Question

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?

Answer 1

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 = ?

Answer:

$\frac{dE}{dt} = 7.346x10^{14}$ $eV/s$

Explanation:

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

$\frac{dE}{dt} =\frac{q^{2}a^{2} }{6\pi{\epsilon}c^{3}}$  

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

We know that centripetal acceleration is given by

$a = \frac{v^2}{R}$

Whereas kinetic energy is given by

$KE = \frac{1}{2}mv^{2}$

$\frac{2KE}{m} = v^2$

Substitute in the equation of acceleration

$a = \frac{2KE}{mR}$

$a=1.13x10^{34}$ $eV/s^{2}$

Therefore, the corresponding energy radiated per second is

$\frac{dE}{dt} =\frac{(1.60x10^{-19})^{2}(1.13x10^{34})^{2} }{6\pi{\epsilon}c^{3}}$

$\frac{dE}{dt} =\frac{3.27x10^{30} }{6\pi8.854x10^{-12}(3x10^{8}^)^{3}}$

$\frac{dE}{dt} = 7.346x10^{14}$ $eV/s$