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A2 · Practice questions · Force on a moving charge

A force that only ever turns.

Six original Cambridge-style questions on F = BQv sin θ, circular paths with r = mv / BQ, the velocity selector and the Hall voltage.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
Keep these straight

Perpendicular force, constant speed, circular path.

01
Calculation
[2 marks]

A proton (charge 1.60 × 10⁻¹⁹ C) moves at 4.5 × 10⁶ m s⁻¹ through a uniform magnetic field of flux density 0.80 T. Calculate the magnetic force on the proton when its velocity is at right angles to the field, and when the velocity makes an angle of 30° with the field.

  • F = BQv = 0.80 × 1.60 × 10⁻¹⁹ × 4.5 × 10⁶ = 5.8 × 10⁻¹³ N. ✓
  • F = BQv sin 30° = 5.76 × 10⁻¹³ × 0.50 = 2.9 × 10⁻¹³ N. ✓
02
Analysis
[3 marks]

A charged particle enters a uniform magnetic field with its velocity at right angles to the field. Explain why it moves in a circle at constant speed, and state the effect on the radius of the path and on the period of the motion if the particle's speed is doubled.

  • The magnetic force is always perpendicular to the velocity, so it does no work and the speed (and kinetic energy) stays constant. ✓
  • A force of constant magnitude that is always perpendicular to the velocity acts as a centripetal force, so the path is a circle. ✓
  • From r = mv / BQ the radius doubles; the period T = 2πr / v = 2πm / BQ does not depend on v, so it is unchanged. ✓
03
Calculation
[3 marks]

An electron (mass 9.11 × 10⁻³¹ kg, charge 1.60 × 10⁻¹⁹ C) moves in a circle of radius 1.2 cm in a uniform magnetic field of flux density 1.5 mT, with its velocity at right angles to the field. Calculate the speed of the electron and the period of its motion.

  • From BQv = mv²/r, v = BQr / m. ✓
  • v = 1.5 × 10⁻³ × 1.60 × 10⁻¹⁹ × 0.012 / 9.11 × 10⁻³¹ = 3.2 × 10⁶ m s⁻¹. ✓
  • T = 2πr / v = 2π × 0.012 / 3.16 × 10⁶ = 2.4 × 10⁻⁸ s (the same as 2πm / BQ). ✓
04
Calculation
[3 marks]

A Hall probe consists of a slice of material of thickness 0.10 mm carrying a current of 2.0 A, placed with its face perpendicular to a field of flux density 0.50 T. The Hall voltage is VH = BI / ntq. Calculate the Hall voltage if the slice is copper, with n = 8.5 × 10²⁸ m⁻³, and if it is a semiconductor with n = 1.0 × 10²⁴ m⁻³. Hence explain why Hall probes are made from semiconductors.

  • Copper: VH = 0.50 × 2.0 / (8.5 × 10²⁸ × 1.0 × 10⁻⁴ × 1.60 × 10⁻¹⁹) = 7.4 × 10⁻⁷ V. ✓
  • Semiconductor: n is 8.5 × 10⁴ times smaller, so VH = 7.35 × 10⁻⁷ × 8.5 × 10⁴ = 6.3 × 10⁻² V (63 mV). ✓
  • VH is inversely proportional to n, so the far smaller carrier density of a semiconductor gives a Hall voltage large enough to measure easily, making the probe sensitive to B. ✓
05
Analysis
[2 marks]

A proton and an electron enter the same uniform magnetic field with the same velocity, at right angles to the field. Compare the directions of the initial forces on the two particles, and compare the radii of their paths. (Proton mass 1.67 × 10⁻²⁷ kg, electron mass 9.11 × 10⁻³¹ kg.)

  • The forces have the same magnitude BQv but opposite directions, because the charges have opposite signs (the electron's conventional current is opposite to its velocity), so the two particles curve in opposite senses. ✓
  • With the same v, B and charge magnitude, r = mv / BQ is proportional to mass, so the proton's radius is 1.67 × 10⁻²⁷ / 9.11 × 10⁻³¹ = 1800 times larger than the electron's. ✓
06
Calculation
[4 marks]

A proton (mass 1.67 × 10⁻²⁷ kg, charge 1.60 × 10⁻¹⁹ C) is accelerated from rest through a potential difference of 2.0 kV and then enters a uniform magnetic field of flux density 0.25 T at right angles to the field.

(a) Calculate the speed of the proton as it enters the field. (b) Calculate the radius of its circular path. (c) An ion with the same charge but twice the mass is accelerated through the same potential difference into the same field. Calculate the radius of its path.

  • (a) QV = ½mv², so v = √(2QV / m) = √(2 × 1.60 × 10⁻¹⁹ × 2000 / 1.67 × 10⁻²⁷). ✓
  • v = 6.2 × 10⁵ m s⁻¹. ✓
  • (b) r = mv / BQ = 1.67 × 10⁻²⁷ × 6.19 × 10⁵ / (0.25 × 1.60 × 10⁻¹⁹) = 2.6 × 10⁻² m (2.6 cm). ✓
  • (c) Doubling m makes v smaller by √2 (same QV), so r = mv / BQ grows by √2: r = 2.58 × √2 = 3.7 cm. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.