How Does a Charged Particle Move in a Magnetic Field?

A charged particle in a uniform magnetic field follows a helical path when its initial velocity is neither entirely parallel nor entirely perpendicular to the field. The path is a combination of circular motion around the field direction and steady motion along it. The magnetic field changes the particle’s direction, but not its speed.

The formulas below assume a constant magnetic field, no electric field, and speeds low enough for nonrelativistic mechanics.

Why the path is helical

The magnetic part of the Lorentz force is

$$\mathbf{F}=q\,\mathbf{v}\times\mathbf{B},$$

where $q$ is the signed charge, $\mathbf{v}$ is velocity, and $\mathbf{B}$ is the magnetic field. This force is perpendicular to the velocity, so it does no work: the particle’s kinetic energy and speed stay constant.

Split the velocity into components parallel and perpendicular to the field. If $\theta$ is the angle between the velocity and the field, their magnitudes are

$$v_{\parallel}=v\cos\theta, \qquad v_{\perp}=v\sin\theta.$$

The parallel component experiences no magnetic force, so it remains constant. The perpendicular component experiences a force of magnitude $|q|v_{\perp}B$, which continually turns it and produces circular motion. Combining these motions gives a helix whose axis is parallel to the magnetic field.

Radius, period, and pitch

The perpendicular magnetic force supplies the centripetal force:

$$|q|v_{\perp}B=\frac{mv_{\perp}^{2}}{r}.$$

For $v_{\perp}>0$, solving for the radius gives

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

The angular cyclotron frequency and period are

$$\omega_c=\frac{|q|B}{m}, \qquad T=\frac{2\pi m}{|q|B}.$$

The period does not depend on the particle’s speed in this nonrelativistic model. In one period, the particle travels along the field by the pitch of the helix:

$$p=v_{\parallel}T.$$

The signed pitch indicates direction along the field; its magnitude is $|p|=|v_{\parallel}|T$. A particle with $v_{\perp}=0$ travels straight along the field, while one with $v_{\parallel}=0$ moves in a circle with zero pitch.

Worked example: a proton in a 0.20 T field

A proton enters a uniform field of $B=0.20\,\text{T}$ at speed $v=3.0\times10^{6}\,\text{m/s}$, at an angle of $60^\circ$ to the field. Use proton mass $m=1.673\times10^{-27}\,\text{kg}$ and charge magnitude $|q|=1.602\times10^{-19}\,\text{C}$.

First split the speed:

$$\begin{aligned} v_{\parallel}&=v\cos 60^\circ=1.50\times10^6\,\text{m/s},\\ v_{\perp}&=v\sin 60^\circ\approx2.60\times10^6\,\text{m/s}. \end{aligned}$$

Then calculate the radius:

$$r=\frac{(1.673\times10^{-27})(2.60\times10^6)}{(1.602\times10^{-19})(0.20)}\approx0.136\,\text{m}.$$

The period is

$$T=\frac{2\pi(1.673\times10^{-27})}{(1.602\times10^{-19})(0.20)}\approx3.28\times10^{-7}\,\text{s}.$$

So the pitch is

$$p=(1.50\times10^6)(3.28\times10^{-7})\approx0.492\,\text{m}.$$

The proton therefore circles with a radius of about $13.6$ cm and advances about $49.2$ cm along the field per turn. Because the proton is positively charged, its rotation direction follows from $\mathbf{v}\times\mathbf{B}$; a negatively charged particle with the same mass and velocity would rotate the other way.

Common mistakes

• Using the total speed for the radius. Only $v_{\perp}$ produces circular motion. Use $v_{\parallel}$ to calculate the advance along the field.
• Leaving out the charge magnitude. Radius and period use $|q|$. The sign of $q$ determines the direction of rotation, not whether the radius is positive or negative.
• Assuming the field changes the speed. A magnetic force alone is perpendicular to the velocity, so it changes direction but not kinetic energy.
• Treating every path as a helix. Purely parallel motion is straight, and purely perpendicular motion is circular.

When the field or speed differs

A uniform-field helix is not an exact description when the magnetic field varies significantly across the particle’s path. The trajectory can then bend or change pitch, and effects such as magnetic mirroring may matter. If the particle is relativistic, replace $m$ in the formulas with $\gamma m$, where $\gamma=1/\sqrt{1-v^2/c^2}$; in particular, the relativistic cyclotron frequency is $|q|B/(\gamma m)$.