If two protons are moving with speed $v = 4.5 \times 10^{5} \, m/s$ parallel to each other,then the ratio of electrostatic and magnetic force between them is:

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:

The magnetic field intensity at the point $O$ of a loop with current $i$,whose shape is illustrated below is

$A$ circular coil connected to a battery of emf $E$ produces a certain magnetic induction field at its centre. The coil is unwound,stretched to double its length,rewound into a coil of $1/3$ of the original radius,and connected to a battery of emf $E^{\prime}$ to produce the same field at the centre. Then $E^{\prime}$ is

Assertion : $A$ charge,whether stationary or in motion,produces a magnetic field around it.
Reason : Moving charges produce only electric field in the surrounding space.

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 $2$.