A current loop in a uniform field feels no net force, but the forces on opposite sides need not act along the same line. Two equal, opposite, offset forces make a twist: a torque. That twist drives every electric motor and the needle of every old-style meter.
Key idea
A flat loop of $N$ turns, current $I$ and area $A$ has a magnetic moment $\vec{\mu} = NIA\,\hat{n}$, along its normal by the right-hand rule (fingers along the current, thumb along $\hat{n}$). In a field:
$$\vec{\tau} = \vec{\mu}\times\vec{B}, \qquad \tau = \mu B\sin\theta, \qquad U = -\vec{\mu}\cdot\vec{B},$$
with $\theta$ the angle between $\vec{\mu}$ (the normal) and $\vec{B}$.
Why it works
Take a rectangular loop with sides $a$ and $b$. The two sides of length $a$ cut across the field and feel forces $IaB$ in opposite directions. Each acts a distance $\tfrac{b}{2}$ from the axis, with lever arm $\tfrac{b}{2}\sin\theta$. Together: $\tau = 2(IaB)\left(\tfrac{b}{2}\sin\theta\right) = I(ab)B\sin\theta = IAB\sin\theta$. The forces on the other two sides point along the axis and only stretch the loop. Since only the area appears, the result holds for any flat loop.
Worked examples
Worked example: Moment and greatest torque
A 50-turn square coil $4.0\ \text{cm}$ on a side carries $2.0\ \text{A}$ in a $0.30\ \text{T}$ field. Find $\mu$ and the largest torque.
- $\mu = NIA = (50)(2.0)(0.040)^2 = 0.160\ \text{A·m²}$.
- Largest at $\theta = 90^\circ$: $\tau_{\max} = \mu B = 0.0480\ \text{N·m}$.
Answer: $\tau_{\max} = 0.0480\ \text{N·m}$
Worked example: An angle given to the plane
The same coil is turned so its plane makes $30^\circ$ with the field. Find the torque.
- The normal is $90^\circ$ from the plane, so $\theta = 90^\circ - 30^\circ = 60^\circ$.
- $\tau = \mu B\sin 60^\circ = 0.0416\ \text{N·m}$.
Answer: $\tau \approx 0.0416\ \text{N·m}$
Worked example: Turning it over
How much work does it take to turn the coil from $\theta = 0$ (moment along the field) to $\theta = 180^\circ$?
- $U = -\mu B\cos\theta$ goes from $-\mu B$ to $+\mu B$.
- $W = \Delta U = 2\mu B = 0.0960\ \text{J}$.
Answer: $W = 0.0960\ \text{J}$
Common mistake
"The torque is greatest when the field passes straight through the loop." That is where the flux is greatest, but the torque is zero there: the forces on opposite sides then pull straight outward along one line, with no lever arm. Check with $\tau = \mu B\sin\theta$: field straight through means $\theta = 0$ and $\sin\theta = 0$. Always measure $\theta$ from the normal, not from the plane.