6.7 Stokes’ theorem  (Page 6/10)

Let C be a closed curve that models a thin wire. In the context of electric fields, the wire may be moving over time, so we write $C(t)$ to represent the wire. At a given time t , curve $C(t)$ may be different from original curve C because of the movement of the wire, but we assume that $C(t)$ is a closed curve for all times t . Let $D(t)$ be a surface with $C(t)$ as its boundary, and orient $C(t)$ so that $D(t)$ has positive orientation. Suppose that $C(t)$ is in a magnetic field $\text{B}(t)$ that can also change over time. In other words, B has the form

where P, Q, and R can all vary continuously over time. We can produce current along the wire by changing field $\text{B}(t)$ (this is a consequence of Ampere’s law). Flux $\varphi (t)={\displaystyle {\iint }_{D(t)}\text{B}(t)·d\text{S}}$ creates electric field $\text{E}(t)$ that does work. The integral form of Faraday’s law states that

In other words, the work done by E is the line integral around the boundary, which is also equal to the rate of change of the flux with respect to time. The differential form of Faraday’s law states that

Using Stokes’ theorem, we can show that the differential form of Faraday’s law is a consequence of the integral form. By Stokes’ theorem, we can convert the line integral in the integral form into surface integral

Since $\varphi (t)={\displaystyle {\iint }_{D(t)}\text{B}(t)·d\text{S}},$ then as long as the integration of the surface does not vary with time we also have

Therefore,

To derive the differential form of Faraday’s law, we would like to conclude that $\text{curl}\text{E}=-\frac{\partial \text{B}}{\partial t}.$ In general, the equation

is not enough to conclude that $\text{curl}\text{E}=-\frac{\partial \text{B}}{\partial t}.$ The integral symbols do not simply “cancel out,” leaving equality of the integrands. To see why the integral symbol does not just cancel out in general, consider the two single-variable integrals ${\displaystyle {\int }_0^{1}xdx}$ and ${\displaystyle {\int }_0^{1}f(x)dx},$ where

Both of these integrals equal $\frac{1}{2},$ so ${\displaystyle {\int }_0^{1}xdx}={\displaystyle {\int }_0^{1}f(x)dx}.$ However, $x\ne f(x).$ Analogously, with our equation ${\displaystyle {\iint }_{D\left(t\right)}-\frac{\partial \text{B}}{\partial t}}·d\text{S}={\displaystyle {\iint }_{D\left(t\right)}\text{curl}\text{E}·d\text{S}},$ we cannot simply conclude that $\text{curl}\text{E}=-\frac{\partial \text{B}}{\partial t}$ just because their integrals are equal. However, in our context, equation ${\displaystyle {\iint }_{D\left(t\right)}-\frac{\partial \text{B}}{\partial t}}·d\text{S}={\displaystyle {\iint }_{D\left(t\right)}\text{curl}\text{E}·d\text{S}}$ is true for any region, however small (this is in contrast to the single-variable integrals just discussed). If F and G are three-dimensional vector fields such that ${\displaystyle {\iint }_s\text{F}·d\text{S}=}{\displaystyle {\iint }_s\text{G}·d\text{S}}$ for any surface S , then it is possible to show that $\text{F}=\text{G}$ by shrinking the area of S to zero by taking a limit (the smaller the area of S , the closer the value of ${\displaystyle {\iint }_s\text{F}·d\text{S}}$ to the value of F at a point inside S ). Therefore, we can let area $D\left(t\right)$ shrink to zero by taking a limit and obtain the differential form of Faraday’s law:

In the context of electric fields, the curl of the electric field can be interpreted as the negative of the rate of change of the corresponding magnetic field with respect to time.