# No emf generated on a loop lying outside changing $\vec B$ (but $\vec E ≠ 0$)

If we establish a changing magnetic field in a region of space, circular electric field lines are induced. This electric field extends even after we exit the region of the magnetic field. If a closed path lies completely outside the magnetic field region, it experiences zero emf ( $$\int \vec E \cdot d \vec s = -\frac{d \phi}{dt}=0$$).

My question is, why does the integral cancel out to zero if loop is outside the magnetic field but is non-zero inside where also the electric field lines are circular (geometry is same)? The only difference is the radial dependence of induced electric field ($$\vec E \propto r$$ inside and $$\vec E \propto 1/r$$ outside). Is there an intuitive explanation for this?

Note: kindly use integral equations instead of differential ones in the explanation. Both would be valid, but as a high school student I don't understand them well.

• I'm not entirely clear on where this loop is when you say it's outside the region of the magnetic field, are you saying the material of the loop is in empty space but the magnetic field is still in the loop or that the loop is physically outside the region and the magnetic field does not go through the loop? Dec 28, 2021 at 21:26
• not only that, the real puzzle is that you can have induced electric field even where $B=\nabla\cdot A=0$ but $\dot A \ne 0$ Dec 28, 2021 at 22:18
• @Triatticus I meant physically outside the definite region where where the magnetic field acts. So, let's say from an arbitrary point there is a magnetic field till a distance of R radially. Then neither the loop nor its corresponding area will be within that region. That's why I said $d \phi / dt = 0$.
– Dodo
Dec 29, 2021 at 10:36
• @hyportnex I apologise but I couldn't find the formula $B=\nabla\cdot A$ on the internet. Could you share what this about? (though I did find this: en.wikipedia.org/wiki/Magnetic_vector_potential)
– Dodo
Dec 29, 2021 at 10:51
• apologies, not $\cdot$ but $\times$ as in $\nabla \times A$; ( $A$ is the so-called vector potential whose curl is $B$) Dec 29, 2021 at 15:32

I believe you are describing the Aharonov–Bohm effect total wrong in reverse.

$$E_{ind}$$ in your illustration is actually the $$A$$ magnetic vector potential:

With $$B=0$$ outside you still get a non-zero induced electric field $$E_{in}$$ (i.e. Induced Electric Vectors in your illustration are actually the magnetic vector potentials $$A$$).