A Level Physics 20.3 Force on a Moving Charge MCQs – 30 Practice Questions with Answers (Cambridge 9702)

Section 20.3 of Cambridge AS & A Level Physics (9702) is an A Level topic on the force that a magnetic field exerts on a moving charge, the circular paths that result, and applications such as the velocity selector, the mass spectrometer and the Hall effect. Work through these 30 original practice questions and open the Show Answer toggle after each one to check your answer and read the full worked explanation.

Formula recap: F = BQv sin θ  |  r = mv / (BQ)  |  T = 2πm / (BQ)  |  velocity selector: v = E / B  |  Hall voltage: VH = BI / (nqt). Constants used: e = 1.6 × 10-19 C, me = 9.11 × 10-31 kg, mp = 1.67 × 10-27 kg. See the full A Level Physics Formula Sheet.

Force on a Moving Charge

Q1. The magnetic force on a charge Q moving at speed v in a magnetic field of flux density B is F = BQv sin θ. In this equation, θ is the angle between

  • (A) the force and the magnetic field
  • (B) the velocity of the charge and the magnetic field
  • (C) the velocity of the charge and the force
  • (D) the charge and the magnetic field
Show Answer

Answer: B – θ is measured between the direction of motion (the velocity) and the direction of the magnetic field. The force itself always acts at right angles to both v and B.

Q2. An electron is at rest inside a strong uniform magnetic field. The magnetic force acting on the electron is

  • (A) equal to Bev
  • (B) the maximum possible value because the field is strong
  • (C) zero
  • (D) equal to the force an electric field would exert on it
Show Answer

Answer: C – F = BQv sin θ, and here v = 0, so F = 0. A magnetic field exerts a force only on a MOVING charge; a stationary charge is unaffected, however strong the field.

Q3. A proton travels exactly parallel to the direction of a uniform magnetic field. The magnetic force on the proton is

  • (A) zero
  • (B) BQv, at right angles to its path
  • (C) BQv, along its path
  • (D) small but not zero
Show Answer

Answer: A – The velocity is parallel to the field, so θ = 0 and sin θ = 0, giving F = 0. The proton continues in a straight line at constant speed.

Q4. For a given charge, speed and field, the magnetic force is greatest when the charge moves

  • (A) parallel to the field
  • (B) at 45° to the field
  • (C) in the opposite direction to the field
  • (D) at right angles (perpendicular) to the field
Show Answer

Answer: D – F = BQv sin θ is maximum when sin θ = 1, i.e. when θ = 90° and the motion is perpendicular to the field. The maximum force is F = BQv.

Q5. An electron moves at 5.0 × 106 m/s at right angles to a uniform magnetic field of flux density 0.20 T. What is the magnitude of the force on the electron?

  • (A) 1.6 × 10-19 N
  • (B) 1.6 × 10-13 N
  • (C) 4.0 × 10-13 N
  • (D) 8.0 × 10-14 N
Show Answer

Answer: B – F = BQv = 0.20 × 1.6 × 10-19 × 5.0 × 106 = 1.6 × 10-13 N (sin 90° = 1).

Q6. A proton moves at 2.0 × 106 m/s at right angles to a uniform magnetic field of flux density 0.50 T. What is the magnitude of the force on the proton?

  • (A) 1.6 × 10-13 N
  • (B) 3.2 × 10-13 N
  • (C) 1.6 × 10-12 N
  • (D) 8.0 × 10-14 N
Show Answer

Answer: A – F = BQv = 0.50 × 1.6 × 10-19 × 2.0 × 106 = 1.6 × 10-13 N. The proton’s much larger mass does not appear in F = BQv.

Q7. An alpha particle (charge +2e = 3.2 × 10-19 C) moves at 1.0 × 106 m/s at an angle of 30° to a magnetic field of flux density 0.40 T. What is the force on the alpha particle?

  • (A) 1.3 × 10-13 N
  • (B) 2.6 × 10-13 N
  • (C) 3.2 × 10-14 N
  • (D) 6.4 × 10-14 N
Show Answer

Answer: D – F = BQv sin θ = 0.40 × 3.2 × 10-19 × 1.0 × 106 × sin 30° = 1.28 × 10-13 × 0.50 = 6.4 × 10-14 N. Option A is the trap value found by forgetting the sin 30° factor.

Q8. Fleming’s left-hand rule is applied using the direction of conventional current, which is the direction of motion of a positive charge. An electron moves through the same field in the same direction as a positive charge would. The force on the electron is

  • (A) in the same direction as the force on the positive charge
  • (B) parallel to the electron’s velocity
  • (C) in the opposite direction to the force on the positive charge
  • (D) in the direction of the magnetic field
Show Answer

Answer: C – For an electron, the conventional current direction is opposite to its velocity. Reversing the “current” in Fleming’s left-hand rule reverses the force, so the electron is pushed the opposite way to a positive charge moving along the same path.

Q9. A charged particle moving in a uniform magnetic field has a magnetic force acting on it, yet its kinetic energy does not change. This is because

  • (A) the force is always at right angles to the velocity, so it does no work on the particle
  • (B) the force is too small to change the energy
  • (C) the field does work but the particle radiates the energy away
  • (D) the force acts along the velocity but is constant
Show Answer

Answer: A – Work done = force × distance moved in the direction of the force. Since the force is always perpendicular to the displacement, the work done is zero: the speed and kinetic energy stay constant and only the direction of motion changes.

Q10. A proton and an alpha particle travel at the same speed, at right angles to the same magnetic field. The force on the alpha particle, compared with the force on the proton, is

  • (A) the same
  • (B) twice as large
  • (C) four times as large
  • (D) half as large
Show Answer

Answer: B – F = BQv depends on charge, not mass. The alpha particle carries charge +2e, twice the proton’s +e, so with the same B and v it experiences twice the force.

Circular Motion in a Magnetic Field

Q11. A charged particle enters a uniform magnetic field at right angles to the field and follows a circular path at constant speed. The motion is circular because

  • (A) the force is constant in direction
  • (B) the force acts parallel to the velocity
  • (C) the force has constant magnitude and is always at right angles to the velocity, so it acts as a centripetal force
  • (D) the magnetic field cancels out the particle’s momentum
Show Answer

Answer: C – F = BQv is constant in magnitude (B, Q and v are all constant) and always perpendicular to v, pulling the particle towards the centre of a circle. That is exactly the requirement for uniform circular motion.

Q12. A particle of mass m and charge Q moves in a circle of radius r at speed v in a field B. Equating the magnetic force BQv to mv2/r gives the radius as

  • (A) r = BQ / (mv)
  • (B) r = mv2 / (BQ)
  • (C) r = m / (BQv)
  • (D) r = mv / (BQ)
Show Answer

Answer: D – BQv = mv2/r, so r = mv2 / (BQv) = mv / (BQ). The radius is proportional to the momentum mv of the particle.

Q13. An electron (m = 9.11 × 10-31 kg) moves at 1.0 × 107 m/s at right angles to a field of flux density 5.0 × 10-3 T. What is the radius of its circular path?

  • (A) 1.1 × 10-2 m
  • (B) 1.1 × 10-3 m
  • (C) 5.7 × 10-2 m
  • (D) 8.8 × 10-2 m
Show Answer

Answer: A – r = mv / (BQ) = (9.11 × 10-31 × 1.0 × 107) / (5.0 × 10-3 × 1.6 × 10-19) = 9.11 × 10-24 / 8.0 × 10-22 = 1.1 × 10-2 m (about 1.1 cm).

Q14. A proton (m = 1.67 × 10-27 kg) moves at 5.0 × 105 m/s at right angles to a field of flux density 0.10 T. What is the radius of its path?

  • (A) 5.2 × 10-3 m
  • (B) 5.2 × 10-2 m
  • (C) 1.0 × 10-1 m
  • (D) 5.2 × 10-1 m
Show Answer

Answer: B – r = mv / (BQ) = (1.67 × 10-27 × 5.0 × 105) / (0.10 × 1.6 × 10-19) = 8.35 × 10-22 / 1.6 × 10-20 = 5.2 × 10-2 m (about 5.2 cm).

Q15. The period of revolution of a charge in a magnetic field is T = 2πm / (BQ). If the same particle enters the same field at twice the speed, its period of revolution

  • (A) doubles
  • (B) is halved
  • (C) becomes four times larger
  • (D) stays the same
Show Answer

Answer: D – T contains no v: doubling the speed doubles the radius (r = mv/BQ), so the particle travels twice as far around a circle of twice the radius in exactly the same time.

Q16. What is the period of revolution of an electron (m = 9.11 × 10-31 kg) in a magnetic field of flux density 1.0 × 10-2 T?

  • (A) 3.6 × 10-7 s
  • (B) 5.7 × 10-9 s
  • (C) 3.6 × 10-9 s
  • (D) 1.8 × 10-9 s
Show Answer

Answer: C – T = 2πm / (BQ) = (2π × 9.11 × 10-31) / (1.0 × 10-2 × 1.6 × 10-19) = 5.72 × 10-30 / 1.6 × 10-21 = 3.6 × 10-9 s.

Q17. A proton and an electron enter the same magnetic field at the same speed, at right angles to the field. The radius of the proton’s path compared with the radius of the electron’s path is

  • (A) about the same
  • (B) about 1800 times larger
  • (C) about 1800 times smaller
  • (D) about 43 times larger
Show Answer

Answer: B – At the same speed and with charges of equal magnitude, r = mv/(BQ) is proportional to m. The proton is about 1830 times as massive as the electron (1.67 × 10-27 / 9.11 × 10-31), so its radius is about 1800 times larger.

Q18. A proton and an electron have the SAME kinetic energy and move in the same magnetic field. The radius of the proton’s path is

  • (A) about 43 times the radius of the electron’s path
  • (B) about 1800 times the radius of the electron’s path
  • (C) the same as the radius of the electron’s path
  • (D) about 43 times smaller than the radius of the electron’s path
Show Answer

Answer: A – With E = ½mv2, mv = √(2mE), so at the same kinetic energy r ∝ √m. The ratio is √(mp/me) = √1830 ≈ 43, not 1830 as in the equal-speed case.

Q19. An alpha particle (mass 4u, charge 2e) and a proton (mass 1u, charge e) enter the same field at the same speed, at right angles to the field. The radius of the alpha particle’s path is

  • (A) half the radius of the proton’s path
  • (B) the same as the radius of the proton’s path
  • (C) twice the radius of the proton’s path
  • (D) four times the radius of the proton’s path
Show Answer

Answer: C – r ∝ m/Q. For the alpha particle m/Q = 4u/2e = 2 (in units of u/e), compared with 1 for the proton, so the alpha particle’s radius is twice as large. Both the extra mass and the extra charge must be counted.

Q20. In a bubble chamber placed in a magnetic field, a fast charged particle leaves a spiral track that curls more and more tightly. The radius of curvature decreases along the track because

  • (A) the magnetic field gets stronger along the track
  • (B) the particle’s charge gradually decreases
  • (C) the particle gains energy from the liquid
  • (D) the particle loses kinetic energy in collisions with the liquid, so its speed falls and r = mv/(BQ) gets smaller
Show Answer

Answer: D – The particle ionises the liquid and loses energy, so v decreases. Since B and Q are constant, r = mv/(BQ) shrinks in proportion, producing the inward spiral.


Velocity Selector, Mass Spectrometer & Hall Effect

Q21. In a velocity selector, a uniform electric field E and a uniform magnetic field B are arranged at right angles so that their forces on a charge Q oppose each other. The charge passes through undeflected when

  • (A) v = E / B
  • (B) v = B / E
  • (C) v = EQ / B
  • (D) v = EB
Show Answer

Answer: A – Undeflected motion needs the electric force QE to balance the magnetic force QvB: QE = QvB, so v = E / B. Only charges with exactly this speed travel straight through.

Q22. A velocity selector has an electric field of 4.0 × 104 V/m and a magnetic field of flux density 0.20 T. What speed must an ion have to pass through undeflected?

  • (A) 8.0 × 103 m/s
  • (B) 5.0 × 10-6 m/s
  • (C) 2.0 × 105 m/s
  • (D) 2.0 × 104 m/s
Show Answer

Answer: C – v = E / B = 4.0 × 104 / 0.20 = 2.0 × 105 m/s.

Q23. A velocity selector set up for one particular speed passes positive ions, negative ions and ions of different masses, provided they all have that speed. This is because

  • (A) heavier ions are slowed down to the selected speed
  • (B) both forces are proportional to Q, so Q cancels from the balance condition QE = QvB
  • (C) the fields exert forces only on singly charged ions
  • (D) the magnetic force depends on the mass of the ion
Show Answer

Answer: B – The electric force QE and the magnetic force QvB both contain Q as a factor, so Q cancels and the selected speed v = E/B is the same for every charge and every mass. (For a negative charge both forces reverse, so they still balance.)

Q24. In a mass spectrometer, singly charged ions that have passed through a velocity selector enter a second uniform magnetic field and follow semicircular paths before striking a detector. Ions of larger mass

  • (A) follow paths of smaller radius and strike nearer the entry slit
  • (B) are not deflected at all
  • (C) follow paths of exactly the same radius as lighter ions
  • (D) follow paths of larger radius, because r = mv/(BQ) is proportional to m when v, B and Q are the same
Show Answer

Answer: D – All the ions have the same v (from the selector), the same Q and the same B, so r ∝ m. Heavier ions curve less, travel in wider semicircles and hit the detector further from the entry point, which is how the masses are separated and measured.

Q25. Singly charged carbon-12 and carbon-14 ions enter a mass spectrometer at the same speed. The ratio radius(carbon-14) / radius(carbon-12) is

  • (A) 6/7
  • (B) 7/6
  • (C) 1
  • (D) 49/36
Show Answer

Answer: B – r ∝ m, so the ratio is 14/12 = 7/6 ≈ 1.17. The carbon-14 ions follow the wider path.

Q26. An ion of mass 3.3 × 10-26 kg and charge +e enters a magnetic field of flux density 0.50 T at 1.0 × 105 m/s, at right angles to the field. What is the radius of its path?

  • (A) 4.1 × 10-2 m
  • (B) 4.1 × 10-3 m
  • (C) 2.1 × 10-2 m
  • (D) 8.3 × 10-2 m
Show Answer

Answer: A – r = mv / (BQ) = (3.3 × 10-26 × 1.0 × 105) / (0.50 × 1.6 × 10-19) = 3.3 × 10-21 / 8.0 × 10-20 = 4.1 × 10-2 m.

Q27. A current-carrying metal strip is placed in a magnetic field perpendicular to the current, and a small voltage (the Hall voltage) appears across the width of the strip. The Hall voltage appears because

  • (A) the field induces an EMF along the length of the strip
  • (B) the resistance of the strip changes in the field
  • (C) the magnetic force pushes the moving charge carriers towards one side of the strip, where charge builds up until the electric force from this charge balances the magnetic force
  • (D) the field heats one side of the strip more than the other
Show Answer

Answer: C – Each moving carrier experiences F = BQv towards one edge of the strip. Charge accumulates on that edge, creating a transverse electric field; equilibrium is reached when the electric force on a carrier balances the magnetic force, and the resulting PD across the strip is the Hall voltage.

Q28. The Hall voltage is given by VH = BI / (nqt). A copper strip of thickness t = 1.0 × 10-3 m carries a current I = 2.0 A in a field B = 0.80 T. For copper, n = 8.5 × 1028 m-3 and q = 1.6 × 10-19 C. What is VH?

  • (A) 1.2 × 10-5 V
  • (B) 1.2 × 10-9 V
  • (C) 1.9 × 10-7 V
  • (D) 1.2 × 10-7 V
Show Answer

Answer: D – VH = BI / (nqt) = (0.80 × 2.0) / (8.5 × 1028 × 1.6 × 10-19 × 1.0 × 10-3) = 1.6 / 1.36 × 107 = 1.2 × 10-7 V. Hall voltages in metals are tiny because n is very large.

Q29. A Hall probe is used to measure the flux density of an unknown magnetic field. The measurement works because

  • (A) with a known, steady current through the probe, the Hall voltage is proportional to B, so the probe can be calibrated against known fields
  • (B) the Hall voltage in volts is numerically equal to B in tesla
  • (C) the current through the probe changes in proportion to B
  • (D) the resistance of the probe measures B directly
Show Answer

Answer: A – From VH = BI/(nqt), with I, n, q and t all fixed for the probe, VH ∝ B. Calibration in known fields converts the measured voltage into a value of B.

Q30. In a semiconductor sample, measuring which side of the strip becomes positive in the Hall effect tells us

  • (A) the size of the magnetic field
  • (B) whether the charge carriers are negative electrons or positive holes
  • (C) the temperature of the sample
  • (D) the resistance of the sample
Show Answer

Answer: B – The magnetic force pushes carriers to the same side of the strip whatever their sign, so that side becomes negative for electron conduction and positive for hole conduction. The polarity of the Hall voltage therefore reveals the sign of the charge carriers.

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