Question

If an electron and a proton enter normally into a uniform magnetic field with equal kinetic energies, then:

• A. the electron travels in circular path of less radius
• B. the proton travels in circular path of less radius
• C. both electron and proton travel in circular paths of same radius
• D. both travel in a straight line

Hint:

In a magnetic field, the radius of a charged particle’s path depends on its mass: $r \propto \sqrt{m}$ for equal kinetic energies. The lighter particle (electron) has a smaller radius.

Answer 1

The Correct Option is A

For a charged particle in a magnetic field, radius of circular path $r = \frac{m v}{q B}$.
Kinetic energy $KE = \frac{1}{2} m v^2 \implies v = \sqrt{\frac{2 KE}{m}}$.
So, $r = \frac{m \sqrt{\frac{2 KE}{m}}}{q B} = \frac{\sqrt{2 KE m}}{q B}$.
For electron and proton, $KE$ is same, $q$ is same (magnitude), but $m_e \ll m_p$.
Thus, $r \propto \sqrt{m}$, so $r_e<r_p$ (electron has smaller mass, hence smaller radius).