Question

A proton moves through a magnetic field of magnitude 4.5 T at a speed of 7.08×10^6 m/s

Perpendicular to the field. Find the

a. Centripetal acceleration

b. Radius of the circular path of the proton

Answer 1

$qvB=ma,\implies a=\frac{qvB}m=5.6\cdot 10^{18}~\frac{m}{s^2},$

$r=\frac{mv}{qB}=9\cdot 10^{-6}~m.$