## College Physics (4th Edition)

$F = 7.4 \times 10^{-12}~N$ (directed to the east)
We can find the speed of the proton: $\frac{1}{2}mv^2 = K$ $v^2 = \frac{2K}{m}$ $v = \sqrt{\frac{2K}{m}}$ $v = \sqrt{\frac{(2)(8.0\times 10^{-13}~J)}{1.67\times 10^{-27}~kg}}$ $v = 3.095\times 10^7~m/s$ We can find the magnetic force: $F = q~v\times B$ $F = (1.6\times 10^{-19}~C)(3.095\times 10^7~m/s)(1.5~T)$ $F = 7.4 \times 10^{-12}~N$ By the right hand rule, $v \times B$ is directed east.