Torque on a current loop

Why a motor turns

A toy motor has a coil of wire between two magnets. Join a battery and the coil spins. Nobody pushes it. So what turns it?

A magnetic field pushes on the current in each side of the coil. On a loop, the pushes on opposite sides point opposite ways. The loop does not fly away. It turns. In this lesson we find that turning effect, the torque.

Forces on a rectangular loop

Take a rectangular loop with sides $a$ and $b$. Its area is $A = ab$. It carries a current $I$ and hangs in a uniform field $B$. It can turn about a vertical axis through its middle.

We describe its tilt with one line, the normal. The normal is at right angles to the plane of the loop. Its direction follows the right-hand rule: curl the fingers of your right hand along the current, and the thumb points along the normal. Let $\theta$ be the angle between the normal and the field.

The picture is a view from above, along the axis. The loop is edge-on. The two long sides, of length $b$, show up as the two circles.

• Each long side is at right angles to the field. It feels a force $F = IbB$.
• The currents in the two long sides run opposite ways. So the forces are equal and opposite, and the net force on the loop is zero.
• The two forces do not lie on one line. Their lines are a distance $a\sin\theta$ apart. Two such forces make a couple.
• The two short sides, of length $a$, feel forces along the axis. These cancel and make no torque about the axis.

The torque of the couple is one force times the distance between the lines:

A coil of $N$ turns has $N$ times the torque, because every turn adds the same:

The shape does not matter. The same result holds for a circle or a triangle, as long as $A$ is the area of one turn.

The magnetic moment

The product $NIA$ comes up every time. We give it a name. The magnetic moment of the coil is

It is a vector. Its size is $NIA$ and its direction is along the normal. Its unit is A m². In vector form,

The torque turns the moment towards the field. Watch the angle. In $\sin\theta$, the angle is between the normal and the field. If a problem gives the angle $\varphi$ between the plane of the coil and the field, then $\theta = 90^\circ - \varphi$, and $\sin\theta = \cos\varphi$.

A worked example. A coil has $40$ turns. Each turn is a rectangle $5.0$ cm by $3.0$ cm. It carries $2.5$ A in a field of $0.20$ T. Find the moment and the torque when the normal makes $30^\circ$ with the field.

1. The area. $A = 5.0 \times 3.0\ \text{cm}^2 = 15\ \text{cm}^2 = 15 \times 10^{-4}\ \text{m}^2$.
2. The moment. $m = NIA = 40 \times 2.5\ \text{A} \times 15 \times 10^{-4}\ \text{m}^2 = 0.15\ \text{A m}^2$.
3. The torque. $\tau = mB\sin\theta = 0.15\ \text{A m}^2 \times 0.20\ \text{T} \times \sin 30^\circ = 0.015\ \text{N m}$.
4. The greatest torque. At $\theta = 90^\circ$ the torque is $mB = 0.030\ \text{N m}$. That is twice the value at $30^\circ$.

Turn the angle slowly through a full circle. At which angles is the torque zero? Which of those would a coil go back to if you pushed it a little?

Where the coil rests

• $\theta = 0^\circ$. The moment points along the field, and the torque is zero. This is stable equilibrium. After a small push, the torque returns the coil.
• $\theta = 90^\circ$. The plane of the coil lies along the field. The torque is greatest, $\tau = mB$.
• $\theta = 180^\circ$. The moment points against the field. The torque is zero again, but this is unstable equilibrium. A small push sends the coil away.

The energy of a loop

Because the field turns the coil, the energy of the coil depends on its angle. The potential energy of a magnetic moment in a field is

The lowest energy, $-mB$, is at $\theta = 0^\circ$. The highest, $+mB$, is at $\theta = 180^\circ$. To turn the coil from $\theta_1$ to $\theta_2$ slowly, you do work

For the coil above, turning it from stable to unstable equilibrium costs $2mB = 0.060\ \text{J}$.

Torque, but no force

In a uniform field, the forces on the loop add up to zero. The loop turns but it does not move. In a field that changes from place to place, the forces on the two sides differ, and the loop also feels a net force. That is how a magnet pulls on a loop, and later on a piece of iron.

A loop is a magnetic dipole

Compare two pairs of ideas.

• An electric dipole of moment $\vec{p}$ in a field $\vec{E}$: $\vec{\tau} = \vec{p} \times \vec{E}$ and $U = -\vec{p}\cdot\vec{E}$.
• A current loop of moment $\vec{m}$ in a field $\vec{B}$: $\vec{\tau} = \vec{m} \times \vec{B}$ and $U = -\vec{m}\cdot\vec{B}$.

The rules have the same form. This is why a small current loop is called a magnetic dipole. Far from the loop, on its axis, its field is $B = \dfrac{\mu_0}{4\pi}\dfrac{2m}{z^3}$, which is the shape of the field of a short bar magnet. We meet that field again in the last lesson.

An electron in orbit

An atom is full of tiny loops. An electron going round a nucleus is a moving charge, so it is a current. Let the electron move in a circle of radius $r$ at speed $v$. One round takes $T = 2\pi r/v$, so the current is $I = e/T$.

The moment is the current times the area:

The angular momentum of the electron is $L = m_evr$. Divide one by the other:

The ratio depends on nothing but $e$ and $m_e$. The moment of an electron points against $L$, because its charge is negative. The smallest value of $L$ in the Bohr model is $\hbar = h/2\pi$. It gives a natural unit of magnetic moment, the Bohr magneton:

The magnetism of materials, in the last lesson, comes from moments of this size.

Practise with new numbers

Try it in Python

The program turns the coil of the worked example from $0^\circ$ to $180^\circ$ in steps of $30^\circ$. It prints the torque and the energy at each angle.

Common mistakes

• Using the wrong angle. In $\tau = mB\sin\theta$ the angle is between the normal and the field, not between the plane and the field.
• Forgetting the turns. A coil of $N$ turns has $m = NIA$.
• Leaving the area in cm². $1\ \text{cm}^2 = 10^{-4}\ \text{m}^2$.
• Thinking a loop in a uniform field is pushed along. The net force is zero. Only the torque is left.
• Thinking the torque is greatest when the moment points along the field. There the torque is zero. It is greatest at $90^\circ$.
• Mixing up the signs of the energy. The energy is lowest, $-mB$, when the moment is along the field.

Remember these

• Torque on a coil: $\tau = NIAB\sin\theta$, with $\theta$ between the normal and $\vec{B}$.
• Magnetic moment: $m = NIA$, along the normal. Unit: A m². Also $\vec{\tau} = \vec{m} \times \vec{B}$.
• Energy: $U = -mB\cos\theta$. Work to turn the coil: $W = mB(\cos\theta_1 - \cos\theta_2)$.
• In a uniform field the net force is zero. Stable at $\theta = 0^\circ$, unstable at $180^\circ$.
• An orbiting electron has $m = \dfrac{evr}{2}$ and $\dfrac{m}{L} = \dfrac{e}{2m_e}$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A circular coil of $25$ turns has a radius of $4.0$ cm and carries $0.60$ A. What is its magnetic moment?

2. A coil has a moment of $0.24\ \text{A m}^2$. What is the greatest torque it can feel in a field of $0.30$ T? In what position of the coil does that happen?

3. A current loop hangs in a uniform field. Why does it turn but not move to one side?

4. The coil of question 2 rests with its moment along the field. It is turned through $180^\circ$. How much energy does it gain?