Question

A particle of charge q and mass m moves in a circular orbit of radius r with angular speed $\omega$.The ratio of the magnitude of magnetic moment to that of its angular momentum depends on
a) 0:9
b) 0q:m
c) q:2m
d) o:m

Answer 1

In the question it is asked to determine the ratio of magnetic moment (M) to that of the angular momentum (L) of the charge moving in a circular path. We will first obtain the expression for magnetic moment and angular momentum in terms of the given physical quantities. Further we will take the ratio of the two and accordingly determine on what the ratio depends on.
Formula used:
$M=iA$
$L=m{{r}^{2}}\omega$
$i=fq$

Complete answer:
It is given in the question that the charge ‘q’ of mass ‘m’ moves in a circular path of radius ‘r’ with angular speed $\omega$. Therefore the angular momentum (L) of the charge moving in a circle is given by,
$L=m{{r}^{2}}\omega$
Magnetic moment (M) of the charge moving in the circle is defined as the product of the rate at which the charge moves along the circle i.e. (i)times the area of the circle (A). mathematically this can be represented as,
$M=iA$
The relation between the frequency(f) of the charge and the angular frequency of the charge is given by,
$\begin{aligned} & \omega =2\pi f \\\ & \Rightarrow f=\dfrac{\omega }{2\pi } \\\ \end{aligned}$
The rate at which the charge moves along the circle is given by the product of the charge times the frequency. Hence we can write,
$\begin{aligned} & i=fq \\\ & \Rightarrow i=\dfrac{\omega q}{2\pi } \\\ \end{aligned}$ Substituting this in equation of magnetic moment of the charge we get,
$\begin{aligned} & M=iA \\\ & \Rightarrow M=\dfrac{\omega q}{2\pi }A\text{, A=}\pi {{\text{r}}^{2}} \\\ & \Rightarrow M=\dfrac{\omega q\pi {{\text{r}}^{2}}}{2\pi }=\dfrac{\omega q{{\text{r}}^{2}}}{2} \\\ \end{aligned}$
Now since we know the angular momentum of the charge and the magnetic moment, the ratio of the magnetic moment to that of the angular momentum is,
$\begin{aligned} & \dfrac{M}{L}=\dfrac{\omega q{{\text{r}}^{2}}}{2m{{r}^{2}}\omega } \\\ & \Rightarrow \dfrac{M}{L}=\dfrac{q}{2m} \\\ \end{aligned}$
Hence from the above ratio obtained we can conclude that the ratio depends on the mass and the magnitude charge of the given charge particle.

Hence the correct answer of the above question is option c.

Note:
The rate at which the charge moves in the circle is given as the ratio of magnitude of the charge to that of the time taken for the charge to move in a circle i.e. its time period. Since time period(T) is defined as the reciprocal of the frequency, we got the above equation as a product of the frequency times the charge. The rate at which the charge moves along the circle can also be written as,
$i=fq=\dfrac{q}{T}$.