Two parallel beams of electrons moving in the same direction produce a mutual force:

$A$ conductor lies parallel to the $Z$-axis between $-1.5 \le Z < 1.5 \text{ m}$,carrying a constant current of $10.0 \text{ A}$ in the $-\hat{a}_z$ direction. For the given magnetic field $\vec{B} = 3.0 \times 10^{-4} e^{-0.2x} \hat{a}_y \text{ T}$,find the power required to move the conductor at a constant speed from $x = 0$ to $x = 2.0 \text{ m}$ in a time interval of $5 \times 10^{-3} \text{ s}$. Assume the motion is parallel to the $X$-axis. ........... $\text{W}$

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:

$A$ compass needle free to turn in a horizontal plane is placed at the centre of a circular coil of $30$ turns and radius $12 \;cm$. The coil is in a vertical plane making an angle of $45^{\circ}$ with the magnetic meridian. When the current in the coil is $0.35 \;A$,the needle points west to east.
$(a)$ Determine the horizontal component of the earth's magnetic field at the location.
$(b)$ The current in the coil is reversed,and the coil is rotated about its vertical axis by an angle of $90^{\circ}$ in the anticlockwise sense looking from above. Predict the direction of the needle. Take the magnetic declination at the place to be zero.

Two charged particles $A$ and $B$,each of charge $+e$ and masses $12 \, amu$ and $13 \, amu$ respectively,follow a circular trajectory in chamber $X$ after passing through a velocity selector as shown in the figure. Both particles enter the velocity selector with a speed of $1.5 \times 10^6 \, ms^{-1}$. $A$ uniform magnetic field of strength $1.0 \, T$ is maintained within the chamber $X$ and in the velocity selector,directed into the plane ($-z$ direction).

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 $2$.