Torque on a Current Loop and the DC Motor – Magnetic Field and Magnetic Forces (Senior 5 Physics) | Ganzaa

Magnetic Field and Magnetic Forces

Lessons

Senior 5 · Physics

Magnetic Field and Magnetic Forces

4 lessons

Lesson 4 of 4

Torque on a Current Loop and the DC Motor

25 min

From a push to a spin

A ceiling fan, a phone's vibration motor, the water pump on a solar-powered farm and the motor of an electric moto taxi all turn for the same reason: a coil carrying a current in a magnetic field feels a turning effect. In the last two lessons you found the force on a single wire. Put four wires together in a rectangle and the forces form a couple. This lesson calculates that couple, defines the coil's magnetic moment and shows how a commutator turns a twisting coil into a continuously spinning motor.

The force and torque on a rectangular coil

Consider a rectangular coil of $N$ turns, length $l$ and width $w$ , carrying current $I$ in a uniform field $B$ . It can turn about an axis through its centre, parallel to the sides of length $l$ , with these sides perpendicular to the field. Let $\theta$ be the angle between the field and the normal to the plane of the coil.

The two sides of width $w$ feel equal and opposite forces along the axis, which cancel and have no turning effect.
The two sides of length $l$ are perpendicular to $B$ , so each feels $F = N B I l$ , in opposite directions and along different lines: a couple.

The perpendicular distance between the lines of action of the two forces is $w\sin\theta$ , so the torque is

\tau = F \times w\sin\theta = NBIlw\sin\theta = NIAB\sin\theta

where $A = l w$ is the area of the coil. Although derived for a rectangle, the result holds for a flat coil of any shape. The net force on the coil in a uniform field is zero; only the torque remains.

Torque is greatest, $N I A B$ , when $\theta = 90^\circ$ : the plane of the coil is parallel to the field.
Torque is zero when $\theta = 0^\circ$ : the plane is perpendicular to the field. The coil is then in stable equilibrium.

The magnetic moment

The combination $N I A$ describes how strongly a coil interacts with a field, so it is given its own name, the magnetic dipole moment:

\mu = NIA, \qquad \tau = \mu B\sin\theta, \qquad \vec{\tau} = \vec{\mu} \times \vec{B}

Its direction is along the normal to the coil, given by the right-hand grip rule: curl the fingers of your right hand round the coil in the direction of the current, and your thumb points along $\vec{\mu}$ (towards the coil's north face). A field turns a coil so that $\vec{\mu}$ lines up with $\vec{B}$ , exactly as it turns a compass needle, which is itself a magnetic dipole.

Definition

The magnetic dipole moment of a flat coil is $\mu = NIA$ , a vector along the normal to the coil; unit $\mathrm{A\,m^2}$ . The torque on it in a uniform field is $\tau = \mu B\sin\theta$ , where $\theta$ is the angle between $\vec{\mu}$ and $\vec{B}$ . Example: a single loop of area $1.0 \times 10^{-3}\,\mathrm{m^2}$ carrying 2.0 A has $\mu = 2.0 \times 10^{-3}\,\mathrm{A\,m^2}$ .

Worked example 1: the torque on a coil

A rectangular coil of 100 turns, 4.0 cm by 3.0 cm, carries 0.50 A in a uniform field of 0.20 T. Find (a) its magnetic moment, (b) the greatest torque on it, (c) the torque when its normal is at 30° to the field.

Solution

(a) $A = 0.040 \times 0.030 = 1.2 \times 10^{-3}\,\mathrm{m^2}$ ; $\mu = NIA = (100)(0.50)(1.2 \times 10^{-3}) = 0.060\,\mathrm{A\,m^2}$ .

(b) $\tau_{\max} = \mu B = (0.060)(0.20) = 0.012$ N m, when the plane of the coil is parallel to the field.

Worked example 2: the magnetic moment of an atom

In Bohr's model of hydrogen (Unit 6), the electron moves at $2.19 \times 10^{6}$ m s⁻¹ in a circle of radius $5.29 \times 10^{-11}$ m. Treating the orbit as a current loop, find its magnetic moment.

Solution

The electron passes any point $\frac{v}{2\pi r}$ times per second, so the current is $I = \frac{ev}{2\pi r}$ . The area is $\pi r^2$ , so

\mu = IA = \frac{ev}{2\pi r} \cdot \pi r^2 = \frac{evr}{2} = \frac{(1.60 \times 10^{-19})(2.19 \times 10^{6})(5.29 \times 10^{-11})}{2} = 9.27 \times 10^{-24}\,\mathrm{A\,m^2}

This value, the Bohr magneton, is the natural unit of atomic magnetism. Lined up in their billions in iron, such tiny moments make a permanent magnet.

The direct-current motor

A single coil in a field would turn only until its moment lined up with the field, overshoot, and be pulled back: it would rock, not spin. To keep it turning, the current in the coil must be reversed every half turn, just as the coil passes the position where the torque is zero. A DC motor does this with a split-ring commutator: a ring cut into two halves, each joined to one end of the coil and turning with it, and two fixed carbon brushes that press on it and carry current from the supply.

In the position drawn, side AB is pushed into the page and side DC out of it. After half a turn the two sides have swapped places, but the commutator halves have also swapped brushes, so the current in the coil has reversed. The side now on the left again carries current upwards and is again pushed into the page: the torque always turns the coil the same way.

A one-coil motor gives a torque that varies from zero to $N I A B$ twice per revolution. Practical motors use:

many coils at different angles, each with its own pair of commutator segments, so the torque is large and nearly steady;
a soft-iron core (armature) inside the coils, and curved pole pieces, to make the field strong and radial, so that the plane of each active coil stays parallel to the field and $\sin\theta = 1$ ;
electromagnets (field windings) instead of permanent magnets in large motors.

A motor's speed and torque rise with the current and the field, and the direction of rotation is reversed by reversing either the current or the field. As the coil spins, it also acts as a generator and produces a back e.m.f. that opposes the supply; you will study this in Senior 6.

A small brushed DC motor taken apart. Left: the end cap holding the two brushes. Centre: the rotor (armature), with its copper coils wound on an iron core and the commutator on the shaft. Right: the steel case, lined with permanent magnets. Photo: Ilia Krivoruk (Iliakriv), derivative work by Andy Dingley, CC BY-SA 3.0, via Wikimedia Commons

Worked example 3: output and efficiency of a pump motor

A 12 V DC motor driving a small water pump on a farm takes 6.5 A and turns at 1500 revolutions per minute against a torque of 0.40 N m. Find (a) its mechanical output power, (b) its efficiency.

Solution

(a) $\omega = \frac{2\pi \times 1500}{60} = 157\,\mathrm{rad\,s^{-1}}$ ; $P_{\text{out}} = \tau\omega = 0.40 \times 157 = 63$ W.

(b) $P_{\text{in}} = VI = 12 \times 6.5 = 78$ W; efficiency $= \frac{63}{78} = 0.81$ , or 81%. The rest is lost as heat in the coils ( $I^{2} R$ ), friction at the brushes and bearings, and eddy currents in the iron.

Warning

Watch the angle. In $\tau = NIAB\sin\theta$ , $\theta$ is the angle between the normal to the coil and the field. Many books instead use the angle $\alpha$ between the plane of the coil and the field, giving $\tau = NIAB\cos\alpha$ . Both are right; mixing them is wrong. Check: the torque is greatest when the plane of the coil is parallel to the field lines.

Tip

The coil wants to face the field. A coil turns until its north face (the direction of $\vec{\mu}$ ) points along $\vec{B}$ , like a compass needle. The commutator's job is to keep it from ever arriving.

How does an Electric Motor work? (DC Motor), Jared Owen. Watch the commutator and brushes reverse the current every half turn, so the coil keeps turning in one direction.

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Practice

Show that the net force on a rectangular current loop in a uniform magnetic field is zero, but the torque need not be.
A circular coil of 50 turns and radius 2.0 cm carries 0.80 A. Find its magnetic moment.
That coil is placed in a 0.30 T field. Find the torque when the normal to the coil makes (a) 90°, (b) 45°, (c) 0° with the field.
State the position of a current loop in a uniform field in which (a) the torque is greatest, (b) it is in stable equilibrium.
A square coil of side 5.0 cm and 200 turns experiences a maximum torque of 0.030 N m in a 0.15 T field. Find the current.
Explain the purpose of the split-ring commutator in a DC motor. What would happen without it?
A motor gives a torque of 0.25 N m at 2400 revolutions per minute. Find its output power.
An engineer wants a motor to give more torque without increasing the current. Evaluate three options: more turns on the coil, a larger coil area, and stronger magnets.

Show answersHide answers

The forces on opposite sides are equal in size and opposite in direction (same current, opposite directions in the same field), so they add to zero. But the two forces on the long sides act along different lines, forming a couple whose torque is $NIAB\sin\theta$ .
$\mu = NI\pi r^2 = (50)(0.80)\pi(0.020)^2 = 0.050\,\mathrm{A\,m^2}$ .
(a) $0.050 \times 0.30 = 0.015$ N m; (b) $0.015\sin 45^\circ = 0.011$ N m; (c) 0.
(a) With its plane parallel to the field (normal perpendicular to $B$ ); (b) with its plane perpendicular to the field and $\vec{\mu}$ pointing along $\vec{B}$ .
$I = \frac{\tau}{NAB} = \frac{0.030}{(200)(2.5 \times 10^{-3})(0.15)} = 0.40$ A.
It reverses the current in the coil every half turn, as the coil passes the zero-torque position, so the torque always acts in the same rotational direction. Without it, the coil would turn until its moment lined up with the field, overshoot and swing back, oscillating and stopping instead of rotating.
$\omega = \frac{2\pi \times 2400}{60} = 251\,\mathrm{rad\,s^{-1}}$ ; $P = 0.25 \times 251 = 63$ W.
All three raise $\tau = NIAB$ . More turns: more torque, but more wire, so more resistance and heating at the same current, and a heavier rotor. Larger area: more torque but a bigger, heavier motor with more inertia. Stronger magnets (e.g. neodymium): more torque with no extra heating or size, but higher cost. Stronger magnets are usually the best choice where cost allows; otherwise more turns.

Key points

A current loop in a uniform field feels zero net force but a torque $\tau = NIAB\sin\theta$ .
Magnetic moment $\mu = NIA$ (A m²), along the normal by the right-hand rule; $\vec{\tau} = \vec{\mu} \times \vec{B}$ ; the coil turns to align $\vec{\mu}$ with $\vec{B}$ .
DC motor: a split-ring commutator reverses the coil current every half turn so the torque keeps one direction; many coils, an iron core and radial fields give a steady torque.
Mechanical power $P = \tau\omega$ .

1From a push to a spin

2The force and torque on a rectangular coil

3The magnetic moment

4The direct-current motor

From a push to a spin

The force and torque on a rectangular coil

The magnetic moment

The direct-current motor