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TASPhysicsSyllabus dot point

How does a current in a magnetic field produce motion?

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

The force on a current-carrying conductor and a moving charge in a magnetic field, the right-hand rule, the torque on a current loop, and how a split-ring commutator drives a direct current motor.

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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=BILsinθF = 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.

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=qvBsinθF = 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=NBIAcosϕ\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.

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 sinθ\sin\theta whenever the wire is not perpendicular to the field.

Exam-style practice questions

Practice questions written in the style of TASC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

TCE 20244 marksA 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.

TCE 20225 marksA 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 2.67×103 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 99^\circ. Calculate the force on each side of the coil, and hence the magnetic flux density of the field.
Show worked answer →

At the 99^\circ reading the magnetic force on each side balances the spring restoring force.

Spring force on each side =(2.67×103 N per degree)(9)=2.40×102 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=2.40×102 NF = 2.40 \times 10^{-2}\ \text{N}.

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

2.40×102=(20)(B)(1.00)(0.01)=0.20B,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 99^\circ, then B=FnILB = \dfrac{F}{nIL} using the 1.00 cm1.00\ \text{cm} side length.

TCE 20232 marksStudents 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=BILsinθF = 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, sin0=0\sin 0 = 0) and maximum when perpendicular (θ=90\theta = 90^\circ, sin90=1\sin 90^\circ = 1). Plotting force against sinθ\sin\theta gives a straight line of gradient BILBIL, allowing BB to be found. Markers want F=BILsinθF = BIL\sin\theta and the link between the sinθ\sin\theta factor and the data.

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