$\begin{array}{1 1} \large\frac{V_0ld}{4 \sqrt 2 R} \\ \large\frac{V_0lB}{4 \sqrt 2 R} \\ \large\frac{V_0lAB}{ \sqrt 2 R} \\ \large\frac{V_0l AB}{4 \sqrt 2 R} \end{array} $

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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} $

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