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Larmor radius calculator and worked examples

A charged particle moving across a magnetic field travels in a circle. The radius of that circle is the Larmor radius. It is also called the gyroradius and the cyclotron radius. Those are three names for one number.

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

Here $m$ is the particle mass in kilograms, $v$ is the component of its speed perpendicular to the field in metres per second, $q$ is the magnitude of its charge in coulombs, and $B$ is the field strength in tesla. The radius comes out in metres.

Where the formula comes from

A magnetic field does no work on a moving charge. The force is always perpendicular to the velocity, so the speed never changes and only the direction turns. Constant speed plus constant turning is a circle.

The magnetic force on a charge moving perpendicular to the field has magnitude $qvB$. Holding a mass on a circle of radius $r$ at speed $v$ requires a centripetal force of $mv^2/r$. The magnetic force is the only force acting, so those two are the same thing.

$$\begin{aligned} qvB &= \frac{mv^2}{r} \\ r &= \frac{mv}{qB} \end{aligned}$$

OpenStax University Physics runs this argument and prints the same result.

Two consequences fall straight out. Momentum $mv$ sits on top, so a faster or heavier particle turns wider. Field strength sits on the bottom, so a stronger field turns it tighter.

Calculator

Larmor radius r = -

The defaults are an electron in a half-tesla field. Charge is entered as a magnitude, and a sign on it changes the direction of the circle rather than its size.

Two particles, same field

Take an electron entering a $B = 0.5$ T field at $v = 1.0\times10^{6}$ m/s, perpendicular to the field. The CODATA 2022 values are $m_e = 9.1093837139\times10^{-31}$ kg and $e = 1.602176634\times10^{-19}$ C.

The numerator is $mv = 9.1094\times10^{-31} \times 1.0\times10^{6} = 9.1094\times10^{-25}$ kg m/s. The denominator is $qB = 1.6022\times10^{-19} \times 0.5 = 8.0109\times10^{-20}$ C T. Divide:

$$r = \frac{9.1094\times10^{-25}}{8.0109\times10^{-20}} = 1.14\times10^{-5}\ \text{m}$$

That is 11.4 micrometres, so the whole orbit is 22.7 micrometres across. An electron moving at a million metres per second circles inside the width of a typical human hair.

Now swap in a proton and change nothing else. $m_p = 1.67262192595\times10^{-27}$ kg, same charge magnitude, same speed, same field. The numerator becomes $1.6726\times10^{-21}$ kg m/s and the denominator is unchanged.

$$r = \frac{1.6726\times10^{-21}}{8.0109\times10^{-20}} = 2.09\times10^{-2}\ \text{m}$$

That is 2.09 cm, about 1836 times the electron radius. It has to be. Only the mass changed, the radius is proportional to mass, and $m_p/m_e = 1836.15$.

The scaling with field is just as direct. Drop the electron’s field from 0.5 T to 1.0 mT and the field is 500 times weaker, so the radius is 500 times larger: 5.69 mm instead of 11.4 micrometres. That is a millimetre-scale orbit, large enough to see and measure, and it is the regime the charge-to-mass ratio problem below works in.

One quantity does not scale at all. The period of the orbit is

$$T = \frac{2\pi m}{qB}$$

with no $v$ in it. A fast particle covers a bigger circle in the same time as a slow one. For the electron in 0.5 T that period is $7.14\times10^{-11}$ s, a cyclotron frequency of 14.0 GHz. A cyclotron accelerator depends on that independence, because it drives the gap with one fixed radio frequency while the particle spirals outward.

Two errors account for most wrong answers. The first is feeding in the full speed when the velocity is not perpendicular to the field. Only the perpendicular component bends. The parallel component is untouched, the path is a helix, and $v$ in the formula means $v_\perp$. The second is entering the charge as a multiple of $e$ instead of in coulombs. An alpha particle has $q = 3.2\times10^{-19}$ C, not 2.

Each of these simulations solves the equation of motion and draws the orbit, so you can change one input and watch the radius answer.

References