An electron having charge $e$ is moving with a constant speed $v$ along a circle of radius $r$ . Its magnetic moment will be
(A) $evr$
(B) $\dfrac{{evr}}{2}$
(C) $2\pi rev$
(D) $zero$

Hint: Start with finding current due to circular motion and then we know that magnetic dipole moment is the product of the current flowing and the area so put the value of the current due to circular motion in the formula of the magnetic dipole moment which is the required answer.

Complete answer:
An electron moving in a circular motion having charge e, constant speed v and the radius of circle is R.
Now, current due to the circular motion of the electron will be:
$i = \dfrac{e}{t}$
Where $e$ is charge of electron as given in the question.
And $t$is time period.

We know that, magnetic moment is given by following formula;
$m = i.A$

In case of electron moving with a constant speed in a circle of radius R:
$m = \dfrac{e}{T}\pi {r^2}$

Multiplying and divide by 2 in the above equation, we get;
$m = e{r^2}\dfrac{{2\pi }}{{2T}}$ (equation 1)

Now we know that;
$\omega = \dfrac{{2\pi }}{T}$

Substituting the value in the equation 1, we get;
$m = \dfrac{{ew{r^2}}}{2}$

Also we know:
$\omega = \dfrac{\nu }{r}$

Putting this value in above formula we get;
$m = \dfrac{{e\nu r}}{2}$

Hence the correct answer is Option(B).

Note: The general formula of the magnetic dipole moment is used and then the specific case of the electron having charge e, current I and moving in a circle of radius R is applied in that general formula. The formula of angular velocity is being used here in order to get the required answer.