 # If E(t,x,y,z) and B(t,x,y,z)represent the electric and magnetic fields at point (x,y,z) at time t, a basic principle of electromagnetic theory says th nicekikah 2021-02-21 Answered
If E(t,x,y,z) and B(t,x,y,z)represent the electric and magnetic fields at point (x,y,z) at time t, a basic principle of electromagnetic theory says that $\mathrm{\nabla }×E=\frac{-\partial B}{\partial t}$. In this expression $\mathrm{\nabla }×E$ is computed with t held fixed and $\frac{\partial B}{\partial t}$ is calculated with (x,y,z) fixed.
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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 ${\oint }_{C}E\cdot dr=-\frac{\partial }{\partial t}\int {\int }_{S}B\cdot nd\sigma$.
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 ${\oint }_{C}E\cdot dr$
The basic principle of the electromagnetic theory is $\mathrm{\nabla }×E=-\frac{\partial B}{\partial t}$.
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 $\mathrm{\nabla }×F$ over S:

Calculation:
Assume that S is a surface in which the Stoke's theorem holds. Then ${\oint }_{C}E\cdot dr=\int {\int }_{S}\left(\mathrm{\nabla }×E\right)\cdot nd\sigma$ ...(1)
By the basic principle of the electromagnetic theory, $\mathrm{\nabla }×E=-\frac{\partial B}{\partial t}$.
Then equation(1) becomes,
${\oint }_{C}E\cdot dr=\int {\int }_{S}\left(\mathrm{\nabla }×E\right)\cdot nd\sigma$
$=\int {\int }_{S}-\frac{\partial B}{\partial t}\cdot nd\sigma$
$=-\frac{\partial }{\partial t}\int {\int }_{S}B\cdot nd\sigma$
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 ${\oint }_{C}E\cdot dr=-\frac{\partial }{\partial t}\int {\int }_{S}B\cdot nd\sigma$ is derived by using Stoke's theorem.