Moving Charges & Magnetism

Question

A proton of energy 8ev moving circular path in uniform magnetic field.the energy of an Alfa particle moving in same mag field and along same path will be?

Answer 1

Hi Prashant!

 

The formula for the radius (r) of a circle traced out by a charged particle in a uniform magnetic field turns out to be:

 

r = mv/qB

 

where

 

m = mass of particle

v = speed of particle

q = (electric) charge of particle

B = strength of magnetic field

 

(we can derive this if you need to, but, for now, I will take it as a given)

 

If we rearrange this to put v on one side of the equation by itself, it becomes:

 

v = qBr/m

 

I take the energy of the particle to mean its kinetic energy (KE)

 

KE = (1/2)mv²

 

Putting in the expression for v we got above, for a particle moving in a circle, this becomes:

 

KE = (1/2)m(qBr/m)² --> q²B²r²/2m

 

So this is an expression for kinetic energy of a particle of mass m and charge q moving in a circular path of radius r in magnetic field B.

 

The charge and the mass of a particle are important factors in this energy expression.  For the two particles in the problems, we can compare those quantities qualitatively (without numbers) by a knowledge of what an alpha particle is:

 

proton:  mass = m, charge = e

alpha:  mass = 4m, charge = 2e

 

This is because an alpha particle is a helium nucleus, containing two protons (each with mass m and charge e) and two neutrons (each with, effectively, mass m, and no electric charge).

 

For the proton, we can then write the KE expression for the circular path as:

 

KE_p = e²B²r²/2m

 

For the alpha particle, as I read the problem, we assume that the magnetic field B and path radius r are the same as for the proton.  The kinetic energy for that particle then becomes:

 

KE_a = (2e)²B²r²/2(4m) = 4e²B²r²/8m = e²B²r²/2m

 

This is exactly the same expression as for the proton.  So the alpha particle turns out to have the same kinetic energy, 8 eV, in this case.  This comes from the specific circumstance of having increased the mass by a factor of 4, but the charge only by a factor of 2.

 

I hope this helps!  If you want to discuss any more about this, or look at the details, just let me know.