Question

A Review Constants 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 where c όπερς dt is the speed of light dE Part A If a proton with a kinetic energy of 6.2 MeV is traveling in a particle accelerator in a circular orbit with a radius of 0.620 m, what fraction of its energy does it radiate per second? | ΛΣφ (dE/dt)-1 s E Submit Request Answer

Answer 1

The acceleration of the proton is given by

Here r is the radius of the orbit. The velocity of the proton can be determined by the kinetic energy given as

$\frac{m_p c^2}{\sqrt{1-v^2/c^2}}-m_pc^2 = 6.2$

The rest mass energy of proton m_pc²= 938 MeV

Putting it in above equation and solving for v gives

v= 0.1144 *c

So the rate of energy emitted is

$\frac{\mathrm{d}E }{\mathrm{d} t} = \frac{q^2 \times(0.1144)^4 c}{6 \pi \epsilon_0 r^2}$

putting q = 1.6*10^-19 C

c=3*10^8 m/s

r =0.62 m

$\epsilon_0$ = 8.854 *10^-12 C^2/Nm^2

dE/dt = 2.05113*10^-23 J

E= 6.2 MeV = 6.2 *1.6*10^-13 = 9.92 *10^-13 J

(dE/dt)/E = 2.066 *10^-11 per sec

2) if v=r => a= v = 0.1144*c

$\frac{\mathrm{d}E }{\mathrm{d} t} = \frac{q^2 \times(0.1144)^2}{6 \pi \epsilon_0 c} = 6.69277 \times10^{-39}~J$

(dE/dt)/E = 6.747*10^-27 per sec