# Radius of the circular path described by a proton

Protons having a kinetic energy of 20 keV are moving in a plane perpendicular to a magnetic field B = 0.05 T directed out of the plane.
Find the radius of the circular path.

Solution.

First of all, let’s convert the kinetic energy in keV to the kinetic energy in Joules.
It’s a fact that 1 eV = 1.6*10^-19 J .
Hence,

20 keV = 20*10^3 eV = 20*10^3 * 1.6*10^-19 = 3.2*10^-15 J

Now, we could calculate the velocity of the proton.
Use the formula for the kinetic energy (Ek) of a moving object.

Ek = m*v^2/2 , where m – the mass of the proton (1.67*10^-27 kg) and v – the velocity of the proton.

Hence,

Ek = m*v^2/2

3.2*10^-15 = 1.67*10^-27 * v^2/2

v ≈ 1.96*10^6 m/s

It is important to mention that the magnetic forces do no work, so the protons’ velocity will not change in magnitude.

According to the Newton’s second law, in case of a uniform circular motion, the force acting on an object is given by the formula

F = m*a = m*v^2/r , where a – the radial acceleration and r the radius of the circular path.

The force acting on the proton is the magnetic Lorentz force. The magnitude of the Lorentz force F is given by the formula

F = q*v*B*sin(θ), where q is the charge of the proton (1.6*10^-19 C) and θ is the smallest angle between the directions of the vectors v and B.

From the fact that protons are moving in a plane perpendicular to a magnetic field the θ must be 900.

Using two last formulas

F = m*v^2/r
F = q*v*B*sin(θ)

and fact that θ = 900 (or sin(900) = 1), we are getting

m*v^2/r = q*v*B

Finally,

r = m*v/(q*B)

Substitute the values into the formula and solve

r = m*v/(q*B) = 1.67*10^-27*1.96*10^6/(1.6*10^-19*0.05) ≈ 0.41 m

The answer is r ≈ 0.41 m.