Period of a Charged Particle | CAIE 9702 Paper 4 Notes

Paper 4

Physics9702

Period of a Charged Particle

Imagine the same charged particle entering the same magnetic field at two different speeds.

The faster particle follows a larger circle. We first compare the two journeys, then prove what happens to the period.

Compare two orbits first

1. Speed doubles

The particle moves twice as fast through the field.

2. Radius doubles

The circular path is twice as large, so one lap is twice as long.

3. Time is unchanged

It travels twice the distance at twice the speed. The two factors cancel.

For one particle in one magnetic field, changing the speed changes the orbit size but not the time for one complete orbit.

Write speed as distance divided by time for one complete orbit.
$v=\dfrac{2\pi r}{T}$
Substitute this expression for speed into the radius relation.
$r=\dfrac{m}{B|q|}\left(\dfrac{2\pi r}{T}\right)$
Cancel the radius from both sides, then rearrange for the period.
$T=\dfrac{2\pi m}{B|q|}$

The period is independent of orbit radius and particle speed.
A larger orbit has a larger speed, so the extra distance is covered in the same time (9702/42/F/M/24 Q5(b)).

Your turnquick check

In the same field, an electron enters at twice the speed. State how its orbit radius and period change.

Show worked answer

The radius doubles. The period is unchanged.