$\frac{\sqrt{ m }}{ e }: \frac{\sqrt{2 \,m }}{ e }: \frac{\sqrt{4\, m }}{2 e }$
$1: \sqrt{2}: 1$
Figure: $Image$
$(A)$ The particle enters Region $III$ only if its velocity $V>\frac{q / B}{m}$
$(B)$ The particle enters Region $III$ only if its velocity $\mathrm{V}<\frac{\mathrm{q} / \mathrm{B}}{\mathrm{m}}$
$(C)$ Path length of the particle in Region $II$ is maximum when velocity $V=\frac{q / B}{m}$
$(D)$ Time spent in Region $II$ is same for any velocity $V$ as long as the particle returns to Region $I$

$1.$ 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 $2.$
