Answer
See answers.
Work Step by Step
The kinetic energy is proportional to the square of the radius of curvature. Prove this from the relationship from Example 20-6, page 567, for a charge circling in a uniform B field.
$$r=\frac{mv}{qB} $$
Solve for the speed.
$$v=\frac{rqB}{m} $$
Now write an expression for the KE.
$$KE=\frac{1}{2}mv^2=\frac{1}{2}m(\frac{rqB}{m})^2=\frac{r^2q^2B^2}{2m}$$
Now write an expression for the change in kinetic energy.
$$\Delta KE=\frac{q^2B^2}{2m}(r_f^2-r_i^2)$$
$$\Delta KE=\frac{(1.60\times10^{-19}C)^2(0.010T)^2}{2(1.67\times10^{-27}kg)}((0.0085m)^2-(0.0100m)^2)$$
$$\Delta KE=-2.1\times10^{-20}J=-0.13eV$$
