investigate and analyse theoretically and practically the force on a current-carrying conductor in a magnetic field, $F = n B I L$, and apply this to the operation of a simple DC motor including the role of the split-ring commutator

What this dot point is asking

VCAA wants you to apply $F = n B I L$ to a current-carrying conductor in a magnetic field, find the direction with the right-hand slap rule, and use these ideas to explain how a simple DC motor produces continuous rotation, including the role of the split-ring commutator.

The answer

Force on a straight current-carrying conductor

A straight wire of length $L$ carrying current $I$, with $n$ turns or wires, in a uniform field $B$ perpendicular to the current, feels a force:

(For a single wire, $n = 1$.) If the current is at angle $\theta$ to the field, replace $B$ with $B \sin\theta$. If $I$ is parallel to $B$, the force is zero.

The direction is given by the right-hand slap rule:

• Fingers point in the direction of conventional current $I$.
• Curl them toward the field $B$.
• The thumb points in the direction of the force $F$.

The force is perpendicular to both $I$ and $B$.

Torque on a current loop

Consider a rectangular coil of $n$ turns, width $w$ and length $L$, carrying current $I$ in a uniform field $B$ parallel to the plane of the coil. The forces on the two long sides are equal in magnitude ($F = n B I L$) but opposite in direction (one up, one down). These forces form a couple that rotates the coil about its axis.

The torque is $\tau = n B I L w$ when the coil is parallel to the field, falling to zero when the coil is perpendicular to the field (in the vertical position). The torque varies as $\cos\alpha$ where $\alpha$ is the angle between the coil plane and the field.

The simple DC motor

A simple DC motor consists of:

1. A rectangular coil of $n$ turns mounted on an axle.
2. A pair of permanent magnets producing a roughly uniform field across the coil.
3. A split-ring commutator attached to the axle, with two carbon brushes that touch the commutator and connect it to the external DC supply.

When current flows through the coil:

• The force on one side of the coil is up; the force on the other side is down (right-hand slap rule).
• These forces produce a torque that rotates the coil.
• As the coil passes the vertical position, the two halves of the split-ring commutator swap brush contacts. This reverses the current direction in the coil.
• After the reversal, the force on each side again pushes the coil in the same rotational direction.

The coil continues to rotate continuously in one direction. Without the commutator, the torque would reverse every half turn and the coil would oscillate rather than rotate.

Why a real motor uses many turns and an iron core

A real motor has hundreds of turns ($n$ large) to multiply the torque without needing huge currents. A laminated iron core inside the coil concentrates the field $B$, increasing torque further. Multiple coils at different angles smooth out the torque so the rotation is uniform rather than pulsing.

Examples in context

Example 1. Melbourne tram traction-motor DC armature. A Z-class Melbourne tram uses series-wound DC traction motors. Each armature coil carries $400$ A through a $0.8$ T field, with $L = 0.3$ m and $N = 60$ turns per coil. Force on one coil side is $F = NBIL = 60 \times 0.8 \times 400 \times 0.3 = 5760$ N. With armature radius $0.2$ m, torque on the coil is $\tau = 2 \times F \times r = 2 \times 5760 \times 0.2 = 2304$ N m per coil. With multiple coils contributing during rotation and the split-ring commutator switching current direction every half-turn, the motor delivers continuous unidirectional torque.

Example 2. Australian Synchrotron undulator coil current. A planar undulator at the Australian Synchrotron uses permanent magnets to create alternating fields, but its supplementary trim coils carry DC currents to fine-tune electron-beam steering. A $50$ A trim current in a $0.2$ m long coil placed in a $0.3$ T residual field experiences force $F = BIL = 0.3 \times 50 \times 0.2 = 3.0$ N per turn. With $N = 20$ turns in the coil, total force is $60$ N, used to steer the beam laterally. Reversing the current reverses the direction of $\vec{F} = I\vec{L} \times \vec{B}$, allowing bidirectional fine-tuning.

Try this

Q1. State the force on a current-carrying conductor in a magnetic field and explain the role of the split-ring commutator in a DC motor. [3 marks]

• Cue. $F = nBIL$ perpendicular to both $\vec{I}$ and $\vec{B}$. Commutator reverses current direction every half-turn so the torque remains unidirectional.

Q2. A coil of $50$ turns and length $0.10$ m carries $2.0$ A in a $0.5$ T field. Calculate (a) the force on one side, and (b) the torque if the coil width is $0.08$ m. [4 marks]

• Cue. (a) $F = NBIL = 50 \times 0.5 \times 2 \times 0.1 = 5.0$ N. (b) $\tau = F \times w = 5.0 \times 0.08 = 0.40$ N m.

Q3. Refer to the Melbourne tram traction motor. (a) Calculate the force on a coil side of $L = 0.3$ m at $I = 400$ A, $B = 0.8$ T, $N = 60$. (b) Determine the torque if armature radius is $0.2$ m. (c) Explain how the commutator maintains unidirectional rotation. [2+2+3 marks]

• Cue. (a) $5760$ N. (b) $2304$ N m. (c) Commutator reverses current direction as the coil passes through the field-perpendicular position, keeping torque in one direction.