InterviewSolution
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InterviewSolution
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A narrow beam of identical ions with specific charge `q//m`, possessing different velocities, enters the region of space, where there are unifrom parallel electric and marnetic fields the strength `E` and induction `B`, at the point `O` (see Fig). The beam direction coincides with the `x` axis at the point `O`. A plane screen oriented at right angles to the `x` axis is located at a distance `l` from the point `O`. Find the equation of the trace that the ions leave on the screen. Demonstrate that at `x lt lt l` it is the equaction of a parabola. |
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Answer» The equcation of the trajectory is, `x = (v_(0))/(omega) sin omega t, z = (v_(0))/(omega) (1 - cos omega t), y = (qE)/(2m) t^(2)` as before see (3.384) Now on the screen `x = l`, so `sin omega t = (omega l)/(v_(0))` or, `omega t = sin^(-1) (omega l)/(v_(0))` At that moment, `y = (qE)/(2m omega^(2)) ("sin"^(-1)(omega l)/(v_(0)))^(2)` so, `(omega l)/(v_(0)) = sin sqrt((2m omega^(2) y)/(qE)) = sin sqrt((2q B^(2) y)/(Em))` and `z = (v_(0))/(omega) 2 "sin"^(2) (omega t)/(2) = l "tan" (omega t)/(2)` `= l "tan" (1)/(2) ["sin"^(-1) (omega I)/(v_(0))] = l "tan" sqrt((qB^(2) y)/(2 mE))` For small `z, (qB^(2) y)/(2 mE) = ("tan"^(-1) (z)/(l))^(2) = (z^(2))/(l^(2))` or, `y = (2 mE)/(q B^(2) l^(2)). z^(2)` is a parabola. |
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