Question

Two electrons $e_1$ and $e_2$ of mass $m$ and charge $q$ are injected into the perpendicular direction of the magnetic field $B$ such that the speed of $e_1$ is double that of $e_2$. The relation of their frequencies of rotation, $f_1$ and $f_2$ respectively. Then

Answer 1

$f_1 = f_2$

Answer 2

The frequency of revolution of a charged particle moving in a uniform magnetic field perpendicular to its velocity is given by the formula:

$f = \frac{qB}{2\pi m}$

Where:

• $q$ is the charge of the particle
• $B$ is the strength of the magnetic field
• $m$ is the mass of the particle

In this problem, we have two electrons, $e_1$ and $e_2$.

1. Charge (q): Both particles are electrons, so their charge is the same, $q_1 = q_2 = e$.
2. Mass (m): Both particles are electrons, so their mass is the same, $m_1 = m_2 = m$.
3. Magnetic Field (B): Both electrons are injected into the same magnetic field, so $B_1 = B_2 = B$.

From the formula, it is evident that the frequency of rotation ($f$) depends only on the charge ($q$), the magnetic field strength ($B$), and the mass ($m$) of the particle. It does not depend on the speed ($v$) of the particle.

Therefore, for electron $e_1$: $f_1 = \frac{eB}{2\pi m}$

And for electron $e_2$: $f_2 = \frac{eB}{2\pi m}$

Since all the parameters (charge, mass, and magnetic field) are identical for both electrons, their frequencies of rotation must be equal, regardless of their speeds. The fact that the speed of $e_1$ is double that of $e_2$ does not affect their frequencies of rotation.

Thus, the relation between their frequencies is: $f_1 = f_2$