Question

An electron of charge $e$ moves in a circular orbit of radius $r$ around the nucleus at a frequency $v$ . The magnitude moment associated with the orbital motion of the electron is :
A. $\pi ve{r^2}$
B. $\dfrac{{\pi v{r^2}}}{e}$
C. $\dfrac{{\pi ve}}{{{r^2}}}$
D. \[\dfrac{{\pi e{r^2}}}{v}\]

Answer 1

Hint:We can find the relation between the magnetic moment and orbital motion of the electron by using the expressions for the current and magnetic moment of the electron. When an electron moves in a circular orbit around the nucleus it generates some current that appears in the magnetic dipole moment.

Complete step by step answer:
We are given that, an electron of charge $e$ moves in a circular orbit of radius $r$ around the nucleus at a frequency $v$ , the electron produces current as
$I = \dfrac{e}{T}$
Where, $e$ - charge on electron
And $T$ - time period of the circular motion of electron
The frequency is given as $v = \dfrac{1}{T}$
So, $I = ev - - - - - - - - - - (1)$
Now, the magnetic moment associated with the electron is given by
$M = I \times A$
Where, $I$ - current and $A$ - area of the circular loop
$M = I \times \pi {r^2} - - - - - - - - - - - (2)$
Here, $A = \pi {r^2}$
Thus, Substituting eq $(1)$ in eq $(2)$ , we get
$M = ev \times \pi {r^2}$
$\therefore M = \pi ve{r^2}$
This is the magnitude moment associated with the orbital motion of the electron.

Hence, option A is correct.

Note: As the electron is negatively charged, the direction of the magnetic moment is into the plane of paper and negatively charged electrons moving in an anticlockwise direction, giving rise to the clockwise current. The magnetic dipole moment is also defined as the current flowing through the bar magnet through the area of the bar magnet.