A model for quantized motion of an electron in a uniform magnetic field B states that the flux passing through the orbit of the electron is n(hle) where n is an integer, h is Planck’s constant and e is the magnitude of electron’s charge. According to the model, the magnetic moment of an electron in its lowest energy state will be (m is the mass of the electron)

In a uniform magnetic field the electron moves in a circular orbit. The magnetic moment μ of a current loop is given by

$\frac{I}{A}$, where I is the current and A is the area of the loop.

The current produced by a single electron going around the orbit once every period T is

$I=\frac{e}{T}$, and the period is related to the orbital speed v and radius r by T = 2πr / v.

Thus the magnetic moment becomes

$\mu =\frac{e}{\mathrm{2\pi }}vr$

Now use the quantization condition from the model: the magnetic flux through the orbit is an integer multiple of h / e, i.e.

$\Phi =BA=\frac{h}{e}$

For the lowest‑energy (ground) state we take the integer n = 1, so the orbital area is

$A=\frac{h}{\mathrm{eB}}$

The area of a circular orbit is also A = πr², giving

$r=\sqrt{\frac{h}{\mathrm{\pi eB}}}$

The kinetic momentum of the electron is p = mv = m v, and the angular momentum is L = mvr.

From the same quantization condition we have a discrete angular momentum

$L=\frac{h}{2}$

Solving for v r gives

$vr=\frac{L}{m}=\frac{h}{2}$

Insert this result back into the expression for μ:

$\mu =\frac{e}{2}\frac{h}{2}$

Simplifying the fractions yields

$\mu =\frac{h}{e}/\frac{2}{\pi }$

or more compactly

$\mu =\frac{h}{e}/\frac{2}{\pi }$

which is exactly

$\frac{he}{2\pi m}$

This is the magnetic moment of the electron in its lowest‑energy state according to the given quantized‑flux model.