Consider an electron to be a uniformly dense sphere of charge, with a total charge of \(-e=-1.60 \cdot 10^{-19} \mathrm{C}\) spinning at an angular frequency, \(\omega\). a) Write an expression for its classical angular momentum of rotation, \(L\) b) Write an expression for its magnetic dipole moment, \(\mu\). c) Find the ratio, \(\gamma_{e}=\mu / L\), known as the gyromagnetic ratio.

For a rotating solid sphere, the angular momentum can be found using the formula: \(L=I\omega\) where \(I\) is the moment of inertia and \(\omega\) is the angular frequency. For a solid sphere of mass \(m\) and radius \(r\), the moment of inertia is given by: \(I=\frac{2}{5}mr^2\) However, we are not given the mass nor the radius of the electron. Instead, we are given its charge and told that it's uniformly dense. We can relate the mass to the charge by density, noting that for a sphere of charge density \(\rho\) and volume \(V\): \(Q=\rho V\) where \(Q = -e\) is the total charge and \(V = \frac{4}{3}\pi r^3\) is the volume of the sphere. We can rearrange this to find the radius in terms of the charge density: \(r=\left(\frac{3Q}{4\pi\rho}\right)^\frac{1}{3}\) Now we can write the moment of inertia in terms of charge density: \(I=\frac{2}{5}m\left(\frac{3Q}{4\pi\rho}\right)^\frac{2}{3}\) Finally, we can write the angular momentum: \(L=I\omega=\frac{2}{5}m\left(\frac{3Q}{4\pi\rho}\right)^\frac{2}{3}\omega\)

For a spinning charged sphere with total charge Q and angular frequency \(\omega\), the magnetic dipole moment can be found using the formula: \(\mu=\frac{1}{2}QR\omega\) where \(R\) is the effective radius (since it's uniformly distributed, this is just the radius \(r\) of the sphere). Using the expression for \(r\) found in step 1, we get: \(\mu=\frac{1}{2}Q\left(\frac{3Q}{4\pi\rho}\right)^\frac{1}{3}\omega\)

To find the gyromagnetic ratio, we need to divide the magnetic dipole moment \(\mu\) by the angular momentum \(L\): \(\gamma_e=\frac{\mu}{L}=\frac{\frac{1}{2}Q\left(\frac{3Q}{4\pi\rho}\right)^\frac{1}{3}\omega}{\frac{2}{5}m\left(\frac{3Q}{4\pi\rho}\right)^\frac{2}{3}\omega}\) The angular frequency \(\omega\) and charge \(Q\) cancel out in the ratio, giving: \(\gamma_e=\frac{5}{2}\frac{m}{\left(\frac{3Q}{4\pi\rho}\right)^\frac{1}{3}}\) This is the expression for the gyromagnetic ratio of the spinning electron.