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Here is the derivation (http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/toroid.html) of magnetic field produced by a toroid. He has taken the loop to be a circle, and Ampère’s law says that current should pass through the surface created by this loop but as far as I can think no current is passing through the surface of that circular Ampèrean loop rather the current is going through the coil of toroid i.e. current is following a helical path it’s not coming out from bottom to top. So how can we equate that current with the total current crossing the surface of loop. I apologise for being not so clear but I’m trying my fullest.

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Another statement for Ampere's law, is that the magnetic flux through any closed surface "in that case the Amperian loop which is the dashed blue circle in the given link" is directly proportional to the sum of all the current passing through some surface.

To simplify even more, imagine you have some plastic tube where obviously the area of this tube would be our Amperian loop, and imagine you inserted 100 wires inside this plastic tube, then by knowing the current passing through each wire and the direction of the current "whether it is into the plastic or out of the plastic where we would assign different sign for each" then by finding the sum of all these currents we could actually find the magnetic flux circulating around the plastic tube.

Where knowing the circumference of the plastic tube is $2\pi r$, and knowing the sum of the current to be $I$, then the magnetic flux is $$ B = \dfrac{\mu_o ~ I }{ 2 \pi r}$$

Now, back to the given derivation in your link, you would find even though you have only one wire carrying a current $I$, but this current is keep moving in and out of Amperian loop for $N$ turns, it is like you have N wires inside our Amperian loop, and for each turn the wire is contributing a current of $I$ into our Amperian loop and hence the total current which is coming into our Amperian loop is $N I$, and as before the magnetic flux would be $$ B = \dfrac{\mu_o ~ N ~ I }{ 2 \pi r}$$

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