# Why is the change in flux in Faradays law proportional to the area enclosed by the wire rather than the wire area perpendicular to the magnetic field?

When deriving Faraday's law in high school physics the change of the area enclosed by the loop of wire is said to be proportional to the change of magnetic flux in Faraday's law. Intuitively, it makes little sense that the magnetic field running through the area enclosed by the loop of wire would induce a voltage rather than the magnetic field running through the wire itself.

Why is the change in flux in Faraday's law proportional to the change in area enclosed by the wire rather than the change in wire area perpendicular to the magnetic field?

Intuitively, it makes little sense that the magnetic field running through the area enclosed by the loop of wire would induce a voltage

It's just Stokes' theorem which isn't particularly intuitive. First, recall this point relationship from Maxwell's equations:

$$\nabla \times \vec E = -\frac{\partial\vec B}{\partial t}$$

There is no area involved here; this equation relates the curl of the electric field at a point to the time rate of change of the magnetic field at the same point.

Now, integrate to find the flux of the vector fields through a surface $\Sigma$:

$$\iint_\Sigma \left(\nabla \times \vec E\right) \cdot \vec n\,\mathrm{d}\Sigma = -\iint_\Sigma \frac{\partial\vec B}{\partial t} \cdot \vec n\,\mathrm{d}\Sigma$$

Now, apply Stokes' theorem on the left hand side:

$$\iint_\Sigma \left(\nabla \times \vec E\right) \cdot \vec n\,\mathrm{d}\Sigma = \oint_{\partial\Sigma} \vec E \cdot \mathrm{d}\vec l$$

By Stokes' theorem, the line integral of the electric field along a closed contour equals the surface integral of the curl of the electric field through a surface bounded by the contour. And so

$$\oint_{\partial\Sigma} \vec E \cdot \mathrm{d}\vec l = -\iint_\Sigma \frac{\partial\vec B}{\partial t} \cdot \vec n\,\mathrm{d}\Sigma$$

• Hi, I appreciate your answer, but could you still explain why the result is true? It has been proved to be mathematically true, but it still doesn't really make sense as to why flux through the center would induce current. Could you give a physical explanation as to why we should consider the flux through the center of the coil? Coincidentally, I outline my queries in my (yet to be answered question) :physics.stackexchange.com/questions/445083/… – John Hon Dec 8 '18 at 7:18