Question

Consider a 26-MeV proton moving perpendicularly to a 1.35 T field in a cyclotron. Find the radius of curvature, in meters, of the path of the proton while moving through the cyclotron.

Answer 1

Answer:

0.545 m

Explanation:

K. E = 26 MeV = 26 x 1.6 x 10^-13 J = 4.16 x 10^-12 J

B = 1.35 T

Let r be the radius of curvature

The formula for the kinetic energy of a cyclotron is given by

$K.E. = \frac{B^{2}q^{2}r^{2}}{2m}$

m = 1.67 x 10^-27 kg, q = 1.6 x 10^-19 c

$4.16\times 10^{-12} = \frac{1.35^{2}\times (1.6\times 10^{-19})^{2}r^{2}}{2\times 1.67\times 10^{-27}}$

r = 0.545 m