Question

A charged particle of mass m moving with speed v in a plane perpendicular to a magnetic field experiences a force where q is the amount of charge and $F⃗=(qvB,perpendiculartov⃗)$ , B is the magnetic field strength. Because the force is always perpendicular to the particle’s velocity, the particle undergoes uniform circular motion. Find an expression for the period of the motion. Gravity can be neglected.

Short answer

Period is expressed as $T=\frac{2\pi m}{qB}$

Step-by-step solution

Given information

Given $F⃗=(qvB,perpendiculartov⃗)$

Explanation

As the particle is moving in circular motion, so

$\frac{\left(m{v}^2\right)}{r}=qvB$

Expressed as r

$r=\frac{\left(m{v}^2\right)}{qvB}=\frac{mv}{qB}..........................\left(1\right)$

Time period for this revolutions is

$T=\frac{2\pi r}{v}$

Substitute value of r from equation(1), we get

$$\begin{gathered} T=(2\pi /v)(mv/qB) \\ T=\frac{2\pi m}{qB} \end{gathered}$$