What fraction of its energy does accelerating proton shed?

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SUMMARY

The discussion focuses on calculating the energy radiated by a proton with a kinetic energy of 6 MeV traveling in a circular orbit of radius 0.75m. The relevant formula used is the Larmor formula, expressed as dE/dt = (q^2 a^2) / (6*π*Epsilon_0 *c^3). The user calculated the energy radiated per second to be 6.55*10^-39 J/s but initially neglected the radius in their calculations. It was emphasized that the radius is crucial for determining the acceleration in circular motion.

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  • Understanding of kinetic energy and its conversion to joules.
  • Familiarity with the Larmor formula for radiation energy loss.
  • Knowledge of circular motion dynamics and centripetal acceleration.
  • Basic grasp of electromagnetic constants such as epsilon_0.
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Physics students, particle accelerator engineers, and researchers in high-energy physics seeking to understand energy loss mechanisms in circular motion.

Ahlibaba
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Homework Statement


If a proton with a kinetic energy of 6 MeV is traveling in a particle accelerator in a circular orbit of radius 0.75m, what fraction of its energy does it radiate per second?

m = 1.67 * 10^-27
epsilon_0 = 8.854 * 10^-12
c = 3 * 10^8

Homework Equations


dE/dt = (q^2 a^2) / (6*π*Epsilon_0 *c^3)

The Attempt at a Solution


E_kin = 6*1.6*10^-19*10^6 J

calculation of velocity by means of E_kin = 0.5 * m * v^2
v = 0.3397*10^8 ms^-1
v = distance/time
a = derivative of v = -d/v^2

using the equation (Larmor formula) above I calculated 6.55*10^-39 J/s that are radiated.
What bugs me, is that I didn't use the given radius and I pretty sure I'm supposed to use it but I don't know how
I'd really appreciate your help!
 
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Ahlibaba said:
v = distance/time
a = derivative of v = -d/v^2
The first formula assumes a constant velocity, which is equivalent to zero acceleration. That is not the situation you have here.
I don't understand how you got -d/v2 in the second line, and how you plugged in a value for the distance then.

Anyway, you have circular motion, there is a standard formula for the acceleration in circular motion.
The radius is important.
 

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