Motion of charged particles — OCR A-Level Physics
Test yourself on Motion of charged particles with OCR A-Level practice questions.
7 days Premium · Then free forever · No card, no charge
Motion of charged particles explained
A charged particle moving at right angles to a uniform magnetic field experiences a magnetic force.
Read the full explanation
The field is uniform, so B has the same magnitude and direction everywhere in the region. Because the velocity is perpendicular to B, the force magnitude is F = BQv, where B is flux density in tesla, Q is charge in coulomb and v is speed in m s⁻¹. The force is always perpendicular to both v and B, so it acts as a centripetal force and does no work: speed stays constant while direction changes. Example: an electron with Q = 1.6 × 10⁻¹⁹ C entering at 2.0 × 10⁶ m s⁻¹ through B = 0.50 T feels F = 0.50 × 1.6 × 10⁻¹⁹ × 2.0 × 10⁶ = 1.6 × 10⁻¹³ N.
(b) charged particles moving in a uniform magnetic field; circular orbits of charged particles in a uniform magnetic field
When a charged particle moves at right angles to a uniform magnetic field, the magnetic force is always perpendicular to its velocity. This force provides the centripetal force needed for circular motion, so the particle follows a circular orbit at constant speed. Equating BQv to mv²/r gives r = mv/(BQ), so the radius depends on mass, speed, flux density and charge. A larger B or Q gives a tighter circle; a faster or heavier particle gives a larger radius. Example: an electron with m = 9.11 × 10⁻³¹ kg, v = 3.0 × 10⁶ m s⁻¹, B = 0.20 T and Q = 1.6 × 10⁻¹⁹ C has r = (9.11 × 10⁻³¹ × 3.0 × 10⁶)/(0.20 × 1.6 × 10⁻¹⁹) ≈ 8.5 × 10⁻⁵ m.
(c) charged particles moving in a region occupied by both electric and magnetic fields; velocity selector.
In a velocity selector, a charged particle moves through crossed electric and magnetic fields. The electric force QE and magnetic force BQv act in opposite directions. When they balance, QE = BQv, so v = E/B. Only particles with this speed travel undeflected; slower or faster particles are deflected one way or the other. The selected speed is independent of charge and mass, which is why the device can pick a chosen velocity from a mixed beam. Example: if E = 4.0 × 10⁴ V m⁻¹ and B = 0.20 T, then v = E/B = 2.0 × 10⁵ m s⁻¹. Particles with this speed pass straight through; others curve towards one plate.
Your focus
- State the condition under which a charged particle experiences a magnetic force in a uniform field.
- Apply F = BQv to calculate the force on a charge moving at right angles to a uniform magnetic field.
- Explain why the magnetic force on a charged particle does no work and leaves its speed unchanged.
Show all 9 objectives
- Explain why a charged particle moving at right angles to a uniform magnetic field follows a circular path.
- Derive and apply r = mv/(BQ) for the radius of the circular orbit.
- Describe how changing B, Q, m or v affects the radius of the orbit.
- Describe the arrangement of crossed electric and magnetic fields in a velocity selector.
- Derive and apply v = E/B for the selected speed.
- Explain why the selected speed is independent of the charge and mass of the particle.
Motion of charged particles exam tips
Marking Points
- States that the magnetic force acts only on a moving charge and is zero when v is parallel to B.
- Uses F = BQv for motion at right angles to a uniform magnetic field, with B in tesla, Q in coulomb and v in m s⁻¹.
- Recognises that the force is perpendicular to both v and B, so it changes direction of motion but not speed.
- Applies the equation to a numerical case, keeping charge magnitude and consistent SI units.
- States that the magnetic force on a charge moving at right angles to a uniform field is perpendicular to velocity and acts as the centripetal force.
- Derives or uses r = mv/(BQ) by equating BQv with mv²/r.
- Explains that speed remains constant in the circular orbit because the magnetic force does no work.
- Applies the radius relationship to compare orbits for different masses, charges, speeds or flux densities.
- States that the electric and magnetic forces on the particle act in opposite directions in a velocity selector.
- Uses the balance condition QE = BQv to derive v = E/B.
- Explains that only particles with speed v = E/B pass through undeflected, independent of charge and mass.
- Applies v = E/B to calculate the selected speed from given field values.
Examiner Tips
- 💡Check the angle between v and B before choosing F = BQv; if the motion is not at right angles, use the perpendicular component of velocity.
- 💡Convert all quantities to SI units, for example mT to T and cm to m, before substituting into the equation.
- 💡Sketch the v and B directions and use the sign of the charge to decide the force direction, rather than relying on memory alone.
- 💡Start from the force balance BQv = mv²/r and cancel one v to obtain r = mv/(BQ) quickly and reliably.
- 💡Check that the particle enters at right angles to the field; otherwise the path is helical rather than circular.
- 💡Use consistent SI units and keep mass in kg, speed in m s⁻¹, flux density in T and charge in C.
- 💡Write the force balance QE = BQv first, then cancel Q to obtain v = E/B.
- 💡Check the directions of E and B so that the electric and magnetic forces oppose each other.
- 💡Keep E in V m⁻¹ and B in T so that v comes out in m s⁻¹.
Common Mistakes
- Using the full velocity vector in F = BQv when the motion is not at right angles; the correct approach is to use only the component of velocity perpendicular to B, since the parallel component produces no force.
- Treating the magnetic force as doing work and changing the particle's kinetic energy; the correction is that the force is always perpendicular to velocity, so speed and kinetic energy remain constant.
- Substituting a negative charge value and expecting a negative force magnitude; the correction is that Q gives the magnitude of force while direction is found from the sign of the charge and the field geometry.
- Assuming the particle speeds up as it curves; the correction is that the magnetic force is perpendicular to velocity, so it changes direction only and speed stays constant.
- Using r = mv/(BQ) with the wrong charge sign or ignoring charge magnitude; the correction is to use the magnitude of Q for the radius and use the sign only to determine the sense of rotation.
- Confusing the radius expression with the period expression; the correction is that r = mv/(BQ) gives the radius, while T = 2πm/(BQ) gives the time for one orbit.
- Thinking the selected speed depends on the particle's charge or mass; the correction is that Q cancels in QE = BQv, so v = E/B is the same for all charged particles.
- Adding the electric and magnetic forces instead of recognising they oppose; the correction is that the forces must be equal and opposite for the particle to travel in a straight line.
- Using the magnetic force as BQv when the particle is not moving at right angles to B; the correction is that the velocity selector geometry requires v perpendicular to B.