Question

A thin uniform donut, carrying charge Q and mass M, rotates about its axis as shown in Fig. 5.64.

(a) Find the ratio of its magnetic dipole moment to its angular momentum. This is called the gyromagnetic ratio (or magnetomechanical ratio).

(b) What is the gyromagnetic ratio for a uniform spinning sphere?[This requires no new calculation; simply decompose the sphere into infinitesimal rings, and apply the result of part (a).]

(c) According to quantum mechanics, the angular momentum of a spinning electron is $\frac{\mathbf{1}}{\mathbf{2}}\mathbf{}\mathbf{\hslash }$ , where $\mathbf{}\mathbf{\hslash }$ is Planck's constant. What, then, is the electron's magnetic dipole moment, in localid="1657713870556" $\mathit{A}{\mathit{m}}^{\mathbf{2}}$ ? [This semi classical value is actually off by a factor of almost exactly 2. Dirac's relativistic electron theory got the 2 right, and Feynman, Schwinger, and Tomonaga later calculated tiny further corrections. The determination of the electron's magnetic dipole moment remains the finest achievement of quantum electrodynamics, and exhibits perhaps the most stunningly precise agreement between theory and experiment in all of physics. Incidentally, the quantity(localid="1657713972487" $\mathbf{(}\mathit{e}\mathbf{\hslash }\mathbf{/}\mathbf{2}\mathit{m}\mathbf{}\mathbf{)}$, where e is the charge of the electron and m is its mass, is called the Bohr magneton.]

Short answer

(a) The gyromagnetic ratio of donut is $\frac{Q}{2M}$.

(b) The gyromagnetic ratio of uniform spinning sphere is also$\frac{Q}{2M}$ .

(c) The magnetic dipole moment of electron is $4.61\times {10}^{-24}\hspace{0.33em}A\cdot m^2$.

Step-by-step solution

Determine the gyromagnetic ratio(a)

The time period of rotation of donut is given as:

$\mathit{t}\mathbf{=}\frac{\mathbf{2}\mathbf{\pi }}{\mathbf{\omega }}$

The current in the donut is given as:

$$\begin{gathered} I=\frac{\mathrm{Q}}{T} \\ I=\frac{\mathrm{Q}}{\left(\frac{2\pi }{\omega }\right)} \\ \mathit{I}\mathbf{=}\frac{\mathbf{Q\omega }}{\left(2\pi \right)} \end{gathered}$$

The cross sectional area of the donut is given as:

$A=\pi R^2$

The magnetic dipole moment of donut is given as:

$$\begin{gathered} m=1·A \\ m=\left(\frac{Q\omega }{2\pi }\right)\left(\pi R^2\right) \\ m=\frac{Q\omega R^2}{2} \end{gathered}$$

The angular momentum of donut is given as:

$L=M\omega R^2$

The gyromagnetic ratio of the donut is given as:

$g=\frac{m}{L}$

Substitute all the values in the above equation.

localid="1657721639031" $$\begin{gathered} g=\frac{\frac{Q\omega R^2}{2}}{M\omega R^2} \\ g=\frac{Q}{2M} \end{gathered}$$

Therefore, the gyromagnetic ratio of donut is localid="1657721706283" $\frac{Q}{2M}$.

Determine the gyromagnetic ratio of uniform spinning sphere(b)

The gyromagnetic ratio of the donut does not depend on the geometric feature (radius) of donut so the gyromagnetic ratio of uniform spinning sphere would be same as of donut.

Therefore, the gyromagnetic ratio of uniform spinning sphere is also$\frac{Q}{2M}$ .

Determine the magnetic dipole moment of electron(c)

Consider the expression for the magnetic dipole moment:

$m_{{}_e}=\frac{et}{4m}$

Here, $\hslash$ is Planck’s constant and its value is $1.05\times {10}^{-34}J\cdot s$, $e$ is the charge of electron and its value is $1.6\times {10}^{-19}\hspace{0.33em}C$, $m$ is the mass of electron and its value is $9.11\times {10}^{-31}kg$.

Substitute all the values in the above equation.

$$\begin{gathered} m_e=\frac{\left(1.6\times {10}^{-19}\mathrm{C}\right)\left(1.05\times {10}^{-34}\mathrm{J}·\mathrm{s}\right)}{4\left(9.11\times {10}^{-31}\mathrm{kg}\right)} \\ {m}_{\mathrm{e}}=4.61\times {10}^{-24}\mathrm{A}·{\mathrm{m}}^2 \end{gathered}$$

Therefore, the magnetic dipole moment of electron is $4.61\times {10}^{-24}\mathrm{A}·{\mathrm{m}}^2$.