A proton with KE $6 M e V$ is travelling in a particle accelerator of circular orbit $0.75 m$ . What fraction of energy does it radiate per sec?

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Solution

Using Larmor's Formula
$P = \frac{2 k q^{2} a^{2}}{3 c^{3}} w h e r e, P = P o w e r r a d i a t e d, k = C o u l o m b^{'} s c o n s t a n t, a = a c c e l e r a t i o n o f p a r t i c l e, c = s p e e d o f l i g h t n o w a = \frac{2 K}{m r}, w h e r e K = K i n e t i c e n e r g y, r = r a d i u s o f c i r c l e n o w p u t t i n g i t i n t h e e q u a t i o n w e g e t \frac{P}{K} = \frac{8 k q^{2} K}{3 m^{2} c^{3}}$
on solving and putting values in correct units we get,
$fraction(PK)=0.047$

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