Finding a minimum value of a variable that makes an inequality true

Questions
, ,
1

Hello all,

This is probably a basic one, but I’m new to this!

I have the following variables, wz, wc and wp, as defined below;

qe = 1.602176462 * 10^-19 ; (* elementary charge, C *);
melectron = 9.10938188 * 10^-31 (* electron's mass, kg *);
c2 = 0.06 * 10^6(* Curvature of potential well, V\:f00c m2*);
u0 = -1(* voltage, V*);

wz=\[Sqrt]((2(-qe)*c2*u0) /( melectron))

qe = 1.602176462 * 10^-19 ; (* elementary charge, C *)
melectron = 9.10938188 * 10^-31 (* electron's mass, kg *);
B0 = 0.9

wc = (qe / melectron) * B0

wp = (wc / 2) +\[Sqrt]((wc / 2)^2 - ((wz)^2 / 2))

This is for a penning trap design, and the condition that the penning trap works is that;

wp > wc

or alternatively put;

(qe*B0/melectron) < wp

Now, I need to find the minimum value of B0 that will make this inequality work.

I have tried;

Solve[((qe*B0/melectron))<wp, B0]

—But this just gives me an error.

Any advice as to how to get this to work would be much appreciated!

Thanks
Izzy

2

Please check this very carefully to see if I have made any mistakes

qe = 1.602176462 * 10^-19 ; (* elementary charge, C *);
melectron = 9.10938188 * 10^-31 (* electron's mass, kg *);
(* B0 = 0.9 remove that assignment to B0 so that we can solve for B0*)
c2 = 0.06 * 10^6(* Curvature of potential well, V\:f00c m2*);
u0 = -1(* voltage, V*);
wz=\[Sqrt]((2(-qe)*c2*u0) /( melectron));
qe = 1.602176462 * 10^-19 ; (* elementary charge, C *)
melectron = 9.10938188 * 10^-31 (* electron's mass, kg *);
wc = (qe / melectron) * B0;
wp = (wc / 2) +\[Sqrt]((wc / 2)^2 - ((wz)^2 / 2));
Reduce[((qe*B0/melectron))<wp, B0]

which cautions about the method that it used to do this, but that can be ignored here

Reduce was unable to solve the system with inexact coefficients.
The answer was obtained by solving a corresponding exact system
 and numericizing the result

and returns

B0<=-0.00116814
3

Beware of the domain of wp:

Plot[{((qe*B0/melectron)), wp},
 {B0, -.002, .002}]
4

Thank you!

5

Thank you!