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 wire are carrying the same current `I`. The current in the loop is in the counterclockwise direction if seen from above.
(q) The magnetic fields(B) at `P` due to the currents in the wires are in opposite directions.
(r) There is no magnetic field at `P`.
(s) The wires repel each other.

(4) When `d~~a` but wires are not touching the loop , it is found that the net magnetic field on the axis of the loop . In that case

Options

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`
D. current in wire 1 and wire 2 is the direction PQ and RS, respectively and `h approx 1.2 a`

Answer

c

Explanation

Magnetic field due to one of the wires is shown in the following figure.

Downward components of magnetic fields due to both the wires are added together. Net downward magnetic field due to both the wires can be written as follows:
` B_("wires " ) = 2B cos theta`
`B_("wires") = 2 (mu_(0)i)/(2 pi sqrt(h^(2) + d^(2)) )d/sqrt(h^(2) + d^(2)) = (mu_(0)i)/pi d/(h^(2) + d^(2))`
Magnetic field due to loop can be written as follows:
` B_("loop") = (mu_(0)i a^(2))/(2(a^(2) + h^(2))^(3//2)) `
For net field to become zero, above two calculated magnetic fields must be equal and opposite of each other.
` B_("loop") = B_("wires") `
`rArr (mu_(0)ia^(2))/(2(a^(2) + h^(2))^(3//2) ) = (mu_(0)i)/pi d/(h^(2) + d^(2))`
`rArr a^(2)/(2(a^(2) + h^(2))^(1//2)) = d/pi`
Value of d is given to be approximately a, hence above equation can be written as follows:
`rArr a^(2)/(2(a^(2) + h^(2))^(1//2)) = a/pi`
` rArr a^(2)/(a^(2) + h^(2)) = 4/pi^(2)`
`rArr pi^(2) a^(2) = 4a^(2) + 4h^(2)`
` rArr 4h^(2) = pi^(2)a^(2) - 4a^(2)`
`rArr h = a/2 sqrt(pi^(2) - 4) approx 1.2 a`
According to direction of current we can see that option (c) is correct.