Q.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 $\tau$, the torque exerted by the magnetic field on the loop about an axis through the centres of rods.

Concept: force on a current-carrying wire, $\vec{F}=I\,\vec{l}\times\vec{B}$.

The two length-$l$ wires sit in parallel across the low-resistance rods, so each has the full $V_0$ across it. The thick wire has resistance $R$; the thin wire (half the cross-section, same material and length) has resistance $2R$:

Only the in-plane component of $\vec{B}$, $B\cos45^\circ=\dfrac{B}{\sqrt{2}}$ (taken perpendicular to the length-$l$ wires), gives a force out of the plane that can torque the loop about the axis through the rod centres. The out-of-plane force on each wire is

$F=I\,l\,\frac{B}{\sqrt{2}}.$ …

Set-up

The loop is a rectangle: two wires of length $l$ (opposite sides) joined by two rods of length $d$. The rods have negligible resistance and are the terminals connected to $V_0$, so the two long wires hang in parallel across the source. Same material and length, but cross-sections differ by a factor of $2$, so since $R\propto 1/A$:

Which forces produce a torque

Take the loop in a plane, the length-$l$ wires along $\hat y$, separated along $\hat x$ by $d$; the axis through the rod centres is then along $\hat y$, midway between the wires. Resolve $\vec B$ (at $45^\circ$ to the plane) into an in-plane part $B_{\parallel}=B\cos45^\circ=B/\sqrt2$ (along $\hat x$, i.e. perpendicular to the wires) and an out-of-plane part $B_{\perp}=B\sin45^\circ$.

For a wire carrying current along $\hat y$, $\vec F=I\,l\,\hat y\times\vec B$. The term $I l\,\hat y\times B_{\parallel}\hat x=-I l B_{\parallel}\hat z$ is the out-of-plane force; only this force, acting at lever arm $d/2$ from the $\hat y$-axis, torques the loop about that axis. (The in-plane force component exerts no torque about this axis.)

Torque …

Method: Torque From Unequal Currents in Parallel Wires Under a Tilted Field

This method applies when two (or more) parallel current-carrying wires, possibly carrying different currents, sit in a magnetic field that is inclined to their plane, and you must find the torque about a specific axis.

Steps

Step 1: Find the current in each wire separately

If the wires are separate parallel branches of a circuit (not a single series loop), do not assume they share the same current. Apply Ohm's law branch-by-branch using each wire's own resistance and the common voltage across them:

Remember resistance scales with the wire's own length and cross-section — a thinner wire of the same material and length has a larger resistance and hence a smaller current.

Step 2: Resolve the field relative to the rotation axis

A field inclined at angle $\theta$ to the plane containing the wires splits into an in-plane component $B\cos\theta$ and an out-of-plane component $B\sin\theta$. Only the component that is perpendicular to BOTH the current direction and the torque axis produces a force that can rotate the system about that axis — work out which component that is from the geometry before writing any force.

Step 3: Compute the out-of-plane force on each wire

Using $\vec F = I\vec l \times \vec B$, the relevant force on a wire of length $l$ is

where $B_\perp$ is the field component identified in Step 2.

Step 4: Take moments about the axis and combine with sign …