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A rectangular loop with a sliding connector CD of length $l = 1.0 m$ is situated in uniform and constant magnetic field B = 2T perpendicular to the plane of loop. Resistance of connector CD is $r = 2 Ω .$ Two resistances of $6 Ω$ and $3 Ω$ are connected as shown in figure. The find the magnitude external force (in newtons) required to keep the connector moving with a constant velocity v = 2 m/s perpendicular to CD and in the plane of the loop.

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Detailed Solution

The equivalent resistance across the rod $R = \frac{6 \times 3}{6 + 3} = 2 Ω .$

Then current through rod of resistance $r = \frac{B l v}{R + r} = 1 a m p e r e .$

Hence the force on rod CD is $B i l = 2 \times 1 \times 1 = 2 N .$

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