As you say, there is no standard definition of the flux through a nonorientable surface. I will try to convince you that you shouldn't want to define this.
There are two standard facts about flux. The first is that, if V is a three dimensional volume, with boundary ∂V, and F is a vector field, then ∫∂VF⋅n=∫V∇⋅F. But a Mobius band cannot be embedded in a closed surface which contains a volume V, so this formula doesn't apply. Physically speaking, what I am saying is that you would never care about Gauss's law for a Mobius band, because you can't make sense of the notion of the band being part of a surface which encloses a charge.
The other key mathematical fact is that, if S is a surface with boundary ∂S and F a vector field, then ∫∂SF=∫S∇×F. I will show that there is no hope of a formula like this for a Mobius strip. Specifically, I will show that there is a vector field F which has ∇×F=0 everywhere on the Mobius strip M, yet ∫∂MF≠0.
Let the z axis pass up through the center of the Mobius strip, with ∂M winding twice around this axis and staying outside a cylinder of radius ϵ around the z-axis. Take the field at point (x,y,z) to be (y/(x2+y2),−x/(x2+y2),0), for any (x,y,z) with x2+y2≥ϵ2, and interpolate however you like within that vertical cylinder. (Physically, this is the B-field from a current along the z-axis. Or, if it isn't, then replace my formula by that one.) Then ∇×F=0 outside the vertical cylinder, and in particular everywhere on the Mobius strip. But ∫∂MF=4π.
I previously had a physical example here, but I think I got some of the details wrong, so it is gone now.
I don't know about the Mobius resistor question. You might get a better answer at physics.stackexchange.com.
This post imported from StackExchange MathOverflow at 2014-12-08 03:46 (UTC), posted by SE-user David Speyer