Know that a current-carrying coil in a magnetic field may experience a turning effect and that the turning effect is increased by increasing: (a) the number of turns on the coil (b) the current (c) the strength of the magnetic field

4.5.5 The d.c. motor

Learning objective (Cambridge IGCSE 0625)

• Explain that a current‑carrying coil placed in a magnetic field experiences a turning effect (torque) and that the torque increases when any of the following are increased:

  1. the number of turns on the coil,
  2. the current through the coil,
  3. the strength of the magnetic field.

• Describe the basic operation of a d.c. motor, including the role of the split‑ring commutator and the brushes.

1. Principle of the turning effect

1.1 Force on a straight conductor

When a straight conductor of length l carrying a current I is placed in a uniform magnetic field B, the magnetic force is

\[

\mathbf{F}=I\,\mathbf{l}\times\mathbf{B}

\]

Direction of F is given by Fleming’s left‑hand rule (thumb = force, first finger = field, second finger = current).

1.2 Torque on a rectangular coil

Consider a rectangular coil of N turns, side lengths l (horizontal) and w (vertical). The area of one turn is

\[

A = l\times w

\]

Opposite sides of the coil experience forces of equal magnitude but opposite direction. These two forces form a couple whose moment about the axis of rotation is the torque.

For a single turn the torque is

\[

\tau_{\text{single}} = B I A \sin\theta

\]

where θ is the angle between the magnetic field direction and the normal to the plane of the coil.

With N identical turns the torques add linearly, giving

\[

\boxed{\tau = N\,B\,I\,A\,\sin\theta}

\]

1.3 Why torque depends on θ

• When the plane of the coil is parallel to the field (θ=0° or 180°) the normal is perpendicular to the field, so sin θ = 0 and the torque is zero – the coil is in a neutral position.
• Maximum torque occurs when the plane of the coil is perpendicular to the field (θ=90°), because sin θ = 1. This is the position used in the “maximum‑torque” formula.

1.4 Assumptions in the formula

• The magnetic field is assumed to be uniform over the whole area of the coil. In a real motor the field is only approximately uniform, especially near the pole edges.
• The coil is rigid so that the forces act at a fixed radius from the axis of rotation.

2. Factors that increase the turning effect (syllabus link)

All three factors appear explicitly in the torque equation \(\tau = NBI A \sin\theta\); increasing any one while keeping the others constant increases the torque in the same ratio, exactly as required by the syllabus.

3. Simple d.c. motor – construction and operation

• Armature (coil): a rectangular coil of fine insulated copper wire mounted on a shaft so it can rotate.
• Magnetic field: produced by two permanent‑magnet poles (or electromagnets) placed on opposite sides of the armature, giving an approximately uniform field across the coil.
• Split‑ring commutator: two insulated copper segments fixed to the ends of the coil. As the coil rotates, the stationary carbon brushes slide over the commutator, swapping the electrical connections every half‑turn.
• Brushes: fixed contacts that maintain electrical contact with the rotating commutator while allowing free rotation.

3.1 Why the commutator is essential

If the current direction were not reversed, the torque would change sign when the coil reaches the positions where the plane of the coil is parallel to the field (θ = 0° or 180°). The split‑ring commutator interchanges the two coil leads each half‑turn, so the direction of current in the coil is reversed. Consequently the torque produced by the two opposite sides of the coil always acts in the same rotational sense, giving continuous rotation.

3.2 Torque direction and Fleming’s left‑hand rule

For each side of the coil, apply Fleming’s left‑hand rule to find the direction of the magnetic force. The two forces are equal in magnitude, opposite in direction, and separated by the coil width w. Their vector sum is a couple that produces a torque about the shaft. The net torque is the vector sum of the two individual torques, which is exactly the value given by \(\tau = NBI A \sin\theta\).

4. Back‑EMF and steady‑state speed

• When the armature rotates, each turn cuts magnetic flux, inducing an emf (the back‑EMF) that opposes the applied supply voltage (Lenz’s law).
• Back‑EMF is proportional to the angular speed ω: \(E_{\text{b}} = k\,B\,N\,A\,\omega\) (where *k* is a constant depending on the motor geometry).
• As the speed increases, the net voltage across the armature (V – E_b) falls, reducing the current and therefore the torque. The motor reaches a steady speed when the motor torque equals the external load torque.
• In the IGCSE exam it is sufficient to state: “Back‑EMF limits the current as the motor speeds up, so the motor runs at a constant speed under a given load.”

5. Practical / safety notes (optional but useful)

• Excessive current heats the coil; the insulation may melt if the motor is overloaded.
• Brushes wear over time; they must be kept clean and properly aligned to avoid sparking.
• Never operate a motor in a strong external magnetic field that could disturb the internal field geometry.

6. Key formulae to remember

• Force on a straight conductor: \(\displaystyle \mathbf{F}=I\,\mathbf{l}\times\mathbf{B}\)
• Torque on a coil: \(\displaystyle \tau = N\,B\,I\,A\,\sin\theta\)
• Maximum torque (θ = 90°): \(\displaystyle \tau_{\max}=N\,B\,I\,A\)
• Back‑EMF (qualitative): \(E_{\text{b}}\propto \omega\) – opposes the applied voltage.

7. Worked example (exam style)

Question: A rectangular coil of 20 turns has dimensions 5 cm × 10 cm and carries a current of 2 A. It is placed in a uniform magnetic field of 0.30 T. Calculate the maximum torque produced by the coil.

Solution:

1. Area of one turn: \(A = 0.05\ \text{m} \times 0.10\ \text{m}=5.0\times10^{-3}\ \text{m}^2\).
2. Maximum torque occurs when \(\sin\theta =1\) (θ = 90°).
3. Apply \(\tau_{\max}=N\,B\,I\,A\):

   \[

   \tau_{\max}=20 \times 0.30\ \text{T} \times 2\ \text{A} \times 5.0\times10^{-3}\ \text{m}^2

   =0.060\ \text{N·m}.

   \]

Hence the coil can produce a maximum torque of 0.06 N·m.

8. Summary checklist

• Torque on a coil is \(\tau = NBI A \sin\theta\); it is zero when the coil is parallel to the field and maximum when the coil is perpendicular.
• Increasing N, I or B increases the torque proportionally – exactly what the syllabus states.
• The split‑ring commutator reverses the current each half‑turn, ensuring the torque always acts in the same rotational direction.
• Back‑EMF, generated by the rotating coil, opposes the supply voltage and limits the current, giving a steady operating speed.
• Use Fleming’s left‑hand rule on each side of the coil to find the direction of the forces; the resulting couple gives the torque.
• Remember practical limits: coil heating, brush wear, and the fact that the magnetic field is only approximately uniform.