Saved Bookmarks
| 1. |
Magentron is a device consisting of a filamentof radiusa and coaxialcylindricalanodeof radiusbwhich are locatedinaunifrom magneticfield parallel to the filament. An acceleratingpotentialdiffernece V is appliedbetweenthe filamentand theanode. Find thevalue of magneticinduction at whichthe electrons leaving the filamnentwith zero velocity reach the anode. |
|
Answer» Solution :This differs from theprevious problemin `(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 chargeof the electronto be `-|E|`, we writethe 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` fromthe 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|))` |
|