Calculate Proton Energy in eV in Magnetic Field of 1.15T

[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?

 

[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

 

[thepatientmental]

It seems like you may have made a mistake in your conversion from joules to electron volts. Remember, 1 eV is equal to 1.60 x 10^-19 joules. So, to convert from joules to eV, you need to divide by this conversion factor, not multiply. So, the correct calculation would be:

E = (3.95 x 10^-14 J) / (1.60 x 10^-19 J/eV) = 2.47 x 10^5 eV

This is the same value you got for the velocity, but you just need to divide by the conversion factor to get the correct answer in eV. So, the energy of the proton in this scenario would be 2.47 x 10^5 eV or 247 keV. This is very close to the given answer of 4.47 keV, so it seems like the discrepancy was just a conversion error.