JEE Advanced 2014 · Paper 2 · Physics Q15

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PhysicsSingle correct+3 / -1Question 15 of 60

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Paragraph For Questions 15 & 16

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.

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.

When $d\approx a$ 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\approx a

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

h\approx a

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

h\approx1.2a

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

h\approx1.2a

Answer and worked solution

Official answer: C

Direction of the wire currents. Take the axis of the loop (perpendicular to the paper through the centre). At a point on it, each wire's field is perpendicular to the line joining the wire to the point. The components along the axis add only if the currents are opposite; the components parallel to the paper then cancel. So the currents must be antiparallel: PQ and SR. (With parallel currents the axial components cancel and nothing can cancel the loop's field.)

Field of the wires on the axis at height $h$ :
$B_w=2\cdot\frac{\mu_0I}{2\pi\sqrt{d^2+h^2}}\cdot\frac d{\sqrt{d^2+h^2}}=\frac{\mu_0Id}{\pi(d^2+h^2)}.$

Field of the loop: $B_\ell=\dfrac{\mu_0Ia^2}{2(a^2+h^2)^{3/2}}$ .

Set them equal with $d=a$ .
$\frac{a}{\pi(a^2+h^2)}=\frac{a^2}{2(a^2+h^2)^{3/2}}\implies\sqrt{a^2+h^2}=\frac{\pi a}2\implies h=a\sqrt{\frac{\pi^2}4-1}\approx1.2a.$

(C)

Topics:Biot-Savart law

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JEE Advanced 2014 · Paper 2, Physics question 15