Question

A proton and an electron travelling along parallel paths enter a region of uniform magnetic field, perpendicular to their paths. Which of them will move in a circular path with higher frequency?

Answer 1

Hint: All the objects in the universe are made up of small undivided particles called atoms. Atoms consist of the proton, electron and the neutron. The flow are the charged particles that make up electricity and their direction is perpendicular to the magnetic field.

Formula used:
The time period of the motion of the particle is given by
\[T = \dfrac{{2\pi m}}{{qB}}\]
Where $T$ is the time period of the charged particle, $m$ is the mass of the particle, $q$ is the charge of the particle and $B$ is the magnetic field.

Complete answer:
It is given in the question that the protons and the neutrons travelling parallel and enters into the magnetic field perpendicularly. By using the formula (1)
\[T = \dfrac{{2\pi m}}{{qB}}\]
We know that the frequency is the reciprocal of the time period.
$\Rightarrow f = \dfrac{1}{T}$
Substituting the value of the time period in the above step.
$\Rightarrow f = \dfrac{{qB}}{{2\pi m}}$
It is known that the electric charge and the magnetic charge for both the electron and the proton are the same. But they differ in the mass. The mass of the electron is much less than the mass of the proton. The particles with the mass less can move faster than the larger mass particles. This is because the frequency is inversely proportional to the mass.
$\Rightarrow f \propto \dfrac{1}{m}$
Thus both the electron and the proton moves in the circular path but the electron moves at a greater frequency than the proton.

Note: The electron is the negatively charged particle. The proton is the positively charged particle with the mass which is $1836$ times the mass of the electron. But these both particles have the same magnitude but differ only in their sign.