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Question

# The wire PQ has mass 0.5 kg and resistance 2 Ω, can slide on the smooth, very long horizontal parallel rails separated by a distance of 1 m. The resistance of the rails is negligible. A uniform magnetic field of magnitude 0.6 T exists in the rectangular region, and a resistance of 3 Ω connects the rails outside the field region. At t=0, the wire PQ is pushed towards the right with a speed v=5 m/s. Find the maximum distance the wire will move.

A
17.5 m
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B
24.4 m
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C
30.5 m
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D
34.7 m
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Solution

## The correct option is D 34.7 mEmf induced across the wire PQ is, E=Bvl=0.6×v×1=0.6v ∴ Current flowing in the circuit is, i=BvlR+r=0.6v5=0.12v So, the force acting on the wire due to magnetic field is, FB=ilB=0.12v×1×0.6=0.072v ∴ Acceleration of the wire is, a=−FBm=−0.072v0.5=−0.144v a is negative because velocity of wire is decreasing. We know that, a=vdvdx, where x is the displacement of the wire. ∴vdvdx=−0.144v ⇒dv=−0.144dx Integrating both sides, ∫05dv=−∫x00.144dx ⇒[0−5]=−0.144x ∴x=34.7 m Hence, (D) is the correct answer.

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