Skip to main content

Electromagnetism and motors - TCE Physics (Tasmania)

Syllabus dot point

“Describe the magnetic force on currents and charges and explain the operation of a DC motor.”

TCEPhysicsUnit 3: Gravity and Electromagnetism9 min read

Quick answer

A current-carrying conductor in a magnetic field feels a force F=BILsin⁡θF = BIL\sin\theta, and a moving charge feels F=qvBsin⁡θF = qvB\sin\theta. In a DC motor these forces produce a torque on a current loop, and a split-ring commutator reverses the current every half turn to keep it spinning.

What this dot point is asking

This dot point is about the motor effect: the force that a magnetic field exerts on moving charge, whether that charge flows as a current in a wire or moves as a free particle.

Force on a current-carrying conductor

When a straight conductor of length LL carrying current II sits in a uniform magnetic field of strength BB, it experiences a force:

F=BILsin⁡θF = BIL\sin\theta

where θ\theta is the angle between the current direction and the field. The force is maximum when the wire is perpendicular to the field (θ=90∘\theta = 90^\circ) and zero when it is parallel to the field. The magnetic field strength BB is measured in tesla.

The direction of the force is perpendicular to both the current and the field. Use the right-hand rule: point your fingers in the direction of conventional current, curl them toward the field BB, and your thumb gives the force. The right-hand slap or palm rule gives the same result.

Direction is always perpendicular

The magnetic force is always at right angles to both the velocity (or current) and the field. This is why a magnetic force on a moving charge does no work and cannot change the charge's speed, only its direction.

Two parallel current-carrying wires illustrate the force directly: each wire sits in the magnetic field of the other, so each feels a force F=BILF = BIL. Currents in the same direction attract, currents in opposite directions repel. This mutual force is in fact how the ampere was historically defined, linking the unit of current to a measurable mechanical force between conductors.

Force on a moving charge

A single charge qq moving at speed vv through a field BB feels:

F=qvBsin⁡θF = qvB\sin\theta

If the charge moves perpendicular to a uniform field, this force is constant in magnitude and always perpendicular to the velocity, so the charge follows a circular path, exactly like centripetal motion. Setting qvB=mv2rqvB = \dfrac{mv^2}{r} gives the radius r=mvqBr = \dfrac{mv}{qB}. The current-on-a-wire force is just the sum of these forces on all the moving charges in the conductor.

Torque on a current loop

A rectangular coil of NN turns and area AA carrying current II in a field BB experiences a torque:

τ=NBIAcos⁡ϕ\tau = N B I A \cos\phi

where ϕ\phi is the angle between the coil plane and the field. Torque is maximum when the coil plane is parallel to the field and zero when the coil plane is perpendicular to it. The two sides of the coil that lie across the field feel forces in opposite directions (because the current runs opposite ways in them), and this couple is what turns the coil. This is the principle of the moving-coil meter and the DC motor.

The DC motor

A direct current motor converts electrical energy into rotational kinetic energy using the motor effect. A current-carrying coil sits in a magnetic field, and the opposing forces on its two sides produce a torque. The key engineering problem is that as the coil passes through the vertical position the torque would reverse and stop the rotation. A split-ring commutator solves this by reversing the current direction in the coil every half turn, so the torque always acts to keep the coil rotating the same way. Practical motors use several coils and a multi-segment commutator to smooth the turning and avoid dead spots.

Worked example: force on a wire

A wire of length 0.25 m0.25\ \text{m} carries a current of 4.0 A4.0\ \text{A} at right angles to a uniform magnetic field of 0.30 T0.30\ \text{T}. Find the magnitude of the force on the wire and state its direction.

Step 1: Write the general formula and identify the angle

The force on a current-carrying conductor depends on how the current direction aligns with the field. The full expression is F=BILsin⁡θF = BIL\sin\theta, where θ\theta is the angle between the current and the field. When the current runs at right angles to the field, θ=90∘\theta = 90^\circ and sin⁡90∘=1\sin 90^\circ = 1, which gives the maximum possible force. This is the most common exam configuration, and recognising it immediately simplifies the formula to:

F=BILF = BIL

Step 2: Substitute the given values and evaluate

With B=0.30 TB = 0.30\ \text{T}, I=4.0 AI = 4.0\ \text{A}, and L=0.25 mL = 0.25\ \text{m}:

F=(0.30)(4.0)(0.25)=0.30 NF = (0.30)(4.0)(0.25) = 0.30\ \text{N}

Step 3: State the direction

The magnetic force on a current-carrying conductor is always perpendicular to both the current and the magnetic field. Use the right-hand rule (or the left-hand rule for conventional current and the direction that produces the force) to identify which way the wire is pushed. The exact direction depends on the orientations given in the diagram, but it is always perpendicular to both the wire and the field.

Final answer: The force on the wire is 0.30 N0.30\ \text{N}, directed perpendicular to both the current and the field.

Forgetting the angle factor

A common error is using F=BILF = BIL when the wire is not perpendicular to the field. If the current makes an angle θ\theta with the field you must include sin⁡θ\sin\theta. When the wire runs parallel to the field, θ=0\theta = 0 and the force is zero, not maximum.

When answering motor questions in the exam, describe the forces on each side of the coil, explain how they create a torque, and clearly state the role of the commutator in maintaining continuous rotation. A labelled diagram strongly supports your explanation, and remember to include sin⁡θ\sin\theta whenever the wire is not perpendicular to the field.

Exam-style questions

Questions in the style of TASC exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice4 marks
A rectangular coil is made of 3030 turns of wire, 5.00 cm5.00\ \text{cm} by 2.00 cm2.00\ \text{cm}, inside a uniform magnetic field of flux density 0.2 T0.2\ \text{T}. A DC current of 5 A5\ \text{A} flows in the coil. Calculate the magnitude of the force on the side AB (the 5.00 cm5.00\ \text{cm} side), and explain why the commutator strips are wrapped only part way around the axle.
Show worked answer →

The force on a current-carrying conductor perpendicular to the field is F=nBILF = nBIL for nn turns:

F=(30)(0.2)(5)(0.05)=1.5 N.F = (30)(0.2)(5)(0.05) = 1.5\ \text{N}.

Each long side experiences 1.5 N1.5\ \text{N}, and the two forces on opposite sides act in opposite directions to produce a torque that turns the coil.

Commutator: the split commutator strips wrap only part way around the axle so that they reverse the current direction in the coil every half turn. As the coil passes the vertical (plane parallel to the field), the gaps break contact, and when contact resumes the current direction has flipped. This keeps the torque acting in the same rotational sense, so the motor spins one way instead of oscillating. Markers want F=nBILF = nBIL and that the commutator reverses current each half turn to maintain rotation.

Practice5 marks
A moving-coil ammeter has a rectangular coil 1.00 cm1.00\ \text{cm} by 1.00 cm1.00\ \text{cm} with 2020 turns rotating in a radial magnetic field. The restoring springs require 2.67×10−3 N2.67 \times 10^{-3}\ \text{N} on each active side per degree of rotation. When 1.00 A1.00\ \text{A} flows, the pointer moves through 9∘9^\circ. Calculate the force on each side of the coil, and hence the magnetic flux density of the field.
Show worked answer →

At the 9∘9^\circ reading the magnetic force on each side balances the spring restoring force.

Spring force on each side =(2.67×10−3 N per degree)(9∘)=2.40×10−2 N= (2.67\times10^{-3}\ \text{N per degree})(9^\circ) = 2.40 \times 10^{-2}\ \text{N}, so the magnetic force on each active side is F=2.40×10−2 NF = 2.40 \times 10^{-2}\ \text{N}.

The magnetic force on each side with 2020 turns is F=nBILF = nBIL:

2.40×10−2=(20)(B)(1.00)(0.01)=0.20 B,B=0.12 T.2.40\times10^{-2} = (20)(B)(1.00)(0.01) = 0.20\,B, \quad B = 0.12\ \text{T}.

The flux density between the magnets is about 0.12 T0.12\ \text{T}. Markers want the spring force at 9∘9^\circ, then B=FnILB = \dfrac{F}{nIL} using the 1.00 cm1.00\ \text{cm} side length.

Practice2 marks
Students pass 10 A10\ \text{A} through a 1.0 cm1.0\ \text{cm} wire in a magnetic field and read the force from a balance as the wire is rotated through an angle θ\theta. State which force equation applies and explain the shape of the force-against-angle data.
Show worked answer →

The force on a straight current-carrying wire in a magnetic field is F=BILsin⁡θF = BIL\sin\theta, where BB is the flux density, II the current, LL the length of wire in the field, and θ\theta the angle between the current and the field.

The force is zero when the wire is parallel to the field (θ=0\theta = 0, sin⁡0=0\sin 0 = 0) and maximum when perpendicular (θ=90∘\theta = 90^\circ, sin⁡90∘=1\sin 90^\circ = 1). Plotting force against sin⁡θ\sin\theta gives a straight line of gradient BILBIL, allowing BB to be found. Markers want F=BILsin⁡θF = BIL\sin\theta and the link between the sin⁡θ\sin\theta factor and the data.

Practise this

Sources & how we know this

ExamExplained