Question 5.58: A thin uniform donut, carrying charge Q and mass M, rotates ...

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{1}{2} \hbar , where \hbar is Planck’s constant. What, then, is the electron’s magnetic dipole moment, in A · m^{2} ? [This semiclassical 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 (e \hbar / 2 m) , where e is the charge of the electron and m is its mass, is called the Bohr magneton.] 

The Blue Check Mark means that this solution has been answered and checked by an expert. This guarantees that the final answer is accurate.
Learn more on how we answer questions.

\text { (a) } I=\frac{Q}{(2 \pi / \omega)}=\frac{Q \omega}{2 \pi} ; a=\pi R^{2} ; m =\frac{Q \omega}{2 \pi} \pi R^{2} \hat{ z }=\frac{Q}{2} \omega R^{2} \hat{ z } . \quad L=R M v=M \omega R^{2} ; L =M \omega R^{2} \hat{ z } .

\frac{m}{L}=\frac{Q}{2} \frac{\omega R^{2}}{M \omega R^{2}}=\frac{Q}{2 M}m =\left(\frac{Q}{2 M}\right) I , and the gyromagnetic ratio is g=\frac{Q}{2 M} .

(b) Because g is independent of R, the same ratio applies to all “donuts”, and hence to the entire sphere (or any other figure of revolution): g=\frac{Q}{2 M} .

\text { (c) } m=\frac{e}{2 m} \frac{\hbar}{2}=\frac{e \hbar}{4 m}=\frac{\left(1.60 \times 10^{-19}\right)\left(1.05 \times 10^{-34}\right)}{4\left(9.11 \times 10^{-31}\right)}=4.61 \times 10^{-24} Am ^{2} .

Related Answered Questions