Explain why a charged particle moving with constant velocity through a region of crossed electric and magnetic fields (a velocity selector) travels in a straight line without deflection, and state the condition on the particle's velocity for this to occur.

IB DP Physics Higher Level (2023 syllabus) — D.3 Motion in electromagnetic fields · Explain · 4 marks · View as Markdown

Written & reviewed by James Millett — Biology (Imperial College London), PGCE Science (University of Cambridge).

A velocity selector consists of uniform electric and magnetic fields directed perpendicular to each other and perpendicular to the initial velocity of a beam of charged particles.

Model answer (4 marks)

The electric field exerts a force F_E=qE on the particle in the direction of the field (or opposite for negative charge). The magnetic field exerts a force F_B=qvB perpendicular to both v and B. In a velocity selector the fields are crossed and the particle’s velocity is perpendicular to both, so F_B is opposite to F_E. When the magnitudes are equal, F_E+F_B=0 and the particle experiences no net force, so it continues in a straight line. The condition for this is qE=qvB, giving v=E/B.

Examiner tips

  • Show the forces with correct vectors and signs
  • State that the net force is zero for straight‑line motion
  • Give the velocity condition v=E/B explicitly
  • Use the symbol q for charge

Common mistakes

  • Forgetting that the magnetic force is perpendicular to v and B
  • Using F=qvB instead of F=qvB×sinθ and assuming θ=90° without justification
  • Not stating the velocity condition explicitly

Mark scheme (4 marks)

  1. The electric field exerts a force on the charged particle (F = qE) in the direction of the field (or opposite, depending on charge sign).
  2. The magnetic field exerts a magnetic force on the moving charged particle (F = qvB) in the opposite direction to the electric force.
  3. The particle travels in a straight line because the electric and magnetic forces are equal in magnitude and opposite in direction, giving a zero net force.
  4. The condition for straight-line travel is qE = qvB, which gives v = E/B.

Key terms in this question

velocity selector

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