Saved Bookmarks
| 1. |
A non-reletivistic proton enters a half-space along the normal to the transverse unifrom magnetic field whose induction equals B = 1.0 T. Find the raatio of the enrgy lost by the proton due to radiation during its motion in the field to its initial kinetic energy. |
|
Answer» Solution :For the semicicirular PATH on the right `(mv^(2))/(R ) = Bev` or `v = (BeR)/(m)`. THUS `K.E. = T = (1)/(2)mv^(2) = (B^(2)e^(2)R^(2))/(2m)` Power radiated `= (1)/(4piepsilon_(0))(2)/(3C^(3)) ((ev^(2))/(R ))^(2)` hence energy radiated `= DeltaW` `= (1)/(4piepsilon_(0)) (2)/(3c^(3)) ((B^(2)e^(3)R)/(m^(2)))^(2).(PIR)/(BeR)m = (B^(3)e^(5)R^(2))/(6epsilon_(0)m^(3)c^(3))` So `(DeltaW)/(T) = (Be^(3))/(3epsilon_(0)c^(3)m^(2)) = 2.06 xx 10^(-18)` (neglecting the change in `v` due to radiation, CORRECT if `DeltaW//T lt lt 1`). 
|
|