Answer to Question #212260 in Electricity and Magnetism for Ani

Question

Derive the expression of the trajectory when a charged particle of 

charge q is subjected to a magnetic field perpendicular to its plane. 

From here comment on the trajectory of the particle.

Answer 1

Force that acts on the particle:

$$F=qvB.$$

Since this force is perpendicular to the velocity vector, the particle experiences centripetal acceleration, which determines the net force:

$$F=m\frac{v^2}{R}.$$

Equiate:

$$qvB=m\frac{v^2}{R},\\\space\\ qB=\frac{mv}{R}.\\\space\\ R=\frac{mv}{qB}$$

The trajectory is a circle with radius R.