The gyro-magnetic ratio of an electron in an H-atom is equal to the ratio of the magnetic moment and the angular momentum of the electron
$$
\mu_{\mathrm{e}}=\frac{\text { magnetic moment of } \mathrm{e}\left(\mathrm{M}_{\mathrm{c}}\right)}{\text { angular momentum of } \mathrm{e}\left(\mathrm{L}_{\mathrm{e}}\right)}
$$
If $I$ is the magnitude of the angular momentum of the electron about the central nucleus (orbital angular momentum). Vectorially,
$$
\mu_{\mathrm{e}}=\frac{-\mathrm{evr}}{2 \mathrm{~m}_{\mathrm{e}} \mathrm{vr}} \quad\left(\because \mathrm{M}_{\mathrm{e}}=\frac{-\mathrm{evr}}{2}, \mathrm{~L}_{\mathrm{e}}=\mathrm{m}_{\mathrm{e}} \mathrm{vr}\right)
$$


So it is independent of velocity or orbit of e depends only on charge.
The negative sign indicates that the angular momentum of the electron is opposite in direction to the magnetic moment.