To understand the quantized motion of an electron in a uniform magnetic field and determine its magnetic moment in the lowest energy state, consider the following derivations and equations:

Magnetic Force and Centripetal Force:

The magnetic force acting on the electron is counterbalanced by the centripetal force necessary for its circular motion:

$ e v B = \frac{m v^2}{r} $

Solving for the velocity $ v $:

$ v = \frac{e B r}{m} $

Flux through the Electron's Orbit:

The model states that the magnetic flux $ \phi $ through the orbit is linked to Planck’s constant $ h $ as:

$ B \pi r^2 = \frac{n h}{e} $

Rearranging gives:

$ B r^2 = \frac{n h}{e \pi} $

Magnetic Moment:

The magnetic moment $ \mu $ is defined as the current $ I $ times the area $ A $ of the orbit:

$ \mu = I A = \frac{e}{T} \pi r^2 $

Considering current due to orbital motion:

$ \mu = \frac{e v}{2 \pi r} \times \pi r^2 = \frac{e v r}{2} $

Substitute for $ v $ from the earlier velocity equation:

$ \mu = \frac{1}{2} e \left(\frac{e B r}{m}\right) r = \frac{1}{2} e^2 \frac{B r^2}{m} $

Using the flux condition $ B r^2 = \frac{n h}{e \pi} $, we find:

$ \mu = \frac{1}{2} e^2 \frac{n h}{e \pi m} = \frac{n e h}{2 \pi m} $

Lowest Energy State:

For the lowest energy state, take $ n = 1 $:

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

Thus, the magnetic moment of an electron in its lowest energy state is $\frac{e h}{2 \pi m}$.