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A non-relativistic electronoriginates at a point A lyingon the axis of a striaghtsolenoidand moveswith velocity v at an angle alphato the axis. The magnetic induction of the field is equal to B. Findthe distance r fromthe axisto the point onscreen intowhichthe electronstrickes. The screeen into which the electronstrikes. Thescreen is orientedat rightangles to the axisand is located at a distance l fromthe point A. |
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Answer» Solution :Let us take the point `A` as the origin`O` and the axis of the solenoid as z-axis. At an arbitary moment of time let us resolvethe VELOCITY of electroninto its two recantgularcomponents, `vec(v_(|\|))` alongthe axis and `vec(v_(_|_))` to the axisof solenoid. We know the magnetic force does no work, so thekineticenergy as well as thespeed of theelectron `|vec(v_(_|_))|` will remain constant in the `x-y` plane. THUS `vec(v_(_|_))` can change only its DIRECTION as shwon in the Fig. `vec(v_(|\|))` will remain constant as it is parallel to `vec(B)`. Thus at `t = t` `v_(x) = v_(_|_) cos omega t = v sin alphacos omega t` `v_(y) = v_(_|_) sin omega t = v sin alpha sin omega t` and `v_(z) = v cos alpha`, where `omega = (eB)/(m)` As at `t = 0`, we have `x = y = z = 0`, so the MOTION law of the electron is. `{:(z=v cos alpha t),(x=(v sin alpha)/(omega) sin omega t),(y=(v sin alpha)/(omega)(cos omega t-1)):}]` (The equcation of the helix) On the SCREEN, `z = l`, so `t = (l)/(v cos alpha)`, Then `r^(2) = x^(2) + y^(2) = (2 v^(2) sin^(2) alpha)/(omega^(2)) (1 - "cos" (omega l)/(v cos alpha))` `r = (2 v sin alpha)/(omega)|"sin"(omega l)/(2 v cos alpha)| = 2 (mv)/(eB) sin alpha| "sin"(l eB)/(2 mv cos alpha)|` |
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