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I am supposed to find the expression of the magnetic field on the axis of a solenoid of radius $a$, length $L$, and with $N$ the number of turns.

So in order to find $dB$ (to integrate over dx), I was expecting something like this :

$dB = \dfrac{\mu_0 I a^2}{2 r^3} dx$, where $r$ is the length between a random point A and the turn considered,

but it turns out in the solution that the expression is something like :

$dB = \dfrac{\mu_0 I a^2}{2 r^3}\cdot \dfrac{N}{L} dx$

And I really don't understand what is this $\frac{N}{L}$ factor doing here :/

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  • $\begingroup$ Don't worry, your English is fine in this case :-) $\endgroup$
    – David Z
    Commented Mar 31, 2014 at 19:52
  • $\begingroup$ Thanks for the first answer. I have think a little more about the problem and I would know if this is right : For the total B in a solenoid, we consider B for a single turn times the number of turns, so for a little dB, the number of turns considered in a little $dx$ interval becomes $\frac{N\cdot dx}{L}$. $\endgroup$
    – dcholleton
    Commented Mar 31, 2014 at 20:47
  • $\begingroup$ Take a look at this question. It will clear your doubt physics.stackexchange.com/questions/113232/… $\endgroup$
    – noir1993
    Commented Dec 15, 2014 at 4:29

3 Answers 3

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For a gut intuition: your expected result looks suspiciously similar to the magnetic field created by a single loop. A solenoid can be approximated by a bunch of loops together, so if you have more loops in the same length (tighter solenoid), you should get a bigger magnetic field.

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The total currect enclosed by the Amperian loop is $I_{enc} = NI$

The division by $L$ comes by applying Ampere's law $\oint{B\cdot{dl}} = BL$

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It is denoted as "surface current sheet" which is equal to $$K=\frac {NI}{L} \ (A-turn)/m$$ So $\frac {NI}{L}dx $ is differential surface current sheet.

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