Result

Circular motion in a magnetic field

(E7.2.1) E7.2 · Motion in a Uniform Field →

r = \frac{m v}{q B}, T = \frac{2 π m}{q B}

Statement

The radius and period of the circular path a charged particle follows in a uniform magnetic field.

Symbols and units

$r$	orbit radius	m
$T$	orbit period	s
$m$	particle mass	kg
$q$	charge	C
$B$	field strength	T
$v$	speed	m/s

When it applies

Uniform field, velocity purely perpendicular to B (a purely circular path).

Don't use when

Requires velocity purely perpendicular to B for a purely circular path — a velocity component along B produces a helical path instead.

Derivation

Set the magnetic force equal to the centripetal force required for circular motion:

q v B = \frac{m v ^{2}}{r}

Solve for the radius, then use the circumference to find the period:

r = \frac{m v}{q B}, T = \frac{2 π r}{v} = \frac{2 π m}{q B}

Worked examples

Effect of speed on the orbit

Doubling the particle's speed doubles the orbit radius:

r^{'} = \frac{m ( 2 v )}{q B} = 2 r

But the period is unchanged, since v cancels out of the period formula entirely:

T = \frac{2 π m}{q B} (no v dependence)

Faster particles trace bigger circles but complete each lap in the same time — the basis of the cyclotron's fixed driving frequency.

Watch out for

The period is independent of speed — doubling a particle's speed doubles the orbit radius but leaves the time per revolution unchanged. This is the principle behind the cyclotron.

Connects to

Centripetal acceleration and force