D.3 · IB Physics
Charge in a magnetic field
A magnetic force always acts at right angles to the velocity, so a charged particle moves in a circle. Change the speed, field, mass and charge and see how the radius and period respond.
Predict first
Start from the default settings (proton-like particle, v = 3.0 ×10⁶ m s⁻¹, B = 1.0 T). If you double the speed to 6.0 ×10⁶ m s⁻¹, what happens to the period T of one circle?
T = 2πm/(|q|B) contains no v. A faster particle moves on a circle of double the radius (3.11 cm to 6.22 cm) but covers it in the same time.
Check it in the model: Move the Speed, v slider from 3.0 to 6.0 (×10⁶ m s⁻¹) and compare the Period T and Radius r readouts.
velocity v magnetic force F = qvB (towards the centre) · crosses: B into the page
Presets
|q| v B = m v² ÷ r
r = m v ÷ (|q| B)
T = 2π m ÷ (|q| B)
m in u (1.66×10⁻²⁷ kg), q in e (1.60×10⁻¹⁹ C)
Try this
- Keep the proton-like defaults and double the speed from 3.0 to 6.0 ×10⁶ m s⁻¹. What happens to the radius and to the period?
- Double the field from 1.0 to 2.0 T. By what factor do the radius and the period change? Then switch the charge to negative and note which way the particle turns.
- Use the Alpha-like preset (mass 4 u, charge 2e) and compare its radius and period with the proton-like preset at the same speed and field.
Exam tip: the magnetic force is perpendicular to the velocity so it does no work and the speed stays constant; equating qvB to mv²/r gives r = mv/(qB), and because T = 2πm/(qB) does not contain v, faster particles simply travel on bigger circles in the same time.