Questions

Figure shows a square loop of side 0.5 m and resistance $10 Ω$ . The magnetic field has a magnitude B = 1.0 T. The work done in pulling the loop out of the field slowly and uniformly in 2.0 s is

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3.125×10−3J

6.25×10−4J

1.25×10−2J

5.0×10−4J

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detailed solution

Correct option is A

Speed of the loop should bev=ℓt=0.52=0.25ms−1Induced emf, ε=Bvℓ=(1.0)(0.25)(0.5)=0.125V∴ Current in the loop, i=εR=0.12510=1.25×10−2AThe magnetic force on the left arm due to the magnetic field isFm=iℓB=1.25×10−2(0.5)(1.0)=6.25×10−3NTo pull the loop uniformly an external force of 6.25 x 10-3N towards right must be applied.∴ W=6.25×10−3N(0.5m)=3.125×10−3J

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Similar Questions

The diagram shows a circuit having a coil of resistance $R = 2.5 Ω$ and inductance L connected to a conducting rod PQ which can slide on a perfectly conducting circular ring of radius 10 cm with its centre at 'P'. Assume that friction and gravity are absent and a constant uniform magnetic field of 5 T exists as shown in figure.

At t = 0, the circuit is switched on and simultaneously a time varying external torque is applied on the rod so that it rotates about P with a constant angular velocity 40 rad/s. Find magnitude of this torque (in milli Nm) when current reaches half of its maximum value. Neglect the self-inductance of the loop formed by the circuit.

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