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.Part AIf 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?(dE/dt)⋅1sE =

Answer 1

Answer:

 P /K = 1,997 10⁻³⁶  s⁻¹

Explanation:

For this exercise let's start by finding the radiation emitted from the accelerator

       $\frac{dE}{dt}$ = $\frac{q^{2} a^{2} }{6\pi \epsilon_{o} c^{2} }$

the radius of the orbit is the radius of the accelerator a = r = 0.530 m

let's calculate

       \frac{dE}{dt} = [(1.6 10⁻¹⁹)² 0.530²] / [6π 8.85 10⁻¹² (3 108)³]

      P= \frac{dE}{dt}= 1.597 10⁻⁵⁴ W

Now let's reduce the kinetic energy to SI units

       K = 5.0 10⁶ eV (1.6 10⁻¹⁹ J / 1 eV) = 8.0 10⁻¹⁹ J

the fraction of energy emitted is

      P / K = 1.597 10⁻⁵⁴ / 8.0 10⁻¹⁹

      P /K = 1,997 10⁻³⁶  s⁻¹