Two identical current carrying coaxial loops, carry current $I$ in an opposite sense. A simple amperian loop passes through both of them once. Calling the loop as $C$,
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,