Two electrons are moving along parallel lines in the same direction with the same velocity. They will:

Match the following and find the correct pairs.

List-$I$	List-$II$
$(A)$ Fleming's left hand rule	$(i)$ Direction of induced current
$(B)$ Right hand thumb rule	(ii) Magnitude and direction of magnetic induction
$(C)$ Biot-Savart law	(iii) Direction of force due to magnetic induction
$(D)$ Fleming's right hand rule	(iv) Direction of magnetic lines due to 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.
$1.$ 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 in the direction $PQ$ and $RS$,respectively and $h \approx a$
$(B)$ current in wire $1$ and wire $2$ is in the direction $PQ$ and $SR$,respectively and $h \approx a$
$(C)$ current in wire $1$ and wire $2$ is in the direction $PQ$ and $SR$,respectively and $h \approx 1.2 a$
$(D)$ current in wire $1$ and wire $2$ is in the direction $PQ$ and $RS$,respectively 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 for question $1$ and $2$.

Two concentric loops $A$ and $B$ of same radius $R = 2 \pi \,cm = 2 \pi \times 10^{-2} \,m$ are placed at right angles to each other. If the currents flowing through $A$ and $B$ are $I_A = 3 \,A$ and $I_B = 4 \,A$ respectively, then the net magnetic field at their common centre is:

$A$ light charged particle is revolving in a circle of radius $r$ due to the electrostatic attraction of a static heavy particle with an opposite charge. How does the magnetic field $B$ at the centre of the circle, produced by the moving charge, depend on $r$?

The dimensional formula of $\frac{1}{2} \mu_0 H^2$ (where $\mu_0$ is the permeability of free space and $H$ is the magnetic field intensity) is: