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Learning about derivation of solenoid as equivalent bar magnet I came across a step raising some doubt. The first step is to take circular element of thickness $dx$ and then to calculate $B$(the magnetic field) at the point due to this element and finally integrating. But my doubt is the derivation says that there are $n×dx$ turns in $dx$ thickness where $n$ is turns per unit length, but how in real world we can have a wire so thin so as to have $n×dx$ turns in infinitesimal length?

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    $\begingroup$ We don't need to have a thin wire as $n×dx$ is infinitesimally small for any finite $n$. So we are not fitting a finite number of turns in an infinitesimal length.. $\endgroup$ – user243267 Dec 1 '19 at 8:31
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The formula assumes the wires form a perfect cylindrical surface with a uniform current. That model is an approximation of a real solenoid, which has a finite number of wires separated by insulation.

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