Question

The CERN particle accelerator is circular with a circumference of 7.0 km.

Required:
a. What is the acceleration of the protons (m=1.67×10^−27kg) that move around the accelerator at of the speed of light? (The speed of light is v=3.00×10^8m/s.)
b. What is the force on the protons?

Answer 1

Step-by-step explanation:

a) centripetal acceleration is the acceleration of a body in a circular path. It is expressed as;

a = mv²/r

m is the mass of proton = 1.67×10^−27kg

v is the velocity = 3.00×10^8m/s

r is the radius

Since C = 2πr

7000m = 2πr

r = 7000/2π

r = 1114.08m

Substitute

a = 1.67×10^−27 (3.00×10^8)²/1114.08

a = 1.67×10^−27 * 9×10^16/1114.08

a = 15.03*10^-11/1114.08

a = 0.001346*10^-11

a = 1.346*10^-14m/s²

b) Force on the proton = mass * acceleration

Force = 1.67×10^−27kg * 1.346*10^-14

Force = 2.246*10^-41N

hence the force on the proton is 2.246*10^-41N