Faraday's Law of electromagnetic induction states that the rate of change of magnetic flux linkage is proportional to the $\mathcal{emf}$ induced. For a conductor, the formula goes

$$\mathcal{emf}=N\frac{\Delta\Phi}{\Delta t}$$

Where $N$ is the number of coils within the wire, $\Phi$ is the magnetic flux linkage, and $t$ is time.

However, what I do not understand is how does $N$ become a variable within the formula. The principle underlying electromagnetic induction is the rate of change of flux linkage, but $N$ is independent of area, though related to $B$, magnetic flux density.

Through the relationship


It becomes intuitive if the $N$ refers to the number of coils of the solenoid $X$ which provides the magnetic field and $B$ is the magnetic flux density provided by a single unit coil of solenoid $X$, but I do not quite understand how to perceive the formula if the circumstances are reversed,

Where $N$ becomes the number of coils of a solenoid $Y$ which will have an $ \mathcal{emf}$ induced, while $B$ refers the magnetic flux density of solenoid $X$ which induces the $\mathcal{emf}.$

I wish for clarification of the variable $N$ within these types of phenomenon, because I cannot see an intuitive way to link the concept of magnetic flux linkage and induced $\mathcal{emf}$ with it.

  • $\begingroup$ Faraday's law is that the emf is equal to $-d\Phi_B/dt$, with $\Phi_B$ being the magnetic flux through some 2D sheet. In the specific case where the magnetic field passes through a coil of $N$ loops, the overall area the field passes through is just $N$ times the area of each loop, so one gets emf = $-N d\Phi_{loop}/dt$. The $N$ is there to make the calculation easier, by considering the area and flux through one loop, multiplying by $N$ to get the total area/flux. $\endgroup$ – PhysicsTeacher Nov 24 '19 at 19:53

I think your problem is why does the emf depends on the number of turns in a coil, isn't it?

Imagine a situation like this (sorry for a bad figure). enter image description here I have tried to make a coil of wire with N turns in total (all the turns are connected but if I would have done that then image would have become messy) and the red arrows represent the magnetic field lines. Since each loop has same area so magnetic flux will be the same through all of the loops (I'm considering magnetic field B to be uniform inside the coil) and any change in magnetic field B will cause the same change in magnetic flux in all the loops. So, emf developed in each loop is given by Faraday's Law as $$ \mathcal{emf} = -\frac{d\phi}{dt}~~~~~~~(1)$$ Now, since we have $N$ loops and in each loop the emf is $-\frac{d\phi}{dt}$ therefore total emf from A to B is the sum of these emf and hence $$ \mathcal{emf_{A ~to~ B} } = -N \frac{d\phi}{dt}$$ Your intuitive problem was that why does the emf depends on number of turns when it has been found experimentally that emf depends only on magnetic flux which in it's turn depends on the area of the loop, well as you have seen from mathematics done above, the emf is caused in each loop according to Faraday's Law but due to connectivity we got a total sum of all emf from one end to the other.

Hope it helps.


The emf around a closed loop is $$\mathscr E = -\frac {d\Phi}{dt}$$ in which $$\Phi =\int_S \vec B.d\vec S$$ the surface integral being evaluated over any surface bounded by the loop.

It's quite possible to apply this as it stands to a coil of N turns. The area S bounded by the helix formed by the wire of the coil is almost equal to N times the area of a single turn treated as if in one plane. Thus $$S=NA_{turn}.$$ An advantage of writing the equations as I did at the beginning, without $N$ appearing explicitly, is that they are formally correct for situations in which different amounts of flux may be linked with different turns, as can be the case for a solenoid. The integration takes care of this.

If the same flux, $\phi$ is linked with each turn of a coil we may write $$\Phi =N\int_{A_{turn}} \vec B.d\vec A_{turn}$$ in conjunction with $$\mathscr E = -\frac {d\Phi}{dt}$$ Here we are still regarding the flux as increasing with the number of turns, according to the topological standpoint explained earlier.

An alternative approach is to write

$$\mathscr E = N\frac {d\phi}{dt}.$$ $\phi$ is the flux linked with a single turn, as given by $\phi =\int_{A_{turn}} \vec B.d\vec A_{turn}$, and one can think of the $N$ turns as being in series, so their emfs add.


N is constant for a given system but it can change from system to system.change of area and rate of change of magnetic field however effects the induce emf in a given system.

  • $\begingroup$ I do not quite understand what do you mean $\endgroup$ – Avid Reader Nov 23 '19 at 8:20

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