Question

A model for quantized motion of an electron in a uniform magnetic field $B$ states that the flux passing through the orbit of the electron is $n(h/e)$ where $n$ is an integer, $h$ is Planck's constant and $e$ is the magnitude of electron's charge. According to the model, the magnetic moment of an electron in its lowest energy state will be ($m$ is the mass of the electron)

• A. $\frac{he}{\pi m}$
• B. $\frac{he}{2\pi m}$
• C. $\frac{heB}{\pi m}$
• D. $\frac{heB}{2\pi m}$

Answer 1

Answer: (B)

$\frac{he}{2\pi m}$

Answer 2

An electron moves in a uniform magnetic field $B$. The magnetic force provides the centripetal force for the circular motion: $evB = \frac{mv^2}{r}$, where $e$ is the magnitude of the electron's charge, $m$ is its mass, $v$ is its speed, and $r$ is the radius of the orbit. From this, we get $v = \frac{eBr}{m}$.

The magnetic flux $\Phi$ passing through the circular orbit of radius $r$ is $\Phi = B \cdot (\text{Area}) = B (\pi r^2)$, assuming the magnetic field is perpendicular to the plane of the orbit.

According to the given model, the magnetic flux is quantized: $\Phi = n(h/e)$, where $n$ is an integer, $h$ is Planck's constant, and $e$ is the magnitude of the electron's charge. So, $B (\pi r^2) = n(h/e)$. This gives $r^2 = \frac{nh}{eB\pi}$.

The magnetic moment $\mu$ of a charged particle moving in a loop is given by $\mu = IA$, where $I$ is the current and $A$ is the area of the loop. The area is $A = \pi r^2$. The current $I$ is the charge $|-e|=e$ passing a point per unit time. The time period of revolution is $T = \frac{2\pi r}{v}$. The frequency of revolution is $f = \frac{1}{T} = \frac{v}{2\pi r}$. The current is $I = e f = e \frac{v}{2\pi r}$.

The magnetic moment is $\mu = I A = \left(e \frac{v}{2\pi r}\right) (\pi r^2) = \frac{1}{2} e v r$.

Substitute the relation $v = \frac{eBr}{m}$ into the magnetic moment formula: $\mu = \frac{1}{2} e \left(\frac{eBr}{m}\right) r = \frac{1}{2} \frac{e^2B}{m} r^2$.

Now substitute the expression for $r^2$ from the quantization condition: $\mu_n = \frac{1}{2} \frac{e^2B}{m} \left(\frac{nh}{eB\pi}\right)$. $\mu_n = \frac{e^2Bnh}{2meB\pi} = \frac{enh}{2m\pi}$.

This is the magnetic moment for the $n$-th state. The lowest energy state corresponds to the lowest possible integer value of $n$. Since the orbit exists (implying $r \neq 0$), $n$ must be a positive integer. The lowest positive integer is $n=1$.

For $n=1$, the magnetic moment in the lowest energy state is: $\mu_1 = \frac{e(1)h}{2m\pi} = \frac{eh}{2\pi m}$.

Comparing this result with the given options, our result matches option (2).