Two particles carrying equal charges move parallel to each other with the speed $150 \ km/s$. If $F_1$ and $F_2$ are magnetic and electric forces between two charged particles,then $\frac{|F_1|}{|F_2|}$ is (Let $\mu_0 \varepsilon_0 = \frac{1}{9 \times 10^{16}} \ s^2/m^2$)

An electron moves straight inside a charged parallel plate capacitor of uniform charge density $\sigma$. The space between the plates is filled with a uniform magnetic field of intensity $B$,as shown in the figure. Neglecting the effect of gravity,the time taken for the straight-line motion of the electron in the capacitor is:

Two parallel beams of protons and electrons,carrying equal currents,are fixed at a separation $d$. The protons and electrons move in opposite directions. $P$ is a point on a line joining the beams,at a distance $x$ from the proton beam. The magnetic field at $P$ is $B$. If $B$ is plotted against $x$,which of the following best represents the resulting curve?

$A$ beam of electrons and protons move parallel to each other in the same direction. Then they:

Which of the following is not a unit of magnetic induction?

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 $2$.