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Charged particles in a field

Electricity and magnetism · level 8, lesson 1 of 6

A question to start

A proton from the Sun reaches Earth's magnetic field. It is not stopped, yet it cannot fly straight on either. What path does it follow?

The idea

A charge \(q\) moving at speed \(v\) at right angles to a field \(B\) feels a force

\[F = qvB,\]

at right angles to both its motion and the field. The force never speeds the particle up or slows it down; it only turns it. So the particle moves in a circle, with the magnetic force as the centripetal force:

\[qvB = \frac{mv^2}{r} \quad\Rightarrow\quad r = \frac{mv}{qB}.\]

This is the gyroradius. One trip round takes \(T = 2\pi m/(qB)\), whatever the speed.

If the particle also moves along the field, that part of its motion is not affected, so it spirals along the field lines. That is how auroral particles are guided down to the poles.

field out of the page: a circlespiral along a field line

Worked examples

1. In the solar wind. A proton (\(m = 1.67\times10^{-27}\) kg, \(q = 1.60\times10^{-19}\) C) at 400 km/s in a 5 nT field: \(r = 1.67\times10^{-27} \times 4\times10^5/(1.60\times10^{-19} \times 5\times10^{-9}) = 835\) km.

2. Near the ground on Earth. The same proton in a 30 µT field: \(r = 6.68\times10^{-22}/(4.8\times10^{-24}) = 139\) m.

3. An electron. It has the same charge but \(9.11\times10^{-31}\) kg, 1,836 times less mass, so at the same speed its circle is 1,836 times smaller.

Try it

  1. Why does a magnetic field not change a particle's speed?
  2. Find the radius for a proton at \(1\times10^6\) m/s in a 0.1 T field.
  3. Find the time for one trip round for a proton in a 30 µT field.
  4. A particle moves straight along a field line. What force does it feel?
Answers
  1. The force is always at right angles to the motion, so it does no work.
  2. \(1.67\times10^{-27} \times 10^6/(1.6\times10^{-19} \times 0.1) = 0.10\) m, about 10 cm.
  3. \(2\pi \times 1.67\times10^{-27}/(1.6\times10^{-19} \times 3\times10^{-5}) = 2.2\times10^{-3}\) s.
  4. None: \(v\) is along \(B\), so there is no part of the motion at right angles to the field.

Common mistake

"A stronger field makes the particle go faster." A stronger field makes a tighter circle, at the same speed.

Where it leads