**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 * 10^{6} eV = 8 * 10^{6} * 1.6 * 10^{-19} = 12.8 * 10^{6-19} = 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 = mv^{2}/r

For the calculation of radius use the second formula

Put the values

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