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How do magnetic fields exert forces, and how is electricity generated by induction?

The force on a current-carrying conductor and on a moving charge, magnetic flux density, and electromagnetic induction with Faraday's and Lenz's laws.

A CCEA A-Level Physics answer on the force on a current-carrying conductor and a moving charge, magnetic flux density and Fleming's left-hand rule, magnetic flux and flux linkage, and electromagnetic induction governed by Faraday's and Lenz's laws.

Generated by Claude Opus 4.811 min answer

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
  4. Try this

What this dot point is asking

CCEA wants you to calculate the force on a current-carrying conductor and on a moving charge, define magnetic flux density and use Fleming's left-hand rule, define magnetic flux and flux linkage, and apply Faraday's and Lenz's laws of electromagnetic induction. A moving-charge or induction calculation appears in every paper.

The answer

Forces in a magnetic field

A charged particle moving across a uniform field follows a circular path, because the magnetic force is always perpendicular to its velocity and so does no work; the radius is r=mvBqr = \frac{mv}{Bq}.

Magnetic flux and flux linkage

Electromagnetic induction

These laws explain generators, transformers and the dynamo effect: moving a magnet into a coil induces a current that opposes the magnet's motion, so work must be done against this force, and that mechanical work is what becomes electrical energy.

Worked example: force on a power line

Examples in context

Example 1. A bicycle dynamo. As the wheel spins, a magnet rotates near a coil, continuously changing the flux linkage and inducing an alternating e.m.f. by Faraday's law. By Lenz's law the induced current opposes the rotation, so you feel a slight extra resistance when the lamp is on, which is the mechanical work being converted into the electrical energy that lights it.

Example 2. A mass spectrometer. Ions of charge qq and speed vv enter a uniform field and follow circular arcs of radius r=mvBqr = \frac{mv}{Bq}. Heavier ions curve less, so measuring the radius separates isotopes by mass. The same r=mv/Bqr = mv/Bq relation governs the beam-steering magnets in particle accelerators.

Try this

Q1. A wire of length 0.20 m0.20\ \text{m} carries a current of 3.0 A3.0\ \text{A} at right angles to a field of flux density 0.50 T0.50\ \text{T}. Find the force on it. [2 marks]

  • Cue. F=BIL=0.50×3.0×0.20=0.30 NF = BIL = 0.50 \times 3.0 \times 0.20 = 0.30\ \text{N}.

Q2. State Lenz's law and the conservation principle it follows from. [2 marks]

  • Cue. The induced current opposes the change that causes it; it follows from conservation of energy.

Q3. An electron moves at 4.0×106 m s−14.0 \times 10^{6}\ \text{m s}^{-1} perpendicular to a field of 0.25 T0.25\ \text{T}. Find the force on it. [2 marks]

  • Cue. F=Bqv=0.25×1.6×10−19×4.0×106=1.6×10−13 NF = Bqv = 0.25 \times 1.6 \times 10^{-19} \times 4.0 \times 10^{6} = 1.6 \times 10^{-13}\ \text{N}.

Exam-style practice questions

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

CCEA 20196 marksA proton travelling at 2.0 times 10 to the 6 m per second enters a uniform magnetic field of flux density 0.35 T at right angles to its velocity. Show that it moves in a circular path and calculate the radius of that path. The mass of a proton is 1.67 times 10 to the minus 27 kg and its charge is 1.6 times 10 to the minus 19 C.
Show worked answer →

The magnetic force F=BqvF = Bqv is always perpendicular to the velocity, so it does no work and cannot change the speed. A constant-magnitude force perpendicular to the velocity provides a centripetal force, so the path is a circle.

Equating the magnetic force to the centripetal force:

Bqv=mv2rBqv = \frac{mv^2}{r}, so r=mvBqr = \frac{mv}{Bq}.

r=1.67×10−27×2.0×1060.35×1.6×10−19=3.34×10−215.6×10−20=0.060r = \frac{1.67 \times 10^{-27} \times 2.0 \times 10^{6}}{0.35 \times 1.6 \times 10^{-19}} = \frac{3.34 \times 10^{-21}}{5.6 \times 10^{-20}} = 0.060 m.

Markers reward the perpendicular-force argument for circular motion, equating BqvBqv to mv2mv^2 over rr, and a radius of about 6 cm.

CCEA 20215 marksA coil of 200 turns and cross-sectional area 2.5 times 10 to the minus 3 m squared is placed with its plane perpendicular to a magnetic field. The field is reduced uniformly from 0.40 T to zero in 0.020 s. State Faraday's law and Lenz's law, and calculate the average e.m.f. induced in the coil.
Show worked answer →

Faraday's law: the magnitude of the induced e.m.f. equals the rate of change of flux linkage.

Lenz's law: the induced current (and e.m.f.) acts in the direction that opposes the change producing it, which is a consequence of conservation of energy.

The initial flux linkage is NΦ=NBA=200×0.40×2.5×10−3=0.20N\Phi = NBA = 200 \times 0.40 \times 2.5 \times 10^{-3} = 0.20 Wb-turns; the final value is zero.

The average induced e.m.f. is

ε=Δ(NΦ)Δt=0.200.020=10\varepsilon = \frac{\Delta(N\Phi)}{\Delta t} = \frac{0.20}{0.020} = 10 V.

Markers reward both law statements, correct flux linkage, and the e.m.f. from the rate of change.

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