Question

A proton makes a (radius = 30.0 cm) circular orbit in a 2.5 T magnetic field. What is the velocity of the proton?

mass of proton = 1.67x10^-27 kg, e = 1.602x10^-19 C

Answer 1

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

$$v=\frac{qBr}{m}=\frac{1.6*10^{-19}*2.5*0.30}{1.67*10^{-27}}=7.2*10^7\:\rm m/s$$