Question

A charged particle moving in a uniform magnetic field penetrates a layer of lead and thereby losses one half of its kinetic energy. The radius of curvature of its path becomes :

• A. $ \frac{1}{\sqrt{2}} $
• B. $ \sqrt{2} $
• C. $ \frac{3}{\sqrt{2}} $
• D. $ \frac{2}{\sqrt{3}} $

Answer 1

Answer: (A)

The radius of circular path is given by
$ r=\frac{\mu \upsilon }{qB} $
If $ {{E}_{K}} $ be kinetic energy of particle then its momentum
$ p=m\upsilon =\sqrt{2m{{E}_{K}}} $
So, the radius is given by
$ r=\sqrt{\frac{2m{{E}_{K}}}{qB}} $
So, $ r\propto \sqrt{{{E}_{K}}} $
Therefore, when kinetic energy is halved the radius is reduced to
$ \frac{1}{\sqrt{2}} $ times its initial value.