Question

The figure shows a circular loop of radius a 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 d. The loop and the wires are carrying the same current I. The current in the loop is in the counterclockwise direction if seen from above.

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Linked Question 1

Consider d >> 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)

$\frac{μ_{0} I^{2} a^{2}}{2 d}$

$\frac{μ_{0} I^{2} a^{2}}{d}$

$\frac{\sqrt{3} μ_{0} I^{2} a^{2}}{2 d}$

$\frac{\sqrt{3} μ_{0} I^{2} a^{2}}{d}$

$τ = \vec{M} \times \vec{B}$

$B = \frac{μ_{0} i}{2 πd} \times 2$

$τ = i π a^{2} \times \frac{μ_{0} i}{πd} sin 15 0 ° = \frac{μ_{0} i^{2} a^{2}}{2 d}$

Linked Question 2

When d ≈ a but wires are not touching the loop, it is found that the net magnetic field on the axis of the loop is zero at a height h above the loop. In that case

current in wire 1 and wire 2 is the direction PQ and RS, respectively and h ≈ a

current in wire 1 and wire 2 is the direction PQ and SR, respectively and h ≈ a

current in wire 1 and wire 2 is the direction PQ and SR, respectively and h ≈ 1.2a

current in wire 1 and wire 2 is the direction PQ and RS, respectively and h ≈ 1.2a

$\frac{µ_{0} i \times a^{2}}{2 {(a^{2} + h^{2})}^{3 / 2}} = \frac{µ_{0} i \times a \times 2}{2 π {(\sqrt{h^{2} + a^{2}})}^{2}}$

$\Rightarrow \frac{πa}{2} = \sqrt{a^{2} + h^{2}}$

$\frac{π^{2}}{4} a^{2} - a^{2} = h^{2}$

$h = \sqrt{1 .5} a \approx 1 . 2 a$