What is the radius of it's path?

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SUMMARY

The radius of the path of a proton with a kinetic energy of 5.6 MeV in a 0.18 T magnetic field is calculated using the formula r = mV/qB. The kinetic energy is converted to joules, yielding a velocity of 3.3 x 107 m/s. Substituting the mass of the proton (1.67 x 10-27 kg), the charge of the proton (1.6 x 10-19 C), and the magnetic field strength (0.18 T) into the radius formula results in a radius of 1.9 m. The solution confirms the calculations are correct and identifies potential calculation errors as the source of previous incorrect answers.

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  • Understanding of kinetic energy and its conversion to joules
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Homework Statement


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?


Homework Equations


kinetic energy: KE = 1/2 mv^2
velocity: v = sqrt(2KE / m)
radius: r = mV/qB

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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If anyone was in the process of helping me on this one, thank you; but I've figured it out.

I was probably making some sort of calculating error.

r = 1.9 m
 

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