D.3.4—Magnetic force on charge

Syllabus
First assessment 2025
Objective
Level
HL

Calculate Magnetic Force on a Charge

Use the magnitude equation

The angle θ\theta is measured between the particle velocity and the magnetic field. Use charge magnitude for the force magnitude; determine direction separately with the right-hand rule and reverse it for a negative charge.

F=|q|vB\sin\theta

Worked example — oblique proton motion

For q=1.60×1019C|q|=1.60\times10^{-19}\,\mathrm{C}, v=3.4×105ms1v=3.4\times10^5\,\mathrm{m\,s^{-1}}, B=5.3×103TB=5.3\times10^{-3}\,\mathrm{T} and θ=32\theta=32^\circ, F=qvBsinθ=1.5×1016NF=|q|vB\sin\theta=1.5\times10^{-16}\,\mathrm{N}. Only the velocity component perpendicular to the field contributes.

Find the direction

The magnetic force is perpendicular to both velocity and field. Use the right-hand rule for a positive charge; reverse the result for a negative charge. This force bends the path but does no work.

Connect to circular motion

For perpendicular motion, set F=qvBF=|q|vB equal to mv2/rmv^2/r to obtain r=mv/(qB)r=mv/(|q|B). Increasing q|q| or BB reduces the radius; increasing mm or vv increases it.

Common trap

Do not use the right-hand rule without reversing for an electron, and do not use θ\theta as the angle between the field and the force. It is the angle between velocity and field.

D.3.4 Exam Analysis

Assessment in practice

1–2 marks
How it is assessed

Questions calculate a force or radius, or determine the force direction on an electron.

Command terms

Show that / State

What earns marks

Use F=|q|vB sinθ, identify θ between velocity and field, and reverse the positive-charge right-hand-rule direction for a negative charge.

Watch for

Using the wrong angle, failing to reverse for negative charge, or confusing magnetic force with a force component parallel to velocity.

Representative question

Question 1

[Maximum number: 2]

There is a potential difference of 2.4 mV between the ends of the copper rod. The distance between the conducting rails is 0.16 m . Determine the magnetic force on a free electron in the copper rod.

Retrieve the D.3 Motion in Electromagnetic Fields Model

D.3 is secure when you can keep electric and magnetic force rules separate and then combine them deliberately.

  • Electric fields give F=qE and constant acceleration in a uniform field
  • Magnetic fields bend moving charges without changing kinetic energy
  • Crossed fields can cancel at v=E/B
  • Moving-charge magnetic force is F=|q|vB sinθ
  • Conductor force is F=BIL sinθ
  • Parallel currents attract in the same direction and repel in opposite directions