Electromagnetic radiation is emitted by accelerating charges. The rate at which energy is emitted from an accelerating charge that has charge q and acceleration a is given by dEdt=q2a26πϵ0c3 where c is the speed of light. If a proton with a kinetic energy of 5.0 MeV is traveling in a particle accelerator in a circular orbit with a radius of 0.530 m, what fraction of its energy does it radiate per second?

Given Information:

Kinetic energy of proton = KE = 5 MeV

Radius = R = 0530 m

Speed of light = c = 3x10⁸ m/s

Permittivity of free space = ε = 8.854×10⁻¹² F/m

Mass of proton m = 1.67x10⁻²⁷ kg

Required Information:

Energy radiated per second = dE/dt = ?

Explanation:

The rate at which energy is emitted from an accelerating charge with charge q and acceleration a is given by

Where ε is the permittivity of free space and c is the speed of light

We know that centripetal acceleration is given by

Whereas kinetic energy is given by

Substitute in the equation of acceleration

Therefore, the corresponding energy radiated per second is

a person with a mass of 106kg has a weight of 1060n