Question

Two electrons, \(e_1\) and \(e_2\), of mass \(m\) and charge \(q\) are injected in the perpendicular direction of the magnetic field \(B\) such that the kinetic energy of \(e_1\) is double than that of \(e_2\). The relation of their frequencies of rotation, \(f_1\) and \(f_2\), is

• A. \(f_1=f_2\)
• B. \(f_1=2f_2\)
• C. \(2f_1=f_2\)
• D. \(4f_1=f_2\)

Hint:

Cyclotron frequency \(f=\frac{qB}{2\pi m}\) is independent of speed and kinetic energy. Kinetic energy affects only the radius of the circular path.

Answer 1

The Correct Option is A

Step 1: Recall circular motion in a field.
A charge moving perpendicular to a uniform field $B$ goes in a circle. Its rotation frequency (cyclotron frequency) is \[ f = \frac{qB}{2\pi m}. \]
Step 2: Spot what the frequency depends on.
The formula contains only $q$, $B$ and $m$. Speed and kinetic energy do not appear.
Step 3: Compare the two electrons.
Both $e_1$ and $e_2$ are electrons, so they share the same charge $q$ and mass $m$, and they sit in the same field $B$.
Step 4: Account for the energy difference.
$e_1$ has twice the kinetic energy of $e_2$, meaning it moves faster. A faster particle traces a bigger circle, but in just the right way to keep the time per revolution the same.
Step 5: Compare frequencies.
Since every factor in $f = \dfrac{qB}{2\pi m}$ is identical for both, \[ f_1 = f_2. \]
Step 6: Conclude.
The extra energy changes only the radius, not the frequency, so the two frequencies are equal.
\[ \boxed{f_1 = f_2} \]