$A$ rectangular conducting loop consists of two wires on two opposite sides of length $l$ joined together by rods of length $d$. The wires are each of the same material but with cross-sections differing by a factor of $2$. The thicker wire has a resistance $R$ and the rods are of low resistance,which in turn are connected to a constant voltage source $V_{0}$. The loop is placed in a uniform magnetic field $\vec{B}$ at $45^{\circ}$ to its plane. Find the torque exerted by the magnetic field on the loop about an axis through the centers of the rods.

(N/A) Let the resistance of the thick wire be $R$ and the thin wire be $2R$. Both wires are connected in parallel to the voltage source $V_{0}$.
The current in the thick wire is $I_{1} = \frac{V_{0}}{R}$ and the current in the thin wire is $I_{2} = \frac{V_{0}}{2R}$.
The magnetic force on each wire is given by $\vec{F} = I(\vec{l} \times \vec{B})$. Since the wires are perpendicular to the magnetic field component in the plane of the loop,the magnitude of the force is $F = IlB \sin(90^{\circ}) = IlB$.
The forces on the two wires are $F_{1} = I_{1}lB = \frac{V_{0}lB}{R}$ and $F_{2} = I_{2}lB = \frac{V_{0}lB}{2R}$.
The axis of rotation passes through the centers of the rods of length $d$. The perpendicular distance of each wire from this axis is $r_{\perp} = \frac{d}{2} \cos(45^{\circ}) = \frac{d}{2\sqrt{2}}$.
The torque exerted by each force is $\tau = F \cdot r_{\perp}$. Both forces create a torque in the same direction about the axis.
$\tau_{net} = F_{1} \left( \frac{d}{2\sqrt{2}} \right) + F_{2} \left( \frac{d}{2\sqrt{2}} \right) = (F_{1} + F_{2}) \frac{d}{2\sqrt{2}}$.
Substituting the values: $\tau_{net} = \left( \frac{V_{0}lB}{R} + \frac{V_{0}lB}{2R} \right) \frac{d}{2\sqrt{2}} = \left( \frac{3V_{0}lB}{2R} \right) \frac{d}{2\sqrt{2}} = \frac{3V_{0}ldB}{4\sqrt{2}R}$.