Radius of Circular Orbit in Magnetic Field

When v ⊥ B, magnetic force supplies centripetal force. Radius is proportional to momentum mv and inversely proportional to charge and field strength.

Setup

A charge $q$ of mass $m$ enters a uniform magnetic field $\vec{B}$ with velocity $\vec{v} \perp \vec{B}$.

The magnetic force is always perpendicular to $\vec{v}$ — it acts as a centripetal force, bending the path into a circle without changing the speed.

Derivation

Equating magnetic force to centripetal force:

$$qvB = \frac{mv^2}{r}$$ $$\boxed{r = \frac{mv}{qB}}$$

Interpretation

• $r \propto mv$ (momentum): faster or heavier particles trace larger circles.
• $r \propto 1/B$: stronger field → tighter orbit.
• $r$ is independent of the direction of entry (as long as $v \perp B$).

In Terms of Kinetic Energy

If the particle has kinetic energy $K = \frac{1}{2}mv^2$, then $mv = \sqrt{2mK}$:

$$r = \frac{\sqrt{2mK}}{qB}$$

In Terms of Momentum

For relativistic particles, $mv$ is replaced by the relativistic momentum $p$:

$$r = \frac{p}{qB}$$

This is used in particle physics detectors — measuring $r$ in a known $B$ gives $p$.