5.3.5 DC Motor
By the end of this lesson you can:
- Describe the parts of a DC motor
- Explain how torque is produced and why a commutator is needed
- Calculate the torque on a coil
A DC (Direct Current) motor is a fundamental application of the motor effect that converts electrical energy into continuous mechanical rotational energy.
Operating Principles & Physics Theory
When a rectangular current-carrying coil (the armature) is placed in a uniform magnetic field, the opposite sides of the coil experience equal and opposite magnetic forces according to F = BIl sin(θ). Because these forces act along different lines of action on either side of the pivot axis, they produce a net torque (τ), causing the loop to rotate:
Where:
- τ is the torque in Newton-meters (N·m)
- n is the number of turns in the wire coil
- B is the magnetic field strength in Teslas (T)
- I is the electric current in Amperes (A)
- A is the area of the coil in square meters (m2)
- θ is the angle between the surface vector of the coil and the magnetic field lines
The Role of the Split-Ring Commutator
As the coil rotates past the perpendicular position , the direction of torque would naturally reverse, causing the loop to oscillate back and forth rather than spinning continuously. To overcome this, a split-ring commutator reverses the direction of current flowing through the coil every half-rotation (180°). This ensures that the torque remains unidirectional, keeping the motor spinning continuously in one direction.
The following video breaks down these key structural components, force vectors, and mathematical relationships in detail.
We are now ready to try a sample problem
Below is a sample problem with a video that explain how to solve it. It is suggested you try the problem beforehand, as this actually aids understanding, even if you are unsure if you are correct.
Practice Problems: Torque on a DC Motor Coil
The torque on a coil of n turns is τ = nBIA sinθ, where θ is the angle between the area vector of the loop (at right angles to its plane) and the magnetic field. The torque is greatest (θ = 90°) when the plane of the coil is parallel to the field, and zero (θ = 0°) when the plane is at right angles to the field.
Given: n = 50, B = 0.30 T, I = 2.0 A, A = 0.040 × 0.060 = 2.4 × 10-3 m2
Angle: The plane of the coil is parallel to the field, so the area vector (which is at right angles to the plane) is perpendicular to the field. So θ = 90° and sinθ = 1.
Formula: τ = nBIA sinθ
Calculation: τ = 50 × 0.30 × 2.0 × 2.4 × 10-3 × 1 = 7.2 × 10-2 N m
Given: n = 100, A = 2.5 × 10-3 m2, I = 1.5 A, B = 0.40 T, θ = 30°
Formula: τ = nBIA sinθ
Calculation: τ = 100 × 0.40 × 1.5 × 2.5 × 10-3 × sin 30° = 0.15 × 0.50 = 0.075 N m
Given: n = 200, A = 3.0 × 10-3 m2, I = 0.80 A, B = 0.25 T
(a) θ = 0°: sin 0° = 0, so τ = 0 N m. The area vector is parallel to the field, which means the plane of the coil is at right angles to the field.
(b) θ = 90°: τ = 200 × 0.25 × 0.80 × 3.0 × 10-3 × 1 = 0.12 N m, the maximum torque. The plane of the coil is parallel to the field.
Commutator: The torque falls to zero when θ = 0° and would reverse direction just after, which would turn the coil back. The commutator reverses the current at this point, so the torque keeps acting in the same direction and the coil keeps turning.
Given: n = 80, A = 5.0 × 10-3 m2, B = 0.20 T, τmax = 0.16 N m
Step 1: The torque is greatest when θ = 90°, so sinθ = 1 and τmax = nBIA.
Step 2 (Rearrange): I = τmax / (nBA)
I = 0.16 / (80 × 0.20 × 5.0 × 10-3) = 0.16 / 0.080 = 2.0 A
Given: n = 40, A = 6.0 × 10-3 m2, I = 3.0 A, B = 0.50 T, τ = 0.18 N m
(a) Rearrange: sinθ = τ / (nBIA)
nBIA = 40 × 0.50 × 3.0 × 6.0 × 10-3 = 0.36 N m, so sinθ = 0.18 / 0.36 = 0.50 → θ = 30°
(b) Plane of the coil: The area vector is at 90° to the plane of the coil, so the angle between the plane and the field is 90° − 30° = 60°.
Interactive
This animation allows you to manipulate a DC motor , to see its components.
This animation by Tom Walsh shows how a DC motor works including the role of the split ring communicator