Write the expression for Lorentz magnetic force on a particle of charge $q$ moving with velocity $v$ in a magnetic field $B$. Shown that two no work is done by this force on the charged particle.

Hint
Lorentz force, the force exerted on a charged particle $q$ moving with velocity $v$ through an electric field $E$ and magnetic field $B$. The entire electromagnetic force $F$ on the charged particle is called the Lorentz force.

Complete step by step answer
We know that,
Lorentz force = magnetic force + electric force.
So, now we can say,
$F{\text{ }} = {\text{ }}[{\text{ }}qvb{\text{ }}sin\theta \; + {\text{ }}qe{\text{ }}]$
$ \Rightarrow \vec F = q(\vec V \times \vec B)\;d\vec s$
Now, $\vec F$ is perpendicular to both $\vec V$ and $\vec B$.
If $d\vec s$ is the instantaneous displacement of the change-
Then, $d\vec s$ is also perpendicular to $\vec F$
Now, according to work done formula,
$W = \vec F.d\vec s$
$ \Rightarrow W = Fs\cos {90^0 }$
But, the value of $cos 90^0$ is equal to zero.
So, $W = 0$,
That means the work done is zero and the increase in kinetic energy is zero.

Note
The work is done when a force acts upon an object to cause a displacement. Three quantities must be known in order to calculate the amount of work. Those three quantities are force, displacement and the angle between the force and the displacement.