Solution :
Force on axis AB due to magnetic field , $F_1= I_1 l B= \large\frac{V_0}{R_1} $$lB$
$\qquad= \large\frac{V_0}{2R} $$lB$
It is acting perpendicular to the arm AB towards observe .
Torque due to force $F_1$ about axis $EF$
$T_1= F_1 \times \large\frac{d}{2}$$\sin 45^{\circ}$
$\quad= \large\frac{V_0}{2R} $$lB \times \large\frac{d}{2} \times \frac{1}{\sqrt 2}$
$\quad= \large\frac{V_0 l B d}{4 \sqrt 2 R}$
(in the anticlock direction )
Next the force on arm DC due to magnetic field,
$F_2= I_2 lB$
$\quad= \large\frac{V_0}{R_2} $$lB$
$\quad= \large\frac{V_0}{R}$$lB$
This perpendicular to arm DC towards observe
Torque due to force $F_2$ about axis $EF$ is
$T_2 = F_2 \times \large\frac{d}{2}$$ \sin 45^{\circ}$
$\qquad= \bigg( \large\frac{V_0}{r} lB\bigg) \large\frac{d}{2} \times \frac{1}{\sqrt 2}$
$\qquad= \large\frac{V_0 l Bd}{2 \sqrt 2 R}$
(in the clock wise direction)
Net torque $T= T_2-T_1$ (in the clock wise direction )
$\quad= \large\frac{V_0 l Bd}{2 \sqrt 2 R}$$\bigg[ 1-\large\frac{1}{2}\bigg]$
$\quad= \large\frac{V_0 l Bd}{4 \sqrt 2 R}$
But $l \times d=A$
$T=\large\frac{V_0l AB}{4 \sqrt 2 R} $
Answer : $\large\frac{V_0l AB}{4 \sqrt 2 R} $