Confusion regarding Ampere's law

Question

Consider the case in which we have two parallel sheets on the $x$-$y$ plane with a distance $d$ between them. Their orientation is along the $x$ direction. Each sheet has a surface current $I$, where the current on the above sheet is in the $+x$ direction and the current on the below sheet is in the $-x$ direction. If we need to find the magnetic field between these two sheets, we will construct a closed contour on the above sheet and calculate the field as follows according to Ampere's law:

$$\int (B.dl)= \mu I_{enclosed}$$

The closed contour constructed will be a rectangle of approximately zero thickness to enclose the sheet of zero thickness. There will be two sides only of the contour in the positive and negative $y$ directions each with length $w$. Therefore the integration will be: $$B \omega + B\omega = \mu I$$ $$B=\mu I/2\omega$$ The same will be done to the other sheet and the magnetic field between the two sheets will be the same of the contributions from each one. My question is: Why did we consider the four sides of the closed contour while calculating the field for each sheet for the the region between the sheets? Why is it not possible to consider only the side that exists in the region between the sheets and Ampere's law will be: $B\omega=I \mu$??

Moreover, if we need to calculate the field in the region above the two plates it will be zero, since each plate will contribute a magnetic field in opposite directions to each other in the required region. Is this right?

Answer 1

What do you mean by "Why is it not possible to consider only the side that exists in the region between the sheets?" Ampere's law is

$$\oint \vec{B}\cdot d\vec{l}=\mu_0 I_{enc},$$

So Ampere's law requires a closed contour - you can't just integrate over any line and expect it to work. Even worse, if you try to integrate solely over a line between the sheets, how are you expected to calculate the current "enclosed"? There is no current enclosed unless the integral is over a closed contour.

Why the closed contour appears can be tracked back to looking at Maxwell's equations: $\vec{\nabla}\times \vec{B}=\mu_0\vec{J}$. To get Ampere's law from this, integrate over a closed contour and apply Stoke's theorem. If you don't integrate over a closed contour, the result does not hold.

For your second question, yes you are correct, the field outside will be zero. A similar thing happens for solenoids - the field outside of one is calculated to be zero since the contribution from adjacent loops approximately cancels.