Motion in a magnetic field
Going round in circles
Fire a proton into a uniform magnetic field, at right angles to the field. Here the field points into the screen, shown by the crosses.
The magnetic force q v × B is always perpendicular to the velocity. So it never speeds the proton up or slows it down. It only turns it.
A force perpendicular to the motion does no work. So the speed, and the kinetic energy, stay exactly the same. Only the direction keeps changing.
A force of constant size that always acts at right angles to the velocity: that is just what makes uniform circular motion. The proton goes round a circle.
The magnetic force supplies the centripetal force. So m v² over r equals q v B, which gives the radius: r equals m v over q B.
A faster particle moves in a bigger circle. A stronger field, or a larger charge, bends it into a smaller one.
The angular frequency is ω equals v over r, which is q B over m. The frequency is ν equals q B over 2πm, and the time for one turn is T equals 2πm over q B.
Look closely: v has cancelled out! A fast proton runs round a big circle, a slow one round a small circle, and both take exactly the same time.
Some numbers. A proton at 2 × 10⁶ metres per second in a field of 0.5 tesla: r equals 1.67 × 10⁻²⁷ times 2 × 10⁶, over 1.6 × 10⁻¹⁹ times 0.5. That is about 4.2 centimetres.
Its period is 2π times 1.67 × 10⁻²⁷, over 1.6 × 10⁻¹⁹ times 0.5: about 1.3 × 10⁻⁷ seconds. An electron in the same field would circle the other way round.
Helical paths
What if the velocity is not perpendicular to the field? Split it into two parts: v parallel, along B, and v perpendicular, across B.
The parallel part feels no force, since sin 0 is zero. So the particle keeps drifting along the field at a steady speed v cos θ.
The perpendicular part, v sin θ, is turned round in a circle of radius m v sin θ over q B, exactly as before.
Put the two together: the particle goes round and round while drifting along the field. Its path is a helix, like the thread of a screw.
The distance it moves along B in one turn is the pitch: p equals v parallel times T, which is 2πm v cos θ over q B.
Example: our proton, at 2 × 10⁶ metres per second, now enters the 0.5 tesla field at 60° to it. v perpendicular is 1.73 × 10⁶, so the radius is 0.866 times 4.2, about 3.6 centimetres.
v parallel is 1 × 10⁶ metres per second, and T is still 1.3 × 10⁻⁷ seconds. So the pitch is about 0.13 metres: 13 centimetres per turn.
Now an electron. It moves at 3 × 10⁷ metres per second, perpendicular to a field of 6 × 10⁻⁴ tesla. Take its mass as 9 × 10⁻³¹ kilogram.
r equals 9 × 10⁻³¹ times 3 × 10⁷, over 1.6 × 10⁻¹⁹ times 6 × 10⁻⁴. That is 2.7 × 10⁻²³ over 9.6 × 10⁻²³: 0.28 metre, or 28 centimetres.
Its frequency is v over 2πr: 3 × 10⁷ over 2π times 0.28, about 1.7 × 10⁷ hertz, or 17 megahertz. And its energy, ½ m v², is 4 × 10⁻¹⁶ joule, about 2.5 kilo electron volts.
Try it: circles and helices
- At θ = 90° the proton goes round a circle. Press “Look along B” to see it face on.
- Double the speed: the circle doubles in size, but the period T does not change.
- Double the field: the circle halves and each turn takes half the time.
- Set θ = 60°: the path becomes a helix. Compare the pitch with p = v cos θ × T.
- Switch to the electron: it circles the opposite way.
Twice as fast
A proton moves in a circle in a uniform magnetic field. Its speed is doubled, with the field unchanged. What happens to the radius r of its path and its period T?
- r doubles; T doubles
- r doubles; T stays the same
- r doubles; T halves
- r stays the same; T halves
Answer
B. r = m v / q B, so r is proportional to v and doubles. But T = 2πm / q B has no v in it. The proton travels twice as far each turn at twice the speed, so T stays the same.
An electron’s circle
An electron (mass 9 × 10⁻³¹ kg, charge 1.6 × 10⁻¹⁹ C) moves at 3 × 10⁷ m/s perpendicular to a magnetic field of 6 × 10⁻⁴ T. What is the radius of its path?
- 2.8 cm
- 28 cm
- 2.8 m
- 520 m
Answer
B. r = m v / q B = (9 × 10⁻³¹ × 3 × 10⁷)/(1.6 × 10⁻¹⁹ × 6 × 10⁻⁴) = (2.7 × 10⁻²³)/(9.6 × 10⁻²³) ≈ 0.28 m = 28 cm. Using the proton’s mass by mistake gives about 520 m.
Summary
- Radius (v ⊥ B):
r = m v / q B - Angular frequency and period:
ω = q B / m, T = 2πm / q B - Helix (v at θ to B):
r = m v sin θ / q B, p = v cos θ × T
- The magnetic force is perpendicular to v, so it does no work: speed and kinetic energy stay constant.
- With v ⊥ B the path is a circle. Its period does not depend on the speed.
- A velocity component along B is not affected, so the general path is a helix around the field lines.
- Positive and negative charges circle in opposite senses.