A magnetic field exerts force on a moving charge according to

$$\mathbf F_B=q\mathbf v\times\mathbf B.$$

Because this force is perpendicular to the instantaneous velocity, an ideal magnetic field changes the direction of motion but not the particle's kinetic energy or speed.

Velocity perpendicular to the field

Suppose a particle of mass $m$ and charge $q$ moves with speed $v$ perpendicular to a uniform magnetic field of magnitude $B$.

The magnetic-force magnitude is

$$F_B=|q|vB.$$

Since the force is always perpendicular to the velocity, it can supply the inward force for uniform circular motion:

$$|q|vB=m\frac{v^2}{r}.$$

Therefore the orbit radius is

$$\boxed{r=\frac{mv}{|q|B}}.$$

Larger momentum produces a larger orbit, while a stronger magnetic field or larger charge magnitude bends the path more tightly.

The sign of $q$ determines the sense of rotation. Positive and negative charges with the same velocity curve in opposite directions.

Cyclotron angular frequency

Using

$$v=r\omega_c,$$

and substituting the radius,

$$\omega_c=\frac{v}{r} =\boxed{\frac{|q|B}{m}}.$$

The corresponding period is

$$\boxed{T=\frac{2\pi m}{|q|B}}.$$

In the nonrelativistic ideal model, the cyclotron frequency is independent of speed. Faster particles move in larger circles but complete each circle in the same time.

Velocity with a component along the field

Decompose the velocity into components parallel and perpendicular to the magnetic field:

$$\mathbf v=\mathbf v_\parallel+\mathbf v_\perp.$$

The parallel component produces no magnetic force because

$$\mathbf v_\parallel\times\mathbf B=\mathbf0.$$

It therefore remains constant. The perpendicular component produces circular motion.

Combining the two motions gives a helix around the magnetic-field direction.

The radius is determined by $v_\perp$:

$$r=\frac{mv_\perp}{|q|B},$$

while the distance advanced along the field during one revolution, the pitch, is

$$\boxed{p=v_\parallel T =\frac{2\pi m v_\parallel}{|q|B}}.$$

Worked example

A proton moves perpendicular to a uniform field with

$$v=2.0\times10^6,\mathrm{m/s},$$

$$B=0.50,\mathrm T.$$

Using

$$m_p=1.67\times10^{-27},\mathrm{kg},$$

$$e=1.60\times10^{-19},\mathrm C,$$

the orbit radius is

$$r=\frac{(1.67\times10^{-27})(2.0\times10^6)}{(1.60\times10^{-19})(0.50)} \approx4.18\times10^{-2},\mathrm m.$$

Thus

$$\boxed{r\approx4.2,\mathrm{cm}}.$$

The cyclotron angular frequency is

$$\omega_c=\frac{(1.60\times10^{-19})(0.50)}{1.67\times10^{-27}} \approx4.79\times10^7,\mathrm{rad/s}.$$

Magnetic fields as momentum selectors

Because

$$r=\frac{p}{|q|B}$$

for nonrelativistic momentum magnitude $p=mv$ perpendicular to the field, measuring curvature in a known magnetic field can reveal charge-to-momentum information. This principle underlies magnetic spectrometers, cyclotrons, and many charged-particle detectors.

The circular and helical trajectories are therefore direct consequences of Lorentz force plus ordinary radial dynamics; magnetism introduces no new circular-motion kinematics.