Question

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.

Answer 1

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|))`