Answer
See answers.
Work Step by Step
a. Use the relationship from Example 20-6, page 567.
$$r=\frac{mv}{qB} $$
Solve for the speed.
$$v=\frac{rqB}{m} $$
Now write an expression for the KE.
$$KE=\frac{1}{2}mv^2=\frac{1}{2}m(\frac{rqB}{m})^2=\frac{r^2q^2B^2}{2m}$$
The kinetic energy is proportional to the square of the radius of curvature, which was to be shown.
b. For a particle moving in a circular path, the angular momentum is L=mvr.
Use the relationship from Example 20-6, page 567.
$$r=\frac{mv}{qB} $$
Solve for the speed.
$$v=\frac{rqB}{m} $$
Now write an expression for the angular momentum.
$$L=mvr=m(\frac{rqB}{m})r=qBr^2$$
This was to be shown.