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A circular coil carries a current, so it must produce its own magnetic field through it. Would the ‘net’ flux through the area bound by the coil be zero?( because of opposite directions of magnetic field at its two faces?)

If so, Gauss’ Law states that net flux through a closed surface is zero, but the area bound by the coil is not a closed surface.

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Would the ‘net’ flux through the area bound by the coil be zero?( because of opposite directions of magnetic field at its two faces?)

The magnetic field actually is in the same direction on each side of the plane. You can see this in the image below and as explained here. As you can see, there is definitely a magnetic flux through the area bound by the coil itself.

enter image description here

If so, Gauss’ Law states that net flux through a closed surface is zero, but the area bound by the coil is not a closed surface.

Yes, you are right. Gauss's law for magnetic fields tells us $$\oint\mathbf B\cdot\text d\mathbf A=0$$

but this is a surface integral over a closed surface. The area bound by the coil is not a closed surface, so we don't need to worry about this applying here.

So, even though the flux through thiis area is in fact not $0$, I will address a concern you seem to have in linking these two ideas together. You seem to be thinking that a $0$ flux means that the surface integral must have been done over a closed surface. This is not the case. The statement "If the integral is over a closed surface, then the magnetic flux is $0$" is not a biconditional statement. In other words, the statement "If the magnetic flux is $0$ then the integral was done over a closed surface" is a false statement.

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  • $\begingroup$ Thanks! I wasn’t confusing the two statements, just wanted to know if the flux is 0 or not. One more thing; if I consider the two faces of the coil as having their own area vector (pointed in opposite directions), the angle between the two area vectors and magnetic field vector will be 0° and 180°. So one dot product cancels the other. So net flux must be zero? $\endgroup$
    – Tavish
    Mar 3, 2020 at 17:55
  • $\begingroup$ @TavishMusic You can't do fluxes like that. You pick a surface, its orientation, and then you do the integral. $\endgroup$ Mar 3, 2020 at 17:59
  • $\begingroup$ If your two surfaces are separated enough to enclose a volume, then the net flux OUT OF that volume will be zero. $\endgroup$
    – R.W. Bird
    Mar 3, 2020 at 18:12
  • $\begingroup$ @R.W.Bird Ah I hadn't interpreted their comment like that. Yes, if you were considering the enclosed cylinder where those two surfaces were the end, then that would be the case. I thought they were still asking about just those open surfaces, not the closed surface that would include those two surfaces. $\endgroup$ Mar 3, 2020 at 18:14
  • $\begingroup$ What is numeric integration? I came up with an integral for the flux but turns out it can’t be integrated manually. $\endgroup$
    – Tavish
    Mar 4, 2020 at 15:55

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