Question

A proton with a kinetic energy of 4.9×10⁻¹⁶ Jmoves perpendicular to a magnetic field of 0.26 T

Part A What is the radius of its circular path Express your answer using two significant figures. 2 cIn Submit My Answers Give Up

Answer 1

mv²/r = qvB

m is the mass of the proton, v is the velocity, q is the charge and B is the applied magnetic field

r = mv/qB

0.5mv² = 4.9 $\times$ 10^-16

v² = 4.9 $\times$ 10^-16/ (0.5 $\times$ 1.67 $\times$ 10^-27)

v = 0.766045 $\times$ 10⁶ m/s

So, r = mv/qB

r = 1.67 $\times$ 10^-27 $\times$ 0.766045 $\times$ 10⁶ / ( 1.6 $\times$ 10^-19 $\times$ 0.26)

r = 0.030752 m