(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A 5.6 MeV (kinetic energy) proton enters a 0.18 T field, in a plane perpendicular to the field.

What is the radius of its path?

2. Relevant equations

kinetic energy: KE = 1/2 mv^2

velocity: v = sqrt(2KE / m)

radius: r = mV/qB

3. The attempt at a solution

KE = 5.6 MeV = 8.9722x10^-13 J

m(p) = 1.67x10^-27 kg

so v = 3.3x10^7 m/s

r = (m(p))(v) / (q(e))(B)

where m(p) is mass of proton

v is velocity as determined by manipulating the KE equation

q(e) is the charge of the proton which is the same as that of an electron (except positive)

B is the given magnetic field

the homework system I'm on, keeps telling me that my answer is wrong. if you see what i'm doing wrong or where i've made a mistake, please let me know.

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# Homework Help: What is the radius of it's path?

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