# 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, $$\displaystyle\oint_{{C}}{E}\cdot{d}{r}=-\frac{\partial}{{\partial{t}}}\int\int_{{S}}{B}\cdot{n}{d}\sigma$$,

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.

### Relevant Questions

Use Green's Theorem in the form of this equation to prove Green's first identity, where D and C satisfy the hypothesis of Green's Theorem and the appropriate partial derivatives of f and g exist and are continuous. (The quantity grad g · n = Dng occurs in the line integral. This is the directional derivative in the direction of the normal vector n and is called the normal derivative of g.)
$$\displaystyle\oint_{{c}}{F}\cdot{n}{d}{s}=\int\int_{{D}}\div{F}{\left({x},{y}\right)}{d}{A}$$
Use Stokes' theorem to evaluate the line integral $$\displaystyle\oint_{{C}}{F}\cdot{d}{r}$$ where A = -yi + xj and C is the boundary of the ellipse $$\displaystyle\frac{{x}^{{2}}}{{a}^{{2}}}+\frac{{y}^{{2}}}{{b}^{{2}}}={1},{z}={0}$$.
Use Stokes' Theorem to evaluate $$\displaystyle\int\int_{{S}}{C}{U}{R}{L}{f}\cdot{d}{S}$$.
$$\displaystyle{F}{\left({x},{y},{z}\right)}={x}^{{2}}{y}^{{3}}{z}{i}+{\sin{{\left({x}{y}{z}\right)}}}{j}+{x}{y}{z}{k}$$,
S is the part of the cone $$\displaystyle{y}^{{2}}={x}^{{2}}+{z}^{{2}}$$ that lies between the planes y = 0 and y = 2, oriented in the direction of the positive y-axis.
z = x Let be the curve of intersection of the cylinder $$\displaystyle{x}^{{2}}+{y}^{{2}}={1}$$ and the plane , oriented positively when viewed from above . Let S be the inside of this curve , oriented with upward -pointing normal . Use Stokes ' Theorem to evaluate $$\displaystyle\int{S}{c}{u}{r}{l}{F}\cdot{d}{S}{\quad\text{if}\quad}{F}={y}{i}+{z}{j}+{2}{x}{k}$$.
Use the Divergence Theorem to calculate the surface integral $$\displaystyle\int\int_{{S}}{F}·{d}{S}$$, that is, calculate the flux of F across S.
$$\displaystyle{F}{\left({x},{y},{z}\right)}={\left({x}^{{3}}+{y}^{{3}}\right)}{i}+{\left({y}^{{3}}+{z}^{{3}}\right)}{j}+{\left({z}^{{3}}+{x}^{{3}}\right)}{k}$$, S is the sphere with center the origin and radius 2.
Suppose that the plane region D, its boundary curve C, and the functions P and Q satisfy the hypothesis of Green's Theorem. Considering the vector field F = Pi+Qj, prove the vector form of Green's Theorem $$\displaystyle\oint_{{C}}{F}\cdot{n}{d}{s}=\int\int_{{D}}\div{F}{\left({x},{y}\right)}{d}{A}$$
where n(t) is the outward unit normal vector to C.
Let C be the ellipse contained in the xy plane whose equation is $$\displaystyle{4}{x}^{{2}}+{y}^{{2}}={4}$$, oriented clockwise. The force field F described by $$\displaystyle{F}{\left({x},{y},{z}\right)}={x}^{{2}}{i}+{2}{x}{j}+{z}^{{2}}{k}$$, moves a particle along C in the same direction as the curve orientation, performing a W job. C as the surface boundary S: $$\displaystyle{z}={4}-{4}{x}^{{2}}-{y}^{{2}},{z}\ge{0}$$ (with ascending orientation, that is, the component in the z direction equal to 1) and assuming $$\displaystyle\pi={3.14}$$, we can state what:
a) It is not necessary to apply Stokes' Theorem, as C is a closed curve and therefore W = 0.
b) Inverting the orientation of the surface S, we can apply Stokes' Theorem and conclude that W = 12.56.
c) We can apply Stokes' Theorem and conclude that W = 6.28
d) We can apply Stokes' Theorem and conclude that W = 12.56.
Use Stokes' Theorem to compute $$\displaystyle\oint_{{C}}\frac{{1}}{{2}}{z}^{{2}}{\left.{d}{x}\right.}+{\left({x}{y}\right)}{\left.{d}{y}\right.}+{2020}{\left.{d}{z}\right.}$$, where C is the triangle with vertices at(1,0,0),(0,2,0), and (0,0,2) traversed in the order.
Use Stokes' Theorem to evaluate $$\displaystyle\int_{{C}}{F}\cdot{d}{r}$$ where C is oriented counterclockwise as viewed from above.
$$\displaystyle{F}{\left({x},{y},{z}\right)}={\left({x}+{y}^{{2}}\right)}{i}+{\left({y}+{z}^{{2}}\right)}{j}+{\left({z}+{x}^{{2}}\right)}{k}$$,
Use the Divergence Theorem to calculate the surface integral $$\displaystyle\int\int_{{S}}{F}·{d}{S}$$, that is, calculate the flux of F across S.
$$\displaystyle{F}{\left({x},{y},{z}\right)}={\left({\cos{{\left({z}\right)}}}+{x}{y}^{{2}}\right)}{i}+{x}{e}^{{-{{z}}}}{j}+{\left({\sin{{\left({y}\right)}}}+{x}^{{2}}{z}\right)}{k}$$
S is the surface of the solid bounded by the paraboloid $$\displaystyle{z}={x}^{{2}}+{y}^{{2}}$$ and the plane z = 9.