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A rectangular loop with a sliding connector of length $l = 1.0 m$ is situated in a uniform magnetic field $B = 2 T$ perpendicular to the plane of loop. Resistance of connector is r = $2 Ω$ . Two resistances of $6 Ω$ and $3 Ω$ are connected as shown in figure. The power dissipated as Joule heat when the connector is moved with a constant velocity of 2 m/s is

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6 W

4 W

2 W

1 W

answer is B.

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Detailed Solution

$P = Fv = (Bil) v = B (\frac{Blv}{R + r}) l \times v$

$= \frac{2^{2} \times 1^{2} \times 2^{2}}{(\frac{6 \times 3}{6 + 3}) + 2} = 4 W$

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