A cubical region of space is filled with some uniform electric and magnetic fields. An electron enters the cube across one of its faces with velocity $v$ and a positron enters via opposite face with velocity $-v$. At this instant,
A charged particle would continue to move with a constant velocity in a region wherein,
Verify that the cyclotron frequency $\omega=e B / m$ has the correct dimensions of $[T]^{-1}$.
For a charge particle moving perpendicular to the magnetic field, the magnetic Lorentz forces provides necessary centripetal force for revolution.
$$\frac{m v^2}{R}=q v B$$
On simplifying the terms, we have
$\therefore \quad\frac{q B}{m}=\frac{v}{R}=\omega$
Finding the dimensional formula of angular frequency
$$\therefore\quad[\omega]=\left[\frac{q B}{m}\right]=\left[\frac{v}{R}\right]=[T]^{-1}$$
Show that a force that does no work must be a velocity dependent force.
Let no work is done by a force, so we have
$$\begin{aligned} &\begin{aligned} \mathrm{dW} & =\mathbf{F} \cdot \mathrm{dl}=0 \\ \Rightarrow\quad\text { F. } \mathbf{v} d t & =0 \quad\text { (Since, } \mathbf{d l}=\mathbf{v} d t \text { and } \mathrm{dt} \neq 0 \text { ) }\\ \Rightarrow\quad\text { F. } \mathbf{v} & =0 \end{aligned}\\ \end{aligned}$$
Thus, $F$ must be velocity dependent which implies that angle between $F$ and $v$ is $90^{\circ}$. If $v$ changes (direction), then (directions) F should also change so that above condition is satisfied,
The magnetic force depends on $v$ which depends on the inertial frame of reference. Does then the magnetic force differ from inertial frame to frame? Is it reasonable that the net acceleration has a different value in different frames of reference?
Yes, the magnetic force differ from inertial frame to frame. The magnetic force is frame dependent.
The net acceleration which comes into existing out of this is however, frame independent (non -relativistic physics) for inertial frames.