InterviewSolution
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InterviewSolution
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Magentron is a device consisting of a filament of radius `a` and coaxial cylindrical anode of radius `b` which are located in a unifrom magnetic field parallel to the filament. An accelerating potential differnece `V` is applied between the filament and the anode. Find the value of magnetic induction at which the electrons leaving the filamnent with zero velocity reach the anode. |
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Answer» This differs from the previous problem in `(a harr b)` and the magnetic field is along the z-direction. Thus `B_(x) = B_(y) = 0, B_(x) = B` Assuming as usual the charge of the electron to be `-|e|`, we write the equaction of motion `(d)/(dt) mv_(x) = (|e| V_(x))/(rho^(2) In (b)/(a)) = -|e| B doty, (d)/(dt) mv_(y) = (|e| V_(y))/(rho^(2) In (b)/(a)) + |e| B dotx` and `(d)/(dt) mv_(x) = 0 implies z = 0` The motion is confined to the plane `z = 0`. Eliminating `B` from the first two equactions, `(d)/(dt) ((1)/(2) mv^(2)) = (|e| V)/(In b//a) (x dotx + y doty)/(rho^(2))` or, `(1)/(2) mv^(2) = |e| V (In rho//a)/(In b//a)` so, as expected since magnetic forces do not work, `v = sqrt((2|e|V)/(m))`, at `rho = b`. On the other hand, elaminating `V`, we also get, `(d)/(dt)m (xy_(y) - yv_(x)) = |e| B (x dotx + y doty)` `i.e. (xy_(y) - yv_(x)) = (|e| B)/(2m) rho^(2) +` constatn The constant is easily evaluated, since `v` is zero at `rho = a`. Thus, `(xy_(y) - yv_(x)) = (|e| B)/(2m) (rho^(2) - a^(2)) gt 0` At `rho = b, (xv_(y) - yv_(x)) le vb` Thus, `vb ge (|e|B)/(2m) (b^(2) - a^(2))` or, `B le (2 mb)/(b^(2) - a^(2)) sqrt((2 |e|V)/(m)) xx (1)/(|e|)` or, `B le (2b)/(b^(2) - a^(2)) sqrt((2mB)/(|e|))` |
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