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How do magnetic fields exert forces on currents and moving charges?

The magnetic field around a current, the force on a current-carrying conductor and on a moving charge, magnetic flux, and Millikan's experiment.

An SQA Advanced Higher Physics answer on magnetic fields, covering the field around a current-carrying conductor, the force on a current-carrying conductor and on a moving charge, magnetic flux, and Millikan's oil-drop experiment determining the charge on the electron.

Generated by Claude Opus 4.815 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this key area is asking
  2. The field around a current and the force on a conductor
  3. The force on a moving charge
  4. Magnetic flux
  5. Millikan's experiment
  6. Examples in context
  7. Try this

What this key area is asking

The SQA wants you to describe the magnetic field around a current, calculate the force on a current-carrying conductor (F=BILF = BIL) and on a moving charge (F=qvBF = qvB), understand magnetic flux, and describe Millikan's experiment, which showed charge is quantised and measured the charge on the electron.

The field around a current and the force on a conductor

For a straight conductor of length LL carrying current II at right angles to a uniform field of magnetic flux density BB:

F=BIL.F = BIL.

The force is perpendicular to both the current and the field, with its direction given by the right-hand (or left-hand) rule. This is the principle of the electric motor: a current loop in a field experiences forces that turn it. If the conductor makes an angle θ\theta with the field, only the perpendicular component contributes, F=BILsin⁡θF = BIL\sin\theta.

The force on a moving charge

This is the same relationship met in "particles from space": the magnetic force changes direction but not speed, so a charge moving across a uniform field follows a circle of radius r=mvqBr = \frac{mv}{qB}. It is the basis of the cyclotron, the mass spectrometer, and the deflection of cosmic rays.

Magnetic flux

Flux is the quantity that matters for electromagnetic induction: a changing flux through a circuit induces an emf (Faraday's law), which is taken up in the inductors key area. For now, flux links the field strength to the area and prepares the ground for induction.

Millikan's experiment

By suspending a drop so that the upward electric force exactly balanced gravity, Millikan found its charge from Q=mgEQ = \frac{mg}{E}. Repeating for many drops, every charge turned out to be a multiple of the same basic unit, the charge on the electron. This was decisive evidence that electric charge is not continuous but comes in discrete quanta.

Examples in context

The electric motor turns because a current loop in a magnetic field feels forces F=BILF = BIL on its sides. The mass spectrometer uses F=qvBF = qvB to sort ions by their charge-to-mass ratio. Loudspeakers drive a cone with the force on a current-carrying coil in a permanent field. Millikan's experiment remains a landmark, the first direct measurement of the quantum of charge, underpinning all of particle physics.

Try this

Q1. Write the relationship for the force on a straight conductor of length LL carrying current II at right angles to a field BB. [1 mark]

  • Cue. F=BILF = BIL.

Q2. State the direction of the magnetic force relative to the velocity of a moving charge. [1 mark]

  • Cue. Perpendicular to the velocity (so it does no work).

Q3. State what Millikan's experiment showed about electric charge. [1 mark]

  • Cue. That it is quantised, in whole multiples of e=1.6×10−19 Ce = 1.6 \times 10^{-19}\ \text{C}.

Exam-style practice questions

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

SQA AH style4 marksA straight wire of length 0.25 m0.25\ \text{m} carries a current of 4.0 A4.0\ \text{A} at right angles to a magnetic field of flux density 0.30 T0.30\ \text{T}. Calculate the force on the wire.
Show worked answer →

The force on a current-carrying conductor at right angles to the field is F=BILF = BIL.

Substitute: F=0.30×4.0×0.25F = 0.30 \times 4.0 \times 0.25.

Evaluate: F=0.30 NF = 0.30\ \text{N}.

The force is perpendicular to both the current and the field, given by the right-hand (or left-hand) rule.

Markers reward the correct relationship, the value with unit, and that the force is perpendicular to both current and field.

SQA AH style5 marksIn Millikan's experiment, a charged oil drop of weight 3.2×10−14 N3.2 \times 10^{-14}\ \text{N} is held stationary between plates 5.0 mm5.0\ \text{mm} apart with a potential difference of 1000 V1000\ \text{V}. Calculate the charge on the drop and the number of electron charges it carries. Take e=1.6×10−19 Ce = 1.6 \times 10^{-19}\ \text{C}.
Show worked answer →

The drop is in equilibrium, so the electric force balances the weight: QE=mgQE = mg, with E=VdE = \frac{V}{d}.

Field strength: E=10005.0×10−3=2.0×105 V m−1E = \frac{1000}{5.0 \times 10^{-3}} = 2.0 \times 10^{5}\ \text{V m}^{-1}.

Charge: Q=mgE=3.2×10−142.0×105=1.6×10−19 CQ = \frac{mg}{E} = \frac{3.2 \times 10^{-14}}{2.0 \times 10^{5}} = 1.6 \times 10^{-19}\ \text{C}.

Number of electron charges: 1.6×10−191.6×10−19=1\frac{1.6 \times 10^{-19}}{1.6 \times 10^{-19}} = 1.

Markers reward balancing the electric force against the weight, finding the field from V/dV/d, the charge, and that it is a whole number multiple of the electron charge.

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