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What force does a magnetic field exert on a current and on a moving charge?

Magnetic flux density, the force on a current-carrying conductor, the force on a moving charge, and Fleming's left-hand rule.

A focused answer to WJEC A-Level Physics Unit 4 magnetic fields, covering magnetic flux density, the force on a current-carrying conductor (the motor effect), the force on a moving charge, and using Fleming's left-hand rule.

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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

WJEC wants you to define magnetic flux density, use the force on a current-carrying conductor and on a moving charge, and apply Fleming's left-hand rule to find the direction of the force. The motor effect and the force on a moving charge underpin motors, loudspeakers and particle accelerators, and the examiners reliably set both a calculation and a direction question.

The answer

Magnetic flux density

The motor effect

A current-carrying conductor in a magnetic field feels a force, the motor effect, where θ\theta is the angle between the current and the field. The force is maximum when they are perpendicular (θ=90∘\theta = 90^\circ) and zero when parallel (the conductor then lies along the field and cuts no field lines).

Force on a moving charge

The two force expressions are consistent: a current is simply moving charge, and BILBIL summed over all the carriers gives BqvBqv per charge.

Fleming's left-hand rule

Use the left-hand rule for the force on a current (or the conventional current direction of a positive charge): the thumb gives the force (motion), the first finger the field, and the second finger the current.

Examples in context

Example 1. A loudspeaker
A coil attached to the speaker cone sits in the field of a permanent magnet. An audio current through the coil produces a motor-effect force F=BILF = BIL that pushes the cone in and out in step with the signal, recreating the sound. Reversing the current direction (the negative half of the waveform) reverses the force, moving the cone the other way.
Example 2. A mass spectrometer
Charged ions are fired into a magnetic field, where the force BqvBqv bends them into circular arcs of radius r=mv/Bqr = mv/Bq. Heavier ions follow a larger radius, so ions of different mass land at different points on a detector. This lets chemists separate isotopes and identify molecules, all from the circular motion of a charge in a field.
Example 3. The cyclotron
A cyclotron accelerates protons by sending them in widening semicircular paths between two D-shaped electrodes, with a magnetic field bending each pass. Because the radius r=mv/Bqr = mv/Bq grows as the speed increases, the protons spiral outward, gaining energy on each crossing of the gap. The same BqvBqv force that steers a single charge is harnessed here to build a compact particle accelerator used to make medical isotopes.

Try this

Q1. A 0.20 m0.20\,\text{m} wire carries 3.0 A3.0\,\text{A} perpendicular to a 0.50 T0.50\,\text{T} field. Find the force on it. [2 marks]

  • Cue. F=BIL=0.50×3.0×0.20=0.30 NF = BIL = 0.50\times3.0\times0.20 = 0.30\,\text{N}.

Q2. Explain why the magnetic force on a moving charge does no work on it. [2 marks]

  • Cue. The force is always perpendicular to the velocity, so it changes the direction but not the speed, and does no work.

Exam-style practice questions

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

WJEC 20195 marksAn electron travelling at 3.0×107 m s−13.0 \times 10^{7}\,\text{m s}^{-1} enters a uniform magnetic field of 0.012 T0.012\,\text{T} at right angles. Calculate the force on the electron and the radius of its circular path. Take the electron mass as 9.11×10−31 kg9.11 \times 10^{-31}\,\text{kg} and charge magnitude 1.6×10−19 C1.6 \times 10^{-19}\,\text{C}.
Show worked answer →

Force on the moving charge (θ=90∘\theta = 90^{\circ}, so sin⁡θ=1\sin\theta = 1):

F=Bqv=0.012×1.6×10−19×3.0×107=5.8×10−14 NF = Bqv = 0.012 \times 1.6 \times 10^{-19} \times 3.0 \times 10^{7} = 5.8 \times 10^{-14}\,\text{N}.

This magnetic force provides the centripetal force, so Bqv=mv2rBqv = \dfrac{mv^2}{r}, giving r=mvBqr = \dfrac{mv}{Bq}.

r=9.11×10−31×3.0×1070.012×1.6×10−19=2.73×10−231.92×10−21=1.4×10−2 mr = \dfrac{9.11 \times 10^{-31} \times 3.0 \times 10^{7}}{0.012 \times 1.6 \times 10^{-19}} = \dfrac{2.73 \times 10^{-23}}{1.92 \times 10^{-21}} = 1.4 \times 10^{-2}\,\text{m}.

Markers reward F=BqvF = Bqv, recognising it as the centripetal force, and the radius r=mv/Bqr = mv/Bq of about 1.4 cm1.4\,\text{cm}.

WJEC 20213 marksA straight wire carrying a current sits in a magnetic field. Describe how you would use Fleming's left-hand rule to determine the direction of the force, and state what happens to the force if the current is reversed.
Show worked answer →

Hold the thumb and first two fingers of the left hand mutually perpendicular. Point the first finger in the direction of the magnetic field (north to south) and the second finger in the direction of the conventional current (positive to negative). The thumb then points in the direction of the force (the thrust or motion).

If the current is reversed, the second finger points the opposite way, so the thumb (force) also reverses direction. The force is now in the opposite sense. Markers reward correctly assigning field, current and force to the fingers and stating that reversing the current reverses the force.

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