Effect of magnetic field in atomic level can be quantified in a classical level with some assumptions such as, the atomic orbit is circular and electron revolves around the nucleus at radius R and the current produced is assumed to steady.
The current is given by, $$ I = \frac{q}{t} = \frac{-ev}{2\pi{R}} $$ where $$ T = \frac{2\pi}{\omega} = {2\pi{R}}{v} \\~\\ v = R\omega $$
The orbital dipole moment of this configuration is given by, $$ \vec{m} = I \vec{a} = \frac{-ev}{2\pi{R}}\pi{R^2} \hat{z} = \frac{-evR}{2}\hat{z} $$ where $\hat{z} $ points in the direction perpendicular to the area of the loop when current flows by the usual right hand thumb rule direction. When it is placed in the magnetic field, this dipole moment experiences a torque which tries to align it along the magnetic field direction.
Without magnetic field, there is only electrostatic interaction, therefore, the force equation gives, $$ \frac{e^2}{4\pi\epsilon_0R^2} = \frac{m_ev^2}{R}$$
If suppose we assume the magnetic field is in the direction of $\hat{z} $ , then the centripetal force can be written as, $$ \frac{e^2}{4\pi\epsilon_0R^2} + ev'B = \frac{m_e v'^2}{R} $$
where v' is the new velocity. If we assume $v'\simeq v $ then, $$ ev'B = \frac{m_e(v'^2 -v^2)}{R} = \frac{m_e}{R} (v'+v)(v'-v) $$
then, $$ v'-v = \frac{eRB}{2m_e} = \delta{v} $$
e,R,B, m are all positive quantities. So, the electron will speed up when the magnetic field is turned on. Similarly, the change in orbital speed develops a change in magnetic moment by, $$\delta\vec{m} = \frac{-e\delta{v} R} \hat{z} = \frac{-e^2R^2}{4m_e} \vec{B} $$ which directed in the opposite direction of applied magnetic field.
That is it! But all this proved to be wrong with quantum mechanics where better explanation is given!
The current is given by, $$ I = \frac{q}{t} = \frac{-ev}{2\pi{R}} $$ where $$ T = \frac{2\pi}{\omega} = {2\pi{R}}{v} \\~\\ v = R\omega $$
The orbital dipole moment of this configuration is given by, $$ \vec{m} = I \vec{a} = \frac{-ev}{2\pi{R}}\pi{R^2} \hat{z} = \frac{-evR}{2}\hat{z} $$ where $\hat{z} $ points in the direction perpendicular to the area of the loop when current flows by the usual right hand thumb rule direction. When it is placed in the magnetic field, this dipole moment experiences a torque which tries to align it along the magnetic field direction.
Without magnetic field, there is only electrostatic interaction, therefore, the force equation gives, $$ \frac{e^2}{4\pi\epsilon_0R^2} = \frac{m_ev^2}{R}$$
If suppose we assume the magnetic field is in the direction of $\hat{z} $ , then the centripetal force can be written as, $$ \frac{e^2}{4\pi\epsilon_0R^2} + ev'B = \frac{m_e v'^2}{R} $$
where v' is the new velocity. If we assume $v'\simeq v $ then, $$ ev'B = \frac{m_e(v'^2 -v^2)}{R} = \frac{m_e}{R} (v'+v)(v'-v) $$
then, $$ v'-v = \frac{eRB}{2m_e} = \delta{v} $$
e,R,B, m are all positive quantities. So, the electron will speed up when the magnetic field is turned on. Similarly, the change in orbital speed develops a change in magnetic moment by, $$\delta\vec{m} = \frac{-e\delta{v} R} \hat{z} = \frac{-e^2R^2}{4m_e} \vec{B} $$ which directed in the opposite direction of applied magnetic field.
That is it! But all this proved to be wrong with quantum mechanics where better explanation is given!
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