Force on a Moving Charge (OCR A Level Physics): Revision Note

A beta particle is incident at 70° to a magnetic field of flux density 0.5 mT, travelling at a speed of 1.5 × 10⁶ m s^–1.

Calculate: 

a) The magnitude of the magnetic force on the beta particle

b) The magnitude of the maximum possible force on a beta particle in this magnetic field, travelling with the same speed

Answer:

Part (a)

Step 1: Write out the known quantities

• Magnetic flux density B = 0.5 mT = 0.5 × 10⁻³ T

• Speed v = 1.5 × 10⁶ m s^–1

• Angle θ between the flux and the velocity = 70°

Step 2: Substitute quantities into the equation for magnetic force on a charged particle

• A beta particle is an electron

• Therefore, the magnitude of electron charge Q = 1.6 × 10^–19 C

• Substituting values gives:

F = BQv sin θ

F = (0.5 × 10^–3) × (1.6 × 10^–19) × (1.5 × 10⁶) × sin (70)

F = 1.1 × 10^–16 N

Part (b)

Step 1: Write out the known quantities

• Magnetic flux density B = 0.5 mT = 0.5 × 10^–3 T

• Speed v = 1.5 × 10⁶ m s^–1

Step 2: Determine the angle to the flux lines

• Angle θ between the flux and the velocity = 90° if the magnetic force is a maximum

Step 3: Substitute quantities into the equation for magnetic force on a charged particle

• The magnitude of electron charge Q = 1.6 × 10^–19 C

  • Substituting values gives:

F = BQv sin θ = BQv when sin 90 = 1

F = (0.5 × 10^–3) × (1.6 × 10^–19) × (1.5 × 10⁶)

F = 1.2 × 10^–16 N

Remember not to mix this up with F = BIL sin θ!

• F = BIL sin θ is the force on a current-carrying conductor

• F = BQv sin θ is the force on an isolated moving charged particle (which may be inside a conductor)

Another super important fact to remember for typical exam questions is that the magnetic force on a charged particle is centripetal, because it always acts at 90° to the particle's velocity. You should practise using Fleming's Left Hand Rule to determine the exact direction!