Assume that an electron of mass mand charge magnitude emoves in a circular orbit of radius rabout a nucleus. A uniform magnetic fieldis then established perpendicular to the plane ofthe orbit. Assuming also that the radius of the orbit does not change and that the change in the speed of the electron due to fieldis small, find an expression for the change in the orbital magnetic dipole moment of the electron due to the field.

Radius of circular orbit isr

Magnetic field is$\overrightarrow{B}$

Magnitude of charge on the electron is e

Mass of the electron is$m_e$

The electric field is induced due to the changing applied magnetic field and due to the induced electric field, there is a change in velocity of the electron moving in a circular orbit.

Magnetic moment of electron depends on the velocity, as velocity changes magnetic moment also changes. This idea is used to derive the required relation for change in magnetic moment of an electron.

Formula:

Current associated with circulating electron is

$\begin{array}{c}i=\frac{e}{T}\\ =\frac{ev}{2\pi r}\end{array}$

Acceleration,$\frac{d\overrightarrow{v}}{dt}=\overrightarrow{a}$

Magnetic dipole moment,

$\begin{array}{c}\overrightarrow{\mu }=i\overrightarrow{A}\\ =i\cdot \pi r^2\end{array}$

Magnitude of force on electron kept in an electric field$F=qE$

Electric field with the circular field lines is induced due to the applied magnetic field. Suppose the magnetic field increases linearly from zero toin time t, i.e. the change in the magnetic field is

$\begin{array}{c}dB=B-0\\ =B\end{array}$

and

The change in time is

$\begin{array}{c}dt=t-0\\ =t\end{array}$

the magnitude of the electric field at the orbit is given by

$\begin{array}{c}E=\frac{r}{2}\cdot \frac{dB}{dt}\\ =\frac{r}{2}\cdot \frac{B}{t}\end{array}$(1)

Where,is the radius of the orbit.

The induced electric field is tangent to the orbit and changes the speed of the electron.

The change in the speed being given by

$\begin{array}{c}dv=\left|\overrightarrow{a}\right|dt\\ =at\end{array}$

By using Newton’s second law,$F=ma$and the force on the electron in an electric field

$\begin{array}{c}F=qE\\ =eE\end{array}$

Thus,$ma=eE$

$a=\frac{eE}{m_e}$

Thus, thechange in speed is

$\begin{array}{c}dv=at\\ =\frac{eE}{m_e}t\end{array}$

Substituting the value of E from equation (1),

$\begin{array}{c}dv=\frac{et}{m_e}\frac{rB}{2t}\\ =\frac{erB}{2m_e}\end{array}$(2)

Now, the current associated with the circulating electron is

$\begin{array}{c}i=\frac{e}{T}\\ =\frac{ev}{2\pi r}\end{array}$

The magnitude of the magnetic dipole moment,

$\begin{array}{c}\mu =iA\\ =i\cdot \pi r^2\\ =\frac{ev}{2\pi r}\cdot \pi r^2\\ =\frac{evr}{2}\end{array}$

So, the change in the magnetic dipole movement will be

$\text{Δ}\mu =\frac{er}{2}\cdot dv$

From equation$\left(2\right),$

$\begin{array}{c}\text{Δ}\mu =\frac{er}{2}\cdot \frac{erB}{2m_e}\\ =\frac{e^2{r}^2}{4m_e}\cdot B\end{array}$

The expression for the change in the orbital magnetic dipole moment of the electron due to the field is$\text{Δ}\mu =\frac{e^2{r}^{2}B}{4m_e}$