When an electric current flows in a rectangular coil of wire placed in a uniform magnetic field, the magnetic forces on the coil produce torque which tends to rotate the coil about its axis.

Let $PQRS$ be a rectangular coil of length $l$ and breadth $b$ so that its area $A=l×b$, placed in a uniform magnetic field $\overrightarrow{B}$. Let $I$ be the current flowing in the coil in anticlockwise direction.

The coil is free to rotate about the axis $XY$ in such a way that at any instant, the normal to the plane of coil makes and angle $\theta$ with the field. So, $B$ will have two components; $B\sin\theta$ along the plane of the coil and $B\cos\theta$ along perpendicular to the plane lf the coil.

According to Fleming’s left hand rule, due to component $B\cos\theta$ the forces acting on the breadth $SR$ and $PQ$ of the coil have the same value but opposite direction in the same line of action so they cancel each other. Therefore, no torque is formed in the coil. The same condition is with the lengths $QR$ and $SP$.

Due to component $B\sin\theta$ the forces acting on the sides $RS$ and $PQ$ are zero as $\theta=0°$ and $180°$ and from the relation \[F=(B\sin\theta)Il\sin0°\] and \[F=(B\sin\theta)Il\sin180°, F=0\]

But the forces acting on the sides $QR$ and $SP$ are \[F=(B\sin\theta)Il\sin90°=BIl\sin\theta\]

According to Fleming’s left hand rule, the forces acting on $QR$ is inward and $SP$ is outward from the plane of coil. Hence, they produce couple of forces and then torque $(\tau)$ in the coil is given by,

\[\tau=\text{either force}×\text{perpendicular distance}\] \[=(BIl\sin\theta)×b\] \[=(BI\sin\theta)(l×b)\] \[=(BI\sin\theta)A\] \[\therefore \tau=BIA\sin\theta\]

### For the coil having N-turns,

\[\tau=BINA\sin\theta \text{ __(1)}\]

This is the expression for torque acting on a rectangular coil carrying current placed in a magnetic field. From this relation, we can see that torque does not depend on the shape of the coil so this relation can be applied to any shape of closed conductor.

### In vector form,

\[\overrightarrow{\tau}=NI(\overrightarrow{A}×\overrightarrow{B})\]

where $\overrightarrow{A}$ is area vector having the magnitude $A$ and directed along perpendicular to the plane of coil. The direction of torque is perpendicular to both $\overrightarrow{A}$ and $\overrightarrow{B}$.

If the plane of the coil makes an angle $\alpha$ with the magnetic field, then,

\[\theta+\alpha=90°\] \[\therefore \theta=90°-\alpha\]

Hence, equation $\text{(i)}$ can be written as,

\[\tau=BINA\sin(90°-\alpha)\] \[\therefore \tau=BINA\cos\alpha\]

Also, since magnetic moment of the coil $(M)$ is $INA$, equation $\text{(i)}$ can be written as \[\tau=MB\sin\theta\]

### Special Cases

- If plane of the coil is perpendicular to $\overrightarrow{B}$ $(\theta=0°$ and $\alpha=90°)$, then, $\tau=0$ (minimum value of $\tau$).
- If plane of the coil is parallel to $\overrightarrow{B}$ $(\theta=90°$ and $\alpha=0°)$, then, $\tau=BINA$ (maximum value of $\tau$).