Saved Bookmarks
| 1. |
A sytemconsistsofa longcylindricalanode ofradius a anda coxialcylindricalcylindricalcathodeof radiusb(b lt a). A filamentlocatedalong theaxis of the systemcarriesa heatingcurrentI producinga magneticfieldin the surroundingspace. FInd theleast potentilal differnce betweenthe cathodeandanode at which thetherimal electrons leavingthe cathodewithout intital velcitystart reaching the anode. |
|
Answer» Solution :When `a`CURRENT `I` flows along the axis, a magnetic field `B_(varphi) = (mu_(0) I)/(2pi p)` is set up where `rho^(2) = X^(2) + y^(2)`. In terms of COMPONENTS, `B_(x) = - (mu_(0) ly)/(2pi rho^(2)), B_(y) = (mu_(0) lx)/(2pi rho^(2))` and `B_(x) = 0` Suppose a p.d. `V` is set up between the inner cathode and the other anode. This mean a potential function of the form `varphi = V (In rho//b)/(In a//b), a gt rho gt b`, as one can check by solvingLaplace equaction , The electricfield correspondingto this is, `E_(x) = (Vx)/(rho^(2) In a//b), E_(y) = (Vy)/(rho^(2) In a//b), E_(x) = 0`. The equactions of motion are, `(d)/(dt) mv_(x) = + (|e|V_(z))/(rho^(2) In a//b) + (|e| mu_(0) I)/(2pi rho^(2)) x dotz` `(d)/(dt) mv_(y) = + (|e|V_(z))/(rho^(2) In a//b) + (|e| mu_(0) I)/(2pi rho^(2)) y dotz` and`(d)/(dt) mv_(x) = -|e| (mu_(0)I)/(2pi rho^(2)) (x dotx + y doty) = -|e| (mu_(0) I)/(2pi) (d)/(dt) In rho` `(-|e|)` is the charge on the electron. Intergrating the last equaction, `mv_(e) = -|e| (mm_(0) I)/(2pi) In rho//a = m dotz`.  since `v_(z) = 0` where `rho = a`. We know substitude this `dotz` in the other two equactions to get `(d)/(dt) ((1)/(2) mv_(x)^(2) + (1)/(2) mv_(y)^(2))` `= [(|e|v)/(In a//b) - (|e|^(2))/(m) ((mu_(0) I)/(2pi))^(2) In rho//b]. (x dotx + y doty)/(rho^(2))` `= [(|e|V)/(In a/b) - (|e|^(2))/(m) ((mu_(0) I)/(2pi))^(2) In rho/b]. (1)/(2RHO^(52)) (d)/(dt) rho^(2)` `= [(|e|V)/(In a/b) - (|e|^(2))/(m) ((mu_(0) I)/(2pi))^(2) In rho/b] (d)/(dt) In (rho)/(b)` Intergatingand using`v^(2) = 0`, at `rho = b`, we get, `(1)/(2) mv^(2)= = [(|e|V)/(In (a)/(b)) IN (rho)/(b) - (1)/(2m) |e|^(2) ((mu_(0) I)/(2pi))^(2) (In (rho)/(b))]` The `RHS` must be positive, for all `a gt rho gt b`. The condition for this is, `V ge (1)/(2) (|e|)/(m) ((mu_(0) I)/(2pi))^(2) In (a)/(b)` |
|