JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Magnetic Effects of Current

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.

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

(A)current in wire 1 and wire 2 is the direction PQ and RS, respectively and $h \approx a$
(B)current in wire 1 and wire 2 is the direction PQ and SR, respectively and $h \approx a$
(C)current in wire 1 and wire 2 is the direction PQ and SR, respectively and $h \approx 1.2 a$ ✓ Correct
(D)current in wire 1 and wire 2 is the direction PQ and RS, respectively and $h \approx 1.2 a$

Answer: (C)

Solution

The magnetic field on the axis of the loop at height

h

B_{loop} = \frac{μ _{0} I a ^{2}}{2 ( h ^{2} + a ^{2} ) ^{3/2}}

(upward).

For the two wires carrying currents in opposite directions (PQ and SR), their fields add at the axis point. Each wire contributes

B_{wire} = \frac{μ _{0} I}{2 π d ^{2} + h ^{2}} \times \frac{d}{d ^{2} + h ^{2}}

(downward component).

For net field to be zero:

\frac{μ _{0} I a ^{2}}{2 ( h ^{2} + a ^{2} ) ^{3/2}} = \frac{μ _{0} I d}{π ( d ^{2} + h ^{2} )}

.

With

d \approx a

, solving gives

π a = 2 h^{2} + a^{2}

, which yields

h \approx 1.2 a

.

The currents must be in directions PQ and SR (opposite) to create the required field cancellation.

JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Single correct option