{5}{2} W$
C

$\frac{5}{3} W$
D

1 W

Solution

Rate of work $= \frac{W}{t} = P = Fv; also F = Bil = B (\frac{Bvl}{R}) l$
$\Rightarrow P = \frac{B^{2} v^{2} l^{2}}{R} = \frac{{(0 .5)}^{2} \times {(5)}^{2} \times {(2)}^{2}}{10} = \frac{5}{2} W$

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