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← JEE Advanced 2014 · Paper 2
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JEE Advanced 2014 · Paper 2, Physics question 16

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Paragraph For Questions 15 & 16

The figure shows a circular loop of radius aa with two long parallel wires (numbered 1 and 2) all in the plane of the paper. The distance of each wire from the centre of the loop is dd. The loop and the wires are carrying the same current II. The current in the loop is in the counterclockwise direction if seen from above.

Two long vertical parallel wires in the plane of the paper: wire 1 on the left running from P (bottom) to Q (top), wire 2 on the right from R (bottom) to S (top). A circular loop of radius a lies midway, its centre at distance d from each wire, with a vertical dashed diameter and counterclockwise current arrows.

Consider d≫ad\gg a, and the loop is rotated about its diameter parallel to the wires by 30° from the position shown in the figure. If the currents in the wires are in the opposite directions, the torque on the loop at its new position will be (assume that the net field due to the wires is constant over the loop)

Choose one option
Answer and worked solution

Official answer: B

Field at the loop. With opposite currents, both wires' fields at the centre point the same way, perpendicular to the paper:
B=2⋅μ0I2πd=μ0Iπd.B=2\cdot\frac{\mu_0I}{2\pi d}=\frac{\mu_0I}{\pi d}.

Torque. The loop's moment m=Iπa2m=I\pi a^2 was along BB; after a 30° rotation about the diameter the angle between them is 30°:
τ=mBsin⁡30°=Iπa2⋅μ0Iπd⋅12=μ0I2a22d.\tau=mB\sin30°=I\pi a^2\cdot\frac{\mu_0I}{\pi d}\cdot\frac12=\frac{\mu_0I^2a^2}{2d}.

(B)

Topics:Torque on a current loopBiot-Savart law
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