I searched and found this question (Solenoid vs coil of wire), which is essentially what I am wondering, but doesn't seem to explain it satisfactorily for me. I'm still left confused.

My book outlines the inductance of a solenoid as such:

$B_{sol} = \frac{\mu_0NI}{l}$

Inductance through a coil is $L = \frac{\phi_m}{I}$, where $\phi_m = N\phi_{m-per-turn}$.

$--> L_{sol} = \frac{\phi_m}{I} = \frac{N\phi_{m-per-turn}}{I} = \frac{NAB_{sol}}{I} = \frac{NA\frac{\mu_0IN}{l}}{I}$. The I's will cancel, and we can rearrange, but this point is where I had trouble. If we already accounted for the N-turns in expanding $\phi_m$ into $N\phi_{m-per-turn}$, then this inner $\phi$ should only be considering the flux through one turn. Then the $B$ term inside of $\phi_{m-per-term}$ should only be the formula for a $B$-field of 1 turn (because if the $\phi$ it's contained in is stating it's only calculating for one turn, its $B$-field can't be for more than 1 turn). However, the $B$ formula contains $\textit{another}$ $N$, which makes me feel like it's being counted twice -- once by the $N\phi$ and another time by the $B$-field formula.

The final formula, by the way, is $L_{sol} = \frac{\mu_0N^2A}{l}$. I know it's correct, but I can't explain why it's an $N^2$, given my reasoning.


1 Answer 1


A solenoid can be thought of as $N$ single loops of wire all connected in series and the magnetic field due to a solenoid is proportional to the number of loops, $N$.

That magnetic field passes through each loop of the solenoid so the flux linkage with one loop of the solenoid is proportional to the magnetic field produced by the solenoid which is proportional to the number of loops, $N$.

So the emf induced in one loop of the solenoid, which is that rate of change of flux though one loop, is proportional to $N$.

However there are $N$ such loops all connected in series each producing an emf proportional to $N$ so the total emf produced for all $N$ loops connected in series is proportional to $N\times N = N^2$.


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