Question

a proton moves through a magnetic field of magnitude 2.0 t at a speed of 5.00 x 10⁶ m/s perpendicular to the field. Find a.) the centripetal acceleration and b.) radius of the circular path of proton

Answer 1

(a)

$$a=\frac{F}{m}=\frac{qvB}{m}$$

$$a=\frac{1.6*10^{-19}*5.00*10^6*2.0}{1.67*10^{-27}}=9.58*10^{14}\:\rm m/s^2$$

(b)

$$a=\frac{v^2}{R}$$

$$R=\frac{v^2}{a}=\frac{(5.00*10^6)^2}{9.58*19^{14}}=0.026\:\rm m$$