# Magnetism Problem

1. Apr 25, 2005

### blue_soda025

A proton moves in a circular path perpendicular to a 1.15-T magnetic field. The radius of its path is 8.40 mm. Calculate the energy of the proton in eV.

So, I calculated the velocity with:
$$qvB = \frac{mv^2}{r}$$
$$(1.60 \times 10^-19)(1.15) = \frac{(6.27 \times 10^-27)v}{0.0084}$$
$$v = 2.47 \times 10^5 m/s$$
Then calculated the energy:
$$E = qv$$
$$E = (1.60 \times 10^-19)(2.47 \times 10^5)$$
$$E = 3.95 \times 10^-14 J$$
Then divided that by eV (1.60 x 10^-19) but I get something like 247000. The answer says 4.47 keV though. What am I doing wrong?

2. Apr 25, 2005

### learningphysics

Mass of a proton is:$$1.67\times10^{-27}kg$$ not the number you used.

Also you have the wrong equation for energy. I think you took E=qV...where V is voltage not velocity... however that equation does not apply here.

The energy of the proton is kinetic energy... so use $$E=\frac{1}{2}mv^2$$