It is a well known fact that whenever we want to calculate the emf in a solenoid we usually multiply the changing flux for one loop times $N$, which is the number of turns in the solenoid.

But why is this? For example, in the case of Amperes law, I know that it makes sense to add currents because you are considering the line integral, one can think of it like

$$ \oint \mathbf{B} \cdot \mathbf{dl} = \oint (\mathbf{\sum_i B_i}) \cdot \mathbf{dl} = \sum_i ( \oint \mathbf{B} \cdot \mathbf{dl} ) _ i = \sum_i \mu_0 I = \mu_0 \sum_i I_i $$

In that case, currents clearly should add, but I don't see why currents or turns in the solenoid are added in any "deep" sense (when applying Faraday's law).

edit: Faraday's law cares about $\frac{d\Phi}{dt}$. Now, suppose a solenoid has $N$ turns. What is it about these "turns" that makes the emf greater? How do turns affect the emf? What property do they have such that they intensify an emf? Is it simply because $\Delta \Phi$ is greater since I have more loops?

  • $\begingroup$ for the same reason that the total current through N identical parallel wires is twice the current in one single wire. $\endgroup$
    – user65081
    Nov 12, 2014 at 16:54
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    $\begingroup$ You've answered the question yourself. Two loops is $N$ times the current of one loop. Multiplying by $N$ just gives you the total current circulating. $\endgroup$ Nov 12, 2014 at 17:02
  • $\begingroup$ Hmm. I dont see how I've answered my question. It would've answered my question if I was referring to Ampere's law... But I don't see how this is immediately obvious in the case of Faraday's law... Thanks. $\endgroup$
    – DLV
    Nov 12, 2014 at 17:58
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    $\begingroup$ Perhaps you can expand your question not to include what you understand, but rather what is confusing to you. Ampere's Law and Faraday's Law are analogous to each other in many ways, one deals with the E field and the other with B. $\endgroup$
    – user6972
    Nov 12, 2014 at 18:38
  • $\begingroup$ @user6972 I think what I've added on the edit portion sums up my main questions. Thanks. $\endgroup$
    – DLV
    Nov 12, 2014 at 19:16

2 Answers 2


Think of each turn of the coil as a battery with its own emf. You put all those batteries in series, and end up with a larger total emf, since all the potential differences add up.


I'm assuming you are talking about the flux of the magnetic field generated by a solenoid.

It's because the magnetic field of $N$ solenoids, is approximately the $N$ times greater than the magnetic field of a single solenoid. Hence the magnetic flux, will inherit this $N$, since the dependence is linear.

The magnetic field of an infinite solenoid with $n$ turns per length unit: $\mathbf{B} = \mu_0 in \mathbf{\hat z}$. Assuming $\mathbf{\hat z}$ as the unit vector in direction which the "wires go".

Let $n = \frac{N}{L}$, where $L$ is the length of the solenoid. Hence, comparing the magnetic flux with $N$ turns, $\Phi_N$ and the flux of a single turn solenoid $\Phi$, we have: $$ \Phi_N = \iint_S \mathbf B_N\cdot \mathbf {ds} = \iint_S \frac{\mu_0 iN}{L} \mathbf{\hat z}\cdot \mathbf {ds} = N\iint_S \frac{\mu_0 i}{L} \mathbf{\hat z}\cdot \mathbf {ds} = N\iint_S \mathbf B\cdot \mathbf {ds} = N\Phi $$

Therefore, we conclude: $\Phi_N = N\Phi$. Since the temporal derivative is a linear operator, the emf of a $N$ turn solenoid one is $N$ times greater than single-turned solenoide.


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