Question

A proton is in moving in a circular orbit radius 14cm I a uniform 0.35T magnetic field perpendicular to the velocity and proton find speed of proton

Answer 1

Given:

$R=0.14\:\rm m$

$B=0.35\:\rm T$

$m=1.67\times10^{-27}\:\rm kg$

$q=1.60\times10^{-19}\:\rm C$

The Newton's second law says

$$\frac{mv^2}{R}=qvB$$

Hence

$$v=\frac{qBR}{m}\\ v=\frac{1.60\times 10^{-19}\times 0.35\times 0.14}{1.67\times10^{-27}}=4.7\times 10^6\:\rm m/s$$