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 grad xx E = (-del B)/(del t). In this expression grad xx E is computed with t held fixed and (del B)/(del t) is calculated with (x,y,z) fixed. Use Stokes' Theorem to derive Faraday's law, oint_C E*dr=-del/(del t) int int_S B * n d sigma,

Question
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 \(\displaystyle\nabla\times{E}=\frac{{-\partial{B}}}{{\partial{t}}}\). In this expression \(\displaystyle\nabla\times{E}\) is computed with t held fixed and \(\displaystyle\frac{{\partial{B}}}{{\partial{t}}}\) is calculated with (x,y,z) fixed.
Use Stokes' Theorem to derive Faraday's law, \(\displaystyle\oint_{{C}}{E}\cdot{d}{r}=-\frac{\partial}{{\partial{t}}}\int\int_{{S}}{B}\cdot{n}{d}\sigma\),

Answers (1)

2021-02-22
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.
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