Question

A 5.0 MeV (kinetic energy) proton enters a 0.21 T field, in a plane perpendicular to the field. What is the radius of its path?

Answer 1

 The force due to the magnetic field generates a centripetal force on the proton.

F = qvB = mv^2/r 
qB = mv/r 
qB = p/r 

kinetic energy (K) = mv^2/2 = p^2/(2m) 

qB = √(2mK) / r 
r = √(2mK) / (qB) 
r = √[(2)(1.67e-27 kg)(5.0 MeV * 1.60e-13 J/MeV)] / [(1.60e-19 C)(0.20 T)] 
r = 1.6 m

Answer 2

We have that the radius of its path  is mathematically given as

R = 1.130 m

Radius of Proton path

Question Parameters:

A 5.0 MeV (kinetic energy) proton enters a 0.21 T field,

Generally the equation for The radius of its path  is mathematically given as

$R =(1/B) \sqrt{2m*KE(in eV)/q}$

Therefore

$R =(1/0.21) \sqrt{2*1.67*10^-27*(2.7*10^6)/1.6*10^-19}\\\\R = (1/0.21) \sqrt {0.0563625}$

R= (1/0.21)(0.23740)

R = 1.130 m

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