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A charged particle moves uniformaly with velocity v along a circle of radius R in the plane xy (fig.) An observer is located on the x axis at a point P which is removed form the centre of the circle by a distance much exceeding R. Find: (a) the relatiship between the observed valuesof the y projection of the particle's acceleration and the y corrdinate of the particle: (b) the ratio of electromagnetic radiation flow densities S_(1)//S_(2) at the point P at the moments of time when the particle moves, form the standpoint of the observer P, toward him and away from him, as shwon in the figure. |
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Answer» Solution :Along the circule `x = R sin omega t, y = R cos omegat` where `omega =(v)/(R )`. If `t` is the parameter in `x(t), y(t)` and `t'` is the observer time then `t' = t+(l-x(t))/(C )` where we have neglect the effect of the `y-`cordinate which is of second order. then observed cordinate are `x'(t') = x(t),y'(t') = y(t)` then `(dy')/(dt') = (dy)/(dt') = (dt)/(dt')(dy)/(dt) = (-omegaR sin omegat)/(1-(omegaR)/(c ) cos omegat) = (-omegax)/(1-(omegay)/(c )) =(-vx//R)/(1-(vy)/(CR))` and `(d^(2)y')/(dt'^(2)) = (dt)/(dt')(d)/(dt) ((-vx//R)/(1-(vy)/(cR)))` `=(1)/(1-(vy)/(cR)){(-(v^(2))/(R^(2))y)/(1-(vy)/(cR))+((vx)/(R )(v^(2)/(cR^(2))x))/((1-(vy)/(cR))^(2))} = (v^(2))/(R)(((v)/(c )-(y)/(R)))/((1-(vy)/(cR))^(3))` This is the observed acceleration. (B) Energy flow density of `EM` radiation `S` is proportional to the square of the `y`-projecton of the observed acceleration of the partcile (i.e `(d^(2)y')/(dt'^(2)))`. THUS `(S_(1))/(S_(2)) =[(((v)/(c)-1))/((1-(v)/(c))^(3))//(((v)/(c)+1))/((1+(v)/(c))^(3))]^(2) = ((1+(v)/(c))^(4))/((1-(v)/(c))^(4))` 
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