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

0.545 m

Step-by-step 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. = (B^(2)q^(2)r^(2))/(2m)`

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

`4.16* 10^(-12) = (1.35^(2)* (1.6* 10^(-19))^(2)r^(2))/(2* 1.67* 10^(-27))`

r = 0.545 m