Given information:

The functions E(t,x,y,z) nad B(t,x,y,z) represent the electric and magnetic fields at point (x,y,z) at time t.

The Faraday's law is \(\displaystyle\oint_{{C}}{E}\cdot{d}{r}=-\frac{\partial}{{\partial{t}}}\int\int_{{S}}{B}\cdot{n}{d}\sigma\).

Here, C represents a wire loop through which current flows counter clockwise with respect to the surface's unit normal n, giving rise to the voltage \(\displaystyle\oint_{{C}}{E}\cdot{d}{r}\)

The basic principle of the electromagnetic theory is \(\displaystyle\nabla\times{E}=-\frac{{\partial{B}}}{{\partial{t}}}\).

Stokes' Theorem:

Let S be a piecewise cmooth oriented surface having a piece smooth boundary curve C. Let F = Mi+Nj+Pk be a vector field whose components have continuous first partial derivatives on an open region containing S. Then the circulation of F around C in the direction counter clockwise with respect to the surface's unit normal vector n equals the integral of the curl vector field \(\displaystyle\nabla\times{F}\) over S:

\(\displaystyle\oint_{{C}}{F}.{d}{r}=\int\int_{{S}}{\left(\nabla\times{F}\right)}\cdot{n}{d}\sigma\)

Calculation:

Assume that S is a surface in which the Stoke's theorem holds. Then \(\displaystyle\oint_{{C}}{E}\cdot{d}{r}=\int\int_{{S}}{\left(\nabla\times{E}\right)}\cdot{n}{d}\sigma\) ...(1)

By the basic principle of the electromagnetic theory, \(\displaystyle\nabla\times{E}=-\frac{{\partial{B}}}{{\partial{t}}}\).

Then equation(1) becomes,

\(\displaystyle\oint_{{C}}{E}\cdot{d}{r}=\int\int_{{S}}{\left(\nabla\times{E}\right)}\cdot{n}{d}\sigma\)

\(\displaystyle=\int\int_{{S}}-\frac{{\partial{B}}}{{\partial{t}}}\cdot{n}{d}\sigma\)

\(\displaystyle=-\frac{\partial}{{\partial{t}}}\int\int_{{S}}{B}\cdot{n}{d}\sigma\)

Then the voltage around a loop equals the negative of the rate of change of magnetic flux through the loop.

Therefore, the Faraday's law \(\displaystyle\oint_{{C}}{E}\cdot{d}{r}=-\frac{\partial}{{\partial{t}}}\int\int_{{S}}{B}\cdot{n}{d}\sigma\) is derived by using Stoke's theorem.

The functions E(t,x,y,z) nad B(t,x,y,z) represent the electric and magnetic fields at point (x,y,z) at time t.

The Faraday's law is \(\displaystyle\oint_{{C}}{E}\cdot{d}{r}=-\frac{\partial}{{\partial{t}}}\int\int_{{S}}{B}\cdot{n}{d}\sigma\).

Here, C represents a wire loop through which current flows counter clockwise with respect to the surface's unit normal n, giving rise to the voltage \(\displaystyle\oint_{{C}}{E}\cdot{d}{r}\)

The basic principle of the electromagnetic theory is \(\displaystyle\nabla\times{E}=-\frac{{\partial{B}}}{{\partial{t}}}\).

Stokes' Theorem:

Let S be a piecewise cmooth oriented surface having a piece smooth boundary curve C. Let F = Mi+Nj+Pk be a vector field whose components have continuous first partial derivatives on an open region containing S. Then the circulation of F around C in the direction counter clockwise with respect to the surface's unit normal vector n equals the integral of the curl vector field \(\displaystyle\nabla\times{F}\) over S:

\(\displaystyle\oint_{{C}}{F}.{d}{r}=\int\int_{{S}}{\left(\nabla\times{F}\right)}\cdot{n}{d}\sigma\)

Calculation:

Assume that S is a surface in which the Stoke's theorem holds. Then \(\displaystyle\oint_{{C}}{E}\cdot{d}{r}=\int\int_{{S}}{\left(\nabla\times{E}\right)}\cdot{n}{d}\sigma\) ...(1)

By the basic principle of the electromagnetic theory, \(\displaystyle\nabla\times{E}=-\frac{{\partial{B}}}{{\partial{t}}}\).

Then equation(1) becomes,

\(\displaystyle\oint_{{C}}{E}\cdot{d}{r}=\int\int_{{S}}{\left(\nabla\times{E}\right)}\cdot{n}{d}\sigma\)

\(\displaystyle=\int\int_{{S}}-\frac{{\partial{B}}}{{\partial{t}}}\cdot{n}{d}\sigma\)

\(\displaystyle=-\frac{\partial}{{\partial{t}}}\int\int_{{S}}{B}\cdot{n}{d}\sigma\)

Then the voltage around a loop equals the negative of the rate of change of magnetic flux through the loop.

Therefore, the Faraday's law \(\displaystyle\oint_{{C}}{E}\cdot{d}{r}=-\frac{\partial}{{\partial{t}}}\int\int_{{S}}{B}\cdot{n}{d}\sigma\) is derived by using Stoke's theorem.