Let's consider a square loop of wire of edge length $1$ inside a variable magnetic field such that $ \dfrac{dB}{dt} = k $. Faraday's Law tells that $$ \mathcal E = -\frac{d(BA)}{dt} = -A\frac{dB}{dt} = -kA $$ then the area of the square is $1$ and everything's fine.

The problem is that one could also decide to calculate $\mathcal E$ as the sum of multiple EMFs between some points belonging to the square. Let me choose 4 points each at a distance $1/5$ from every vertex, as shown in the picture. Then in this case $$ \mathcal E = \Delta V_{AB} + \Delta V_{BC} + \Delta V_{CD} + \Delta V_{DA} = 4 \Delta V_{AB} = -4kA_{triangle} $$

I have applied Faraday's law on the closed loop described by one of the 4 triangles bounded by the black wire and the green dashed line.

$$ A_{triangle} = \frac 1 2 \cdot \frac 4 5 \cdot \frac 1 5 = \frac 2 {25} $$

Now $ 4A_{triangle}\neq A_{square}$ and thus the two approaches lead to different values for $\mathcal E$. How is that possible?

  • 2
    $\begingroup$ Why $\Delta V_{AB}=-kA_{triangle}$?? $\endgroup$
    – velut luna
    May 14, 2017 at 11:13
  • $\begingroup$ How do you know that proportionality constant of case of Triangle emf is same as that of square's emf? $\endgroup$ May 14, 2017 at 11:14
  • $\begingroup$ The proportionality constant is the time derivative of the magnetic field, and thus does not depend on the loop considered. $\endgroup$ May 14, 2017 at 11:15
  • $\begingroup$ @velutluna it's the general equation $\mathcal E = -kA$ applied in the particular case in which $A=A_{triangle}$ $\endgroup$ May 14, 2017 at 11:18
  • $\begingroup$ @MarcoV But it should be the emf along the whole triangle, not just $AB$. $\endgroup$
    – velut luna
    May 14, 2017 at 11:19

1 Answer 1


Here, you are using implicitely the fact that for any points $P$ and $Q$ and any path $\Gamma$ going from $P$ to $Q$,

$$ \int\limits_\Gamma \mathbf{E}\cdot \textrm{d}\mathbf{l} = C^{te} = \Delta V_{PQ} $$

We could find an even more convincing contradiction: let's consider an arbitrary point $A$ on the wire. Let's call $\Gamma$ the path that travels along the whole loop from $A$ to $A$, then $\mathcal{E}$ is defined by:

$$\mathcal{E} = \int\limits_{\Gamma} \mathbf{E}\cdot \textrm{d}\mathbf{l} = -\frac{\textrm{d}AB}{\textrm{d}t} = -kA \ne 0$$

Now, consider $\Gamma'$ the trivial path going from $A$ to $A$, ie. $\forall t,\,\Gamma'(t) = A$, then according to the first equation,

$$\mathcal{E} = \int\limits_{\Gamma'} \mathbf{E}\cdot\textrm{d}\mathbf{l} = 0$$

since $\Gamma' = A$ (this is an integral over a point). Finally,

$$ 0 \ne 0$$

Well, there's definitely something wrong here... Induction is fondamentally due to the variations of the electromagnetic field: without $\mathbf{B}$ varying, according to Faraday's law, there wouldn't be any electromotive force, so no induction. However, in the previous proof of the non-nullity of zero, we used a result only true in for static fields: the field $\mathbf{E}$ is conservative, or in other words, the first equation is wrong. So, the symbol $\Delta$ doesn't make sense here, and integrating over different parts of the triangles must lead to different results.

  • $\begingroup$ Your explaination is convincing. But in this example physics.stackexchange.com/a/332999/155486, why is it that making the initial assumption (i.e. E is conservative) leads to the correct result? Is it just a fortunate choice for the loop? $\endgroup$ May 14, 2017 at 11:41
  • 1
    $\begingroup$ This answer doesn't make the assumption that E is conservative: indeed, it says that the circulation of E along any closed path equals the derivative of the flux of B through any surface delimited by the path. However, flux depends on the surface, so this doesn't mean that E is conservative. $\endgroup$
    – Spirine
    May 14, 2017 at 11:52
  • $\begingroup$ What allows to calculate the emf between the two open ends of the moving wire if it is not true that $\int_\Gamma \mathbf E\cdot d\mathbf l = \Delta V_{PQ}$? I still can't see the difference bewteen my approach and the one of the link above... $\endgroup$ May 14, 2017 at 12:16
  • 1
    $\begingroup$ I apologize, my previous comment was irrelevant: in the post linked, $\mathbf{B}$ is static, so $\mathbf{E}$ is conservative. $\endgroup$
    – Spirine
    May 14, 2017 at 12:56

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