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.(adsbygoogle = window.adsbygoogle || []).push({});

So, I calculated the velocity with:

[tex]qvB = \frac{mv^2}{r}[/tex]

[tex](1.60 \times 10^-19)(1.15) = \frac{(6.27 \times 10^-27)v}{0.0084}[/tex]

[tex]v = 2.47 \times 10^5 m/s[/tex]

Then calculated the energy:

[tex]E = qv[/tex]

[tex]E = (1.60 \times 10^-19)(2.47 \times 10^5)[/tex]

[tex]E = 3.95 \times 10^-14 J[/tex]

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?

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# Homework Help: Magnetism Problem

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