Answer
See answers.
Work Step by Step
a. For a particle moving in a circular path, the angular momentum is L=mvr.
Use the relationship from Example 20-6, page 567.
$$r=\frac{mv}{qB} $$
If the kinetic energy (and the speed) of the particle remains constant while the magnetic field B doubles, we see that the radius of curvature is halved.
The angular momentum of a particle is given by L=mvr. The speed v remains constant while the radius of curvature 2 is cut in half, so the angular momentum is halved.
b. The magnetic dipole moment is defined as M=NIA. In the case of an electron circling the nucleus, the number of turns, N, is 1.
The current I is the charge per unit time passing a point in space. This is the electron’s charge divided by the period of the (circular) motion, I=e/T.
Assume that the electron is moving in a circular orbit of radius r, so that the area of its orbit is $A=\pi r^2$. The period T is the circumference divided by the speed.
$$T=\frac{2\pi r}{v}$$
Finally, the angular momentum of a particle moving in a circle is L=mvr.
Combine the above relationships to solve for the magnetic moment M in terms of the angular momentum L.
$$M=NIA=(1)\frac{e}{T}\pi r^2=\frac{e}{2 \pi r/v}\pi r^2=\frac{e\pi r^2 v}{2 \pi r}$$
$$=\frac{e r v}{2 }=\frac{emvr}{2m }=\frac{e }{2m }mvr=\frac{e }{2m }L$$
$$M=\frac{e }{2m }L$$
This was to be proven.
