InterviewSolution
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InterviewSolution
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At the momentt = 0 a relativsiticproton files with avelocityv_(0) into theregion where there is a unifromtransverseelectric field of strengthE, withv_(0) _|_ E. Findthe time dependence of (a)theangle theta between the proton's velocity vector v and the initial direction of its motion, (b) the projection v_(x)of the vector v on the initialdirection of motion. |
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Answer» Solution :The EQUACTIONS are, `(d)/(dt) ((m_(0) v_(x))/(sqrt(1 - (v^(2)//C^(2))))) = 0` and `(d)/(dt) ((m_(0) v_(y))/(sqrt(1 - (v^(2)//c^(2))))) = E E` Hence, `(v_(x))/(sqrt(1 - v^(2)//c^(2)))` = CONSTANT `= (v_(0))/(sqrt(1 - (v_(0)^(2)//c^(2))))` Also, by energy conservation, `(m_(0) c^(2))/(sqrt(1 - (v^(2)//c^(2)))) = (m_(0) c^(2))/(sqrt(1 - (v_(0)^(2)//c^(2)))) + e Ey` Dividing `v_(x) = (v_(0) epsilon_(0))/(epsilon_(0) + e Ey), epsilon_(0) = (m_(0) c^(2))/(sqrt(1 - (v_(0)^(2)//c^(2))))` Also, `(m_(0))/(sqrt(1 - (v^(2)//c^(2)))) = (epsilon_(0) + e E y)/(c^(2))` Thus, `(epsilon_(0) + e Ey) v_(y) = c^(2) e E t +` constant Intergating again, `epsilon_(0) y + (1)/(2) e E y^(2) = (1)/(2) c^(2) E t^(2) +` CONSTNAT. "contast" = 0, as `y = 0`, at `t = 0`. Thus, `(ce E t)^(2) = (e y E)^(2) + 2 epsilon_(0) e E y + epsilon_(0)^(2) - epsilon_(0)^(2)` or, `ceEt = sqrt((epsilon_(0) + e Ey)^(2) - epsilon_(0)^(2))` or, `epsilon_(0) + e Ey = sqrt(epsilon_(0)^(2) + c^(2) e^(2) E^(2) t^(2))` Hence, `v_(x) = (v_(0) epsilon_(0))/(sqrt(epsilon_(0)^(2) + c^(2) e^(2) E^(2) t^(2)))` also, `v_(y) = (c^(2) e Et)/(sqrt(epsilon_(0)^(2) + c^(2) e^(2) E^(2) t^(2)))` and`tan theta = (v_(y))/(v_(x)) = (e Et)/(m_(0) v_(0)) sqrt(1 - (v_(0)^(2)//c^(2)))`. |
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