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.$Image$

$1.$ When $d \approx$ a but wires are not touching the loop, it is found that the net magnetic filed 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 $P Q$ and $R S$, 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 $RS$, resepectively and $h \approx 1.2 a$

$2.$ Consider $d \gg a$, and the loop is rotated about its diameter parallel to the wires by $30^{\circ}$ 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)

$(A)$ $\frac{\mu_0 I^2 a^2}{d}$ $(B)$ $\frac{\mu_0 I^2 a^2}{2 d}$ $(C)$ $\frac{\sqrt{3} \mu_0 I^2 a^2}{d}$ $(D)$ $\frac{\sqrt{3} \mu_0 I^2 a^2}{2 d}$

Give the answer question $1$ and $2.$

Answer

$1.$ $\vec{B}_R=\vec{B}$ due to ring

$\vec{B}_1=\vec{B}$ due to wire $-1$

$\vec{B}_2=\vec{B}$ due to wire $-2$

In magnitudes $B _1= B _2=\frac{\mu_0 I }{2 \pi r }$

Resultant of $B _1$ and $B _2=2 B _1 \cos \theta=\frac{\mu_0 Ia }{\pi \pi^2}$

$B _{ R }=\frac{2 \mu_0 I a ^2}{4 \pi r ^3}$

For zero magnetic field at $P$

$\frac{\mu_0 Ia }{\pi r ^2}=\frac{2 \mu I \pi a ^2}{4 \pi r ^3} $

$\Rightarrow h \approx 1.2 a$

$Image$

$2.$ Magnetic field at mid point of two wires $=\frac{\mu_0 I }{\pi d } \otimes $

Magnetic moment of loop $= I a ^2 $

Torque on loop $= MB \sin 150^{\circ} $

$\qquad=\frac{\mu_0 I ^2 a ^2}{2 d }$

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