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Problem 4: A 8Mev proton enters perpendicular into a uniform magnetic field of 2.5 T. Find (a) force of proton (b) what will be the radius of the path of proton?

Solution

When the charged particle (proton) enters perpendicularly into the magnetic field it experiences a magnetic force which causes the charged particle move in a circular path. In order to find the force on the particle and radius of its circular path, however, we should know the velocity of the particle.

Given data               K.E = 8 MeV = 8 * 106 eV = 8 * 106 * 1.6 * 10-19 = 12.8 * 106-19 = 12.8 * 10-13 J

                                   B = 2.5 T

Required                   F =?                  r =?

Now to find the velocity of the particle we know that

Put the values

Here 1.67 * 10-27 is the mass of the proton.

Formulae

F = evB                                F = mv2/r

For the calculation of radius use the second formula

Put the values

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