Force on a Moving Charge
- The magnetic force on an isolated moving charged particle, such as a proton, is given by the equation:
F = BQv
- Where:
- F = magnetic force on the particle (N)
- B = magnetic flux density (T)
- Q = charge of the particle (C)
- v = speed of the particle (m s−1)
- This is the maximum force on the charged particle, when F, B and v are mutually perpendicular
- Therefore if a particle travels parallel to a magnetic field, it will not experience a magnetic force
- Current is the rate of flow of positive charge
- This means that the direction of the 'current' for a flow of negative charge (e.g. an electron beam) is in the opposite direction to its motion
- If the charged particle is moving at an angle θ to the magnetic field lines, then the size of the magnetic force F is given by the equation:
F = BQv sin θ
- This equation shows that:
- The size of the magnetic force is zero if the angle θ is zero (i.e. the particle moves parallel to the field lines)
- The size of the magnetic force is maximum if the angle θ is 90° (i.e. the particle moves perpendicular to field lines)
Worked example
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
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
Exam Tip
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!
