Consider a Mobius strip with a rod coinciding with its central axis. Consider this desmos figure for reference.
Now consider a field having strength which is radially radially outwards about the axis and decreases inversely with the square of the distance from the axis. If one were to find the dot product of the field with the differential area vector, how would one proceed?
This can be related to this question physics.se: Flux through a Möbius strip (I considered asking it on Math-SE as there was no constructive response in the Phy-SE threads and it's more of a math concept)
Another question: If one were given that the field is circular and its strength is inversely proportional to the distance from the centre, how would one proceed?
I am more interested to know the way of solving this problem than the final expression. My teacher suggested me to use angle of twist, if someone could elaborate that, it would be helpful as well.
Best Answer
The issue here is that the Möbius strip is not orientable. The only global differential of area it has is scalar, not vector, so you cannot take the dot product with respect to it.
Now you can take a double cover of the strip and have a vector differential of area over that double cover. But then the integral is always $0$ since you cover every point twice, with the area vector changing signs between the two visits.
Or you can effectively cut the Möbius strip, treating it as just a twisted rectangle in space, and accepting there is a discontinuity at the cutting line. The result you get will vary on where you make that cut.