Saved Bookmarks
| 1. |
A charged particle moves along the y axis according to the law y = a cos omegat, and the point of observation P is located on the x axis at a distance l from the particle (l gt gt a). Find the ratio of electromagnetic radiation flow densities S_(1)//S_(2) at the point P at the moments when the corrdinate of the particle y_(1) = 0 and y_(2) = a. Calculate that ratio if omega = 3.3.10^(6)s^(-1) and l = 190 m. |
|
Answer» Solution :`P` is a fixed point at a distance `l` form the EQUILIBRIUM position of the particle. `l gt a`, to first in `(a)/(l)` the distance `P` and the instantaneous position of the particle is still `l`. For the first case `y = 0` so `t = T//4` the corresponding retarded time is `t' = (T)/(4)-(l)/(c )` Now, `ddot(y) (t') =- omega^(2) a cos omega ((I)/(4)-(l)/(c )) =- omega^(2) a sin((omega l)/(c ))` Thus`ddot(y) (t') =- omega^(2)a cos ((omegal)/(c ))` The radiation FLUXES in the two CASES are proprtional to `(ddot(y)(t'))^(2)` so `(S_(1))/(S_(2)) = tan^(2)((omegal)/(c)) = 3.06` on subsituation. |
|