The quantity [latex](\text{curl }{\bf{F}})(P_0)\cdot{\bf{N}}(P_0)[/latex] is constant, and therefore:

[latex]\displaystyle\iint_{D_r}(\text{curl }{\bf{F}})(P_0)\cdot{\bf{N}}(P_0)dS=\pi{r}^2[(\text{curl }{\bf{F}})(P_0)\cdot{\bf{N}}(P_0)][/latex].

Thus:

[latex]\displaystyle\int_{C_r}{\bf{F}}\cdot{d}{\bf{r}}\approx\pi{r}^2[(\text{curl }{\bf{F}})(P_0)\cdot{\bf{N}}(P_0)][/latex],

and the approximation gets arbitrarily close as the radius shrinks to zero. Therefore Stokes’ theorem implies that:

[latex](\text{curl }{\bf{F}})(P_0)\cdot{\bf{N}}(P_0)=\displaystyle\lim_{r\to0^+}\frac{1}{\pi{r^2}}\displaystyle\int_{C_r}{\bf{F}}\cdot{d}{\bf{r}}[/latex].

This equation relates the curl of a vector field to the circulation. Since the area of the disk is [latex]\pi r^2[/latex], this equation says we can view the curl (in the limit) as the circulation per unit area.

Recall that if [latex]{\bf{F}}[/latex] is the velocity field of a fluid, then circulation [latex]\displaystyle\oint_{c_r}{\bf{F}}\cdot{d}{\bf{r}}=\displaystyle\oint_{C_r}{\bf{F}}\cdot{\bf{T}}ds[/latex] is a measure of the tendency of the fluid to move around [latex]C_r[/latex]. The reason is that [latex]{\bf{F}}\cdot{\bf{T}}[/latex] is a component of [latex]{\bf{F}}[/latex] in the direction of [latex]{\bf{T}}[/latex], and the closer the direction of [latex]{\bf{F}}[/latex] is to [latex]{\bf{T}}[/latex], the larger the value of [latex]{\bf{F}}\cdot{\bf{T}}[/latex].

Remember that if [latex]{\bf{a}}[/latex] and [latex]{\bf{b}}[/latex] are vectors and [latex]{\bf{b}}[/latex] is fixed, then the dot product [latex]{\bf{a}}\cdot{\bf{b}}[/latex] is maximal when [latex]{\bf{a}}[/latex] points in the same direction as [latex]{\bf{b}}[/latex]

Therefore, if [latex]{\bf{F}}[/latex] is the velocity field of a fluid, then [latex]\text{curl }{\bf{F}}\cdot{\bf{N}}[/latex] is a measure of how the fluid rotates about axis [latex]{\bf{N}}[/latex]. The effect of the curl is largest about the axis that points in the direction of [latex]{\bf{N}}[/latex], because in this case [latex]{\bf{F}}\cdot{\bf{N}}[/latex] is as large as possible.

To see this effect in a more concrete fashion, imagine placing a tiny paddlewheel at point [latex]P_0[/latex] (Figure 3). The paddlewheel achieves its maximum speed when the axis of the wheel points in the direction of [latex]\text{curl }{\bf{F}}[/latex]. This justifies the interpretation of the curl we have learned: curl is a measure of the rotation in the vector field about the axis that points in the direction of the normal vector [latex]{\bf{N}}[/latex], and Stokes’ theorem justifies this interpretation.

Faraday’s Law

Now that we have learned about Stokes’ theorem, we can discuss applications in the area of electromagnetism. In particular, we examine how we can use Stokes’ theorem to translate between two equivalent forms of Faraday’s law.

Let [latex]C[/latex] 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 [latex]C(t)[/latex] to represent the wire. At a given time [latex]t[/latex], curve [latex]C(t)[/latex] may be different from original curve [latex]C[/latex] because of the movement of the wire, but we assume that [latex]C(t)[/latex] is a closed curve for all times [latex]t[/latex].

Let [latex]D(t)[/latex] be a surface with [latex]C(t)[/latex] as its boundary, and orient [latex]C(t)[/latex] so that [latex]D(t)[/latex] has positive orientation. Suppose that [latex]C(t)[/latex] is in a magnetic field [latex]\mathbf{B}(t)[/latex] that can also change over time. In other words, [latex]\mathbf{B}[/latex] has the form:

[latex]{\bf{B}}(x,y,z)=\langle{P}(x,y,z),Q(x,y,z),R(x,y,z)\rangle[/latex],

where [latex]P[/latex], [latex]Q[/latex], and [latex]R[/latex] can all vary continuously over time.

We can produce current along the wire by changing field [latex]\mathbf{B}(t)[/latex] (this is a consequence of Ampere’s law). Flux [latex]\phi(t)=\displaystyle\iint_{D(t)}{\bf{B}}(t)\cdot{d}{\bf{S}}[/latex] creates electric field [latex]\mathbf{E}(t)[/latex] that does work.

Faraday’s law: integral form

The integral form of Faraday’s law states that:

[latex]\text{Work }=\displaystyle\int_{C(t)}{\bf{E}}(t)\cdot{d}{\bf{r}}=-\frac{\partial\phi}{\partial{t}}[/latex].

In other words, the work done by [latex]\mathbf{E}[/latex] is the line integral around the boundary, which is also equal to the rate of change of the flux with respect to time.

Faraday’s law: differential form

The differential form of Faraday’s law states that:

[latex]\text{curl }{\bf{E}}=-\frac{\partial{\bf{B}}}{\partial{t}}[/latex].

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:

[latex]-\dfrac{\partial\phi}{\partial{t}}=\displaystyle\int_{C(t)}{\bf{E}}(t)\cdot{d}{\bf{r}}=\displaystyle\iint_{D(t)}\text{curl }{\bf{E}}(t)\cdot{d}{\bf{S}}[/latex].

Since [latex]\phi(t)=\displaystyle\iint_{D(t)}{\bf{B}}(t)\cdot{d}{\bf{S}}[/latex], then as long as the integration of the surface does not vary with time we also have:

[latex]-\dfrac{\partial\phi}{\partial{t}}=\displaystyle\iint_{D(t)}-\dfrac{\partial{\bf{B}}}{\partial{t}}\cdot{d}{\bf{S}}[/latex].

Therefore:

[latex]\displaystyle\iint_{D(t)}-\dfrac{\partial{\bf{B}}}{\partial{t}}\cdot{d}{\bf{S}}=\displaystyle\iint_{D(t)}\text{curl }{\bf{E}}(t)\cdot{d}{\bf{S}}[/latex].

To derive the differential form of Faraday’s law, we would like to conclude that [latex]\text{curl }{\bf{E}}=-\frac{\partial{\bf{B}}}{\partial{t}}[/latex]. In general, the equation above is not enough to conclude this. The integral symbols do not simply “cancel out,” leaving equality of the integrands.

Why Can’t We Just Cancel the Integrals? 

To see why the integral symbol does not just cancel out in general, consider the two single-variable integrals [latex]\int_0^1 x , dx[/latex] and [latex]\int_0^1 f(x) , dx[/latex], where:

[latex]f(x)=\begin{array}{cc} \{ & \begin{array}{cc} 1, & 0\leq{x}\leq1/2 \\ 0,& 1/2\leq{x}\leq1 \end{array} \end{array}[/latex].

Both of these integrals equal [latex]\frac{1}{2}[/latex], so [latex]\int_0^1 x , dx = \int_0^1 f(x) , dx[/latex]. However, [latex]x \neq f(x)[/latex].

In our context, the equation [latex]\displaystyle\iint_{D(t)}-\frac{\partial{\bf{B}}}{\partial{t}}\cdot{d}{\bf{S}}=\displaystyle\iint_{D(t)}\text{curl }{\bf{E}}(t)\cdot{d}{\bf{S}}[/latex] is true for any region, however small. If [latex]\mathbf{F}[/latex] and [latex]\mathbf{G}[/latex] are three-dimensional vector fields such that [latex]\displaystyle\iint_S{\bf{F}}\cdot{d}{\bf{S}}=\displaystyle\iint_S{\bf{G}}\cdot{d}{\bf{S}}[/latex] for any surface [latex]S[/latex], then it is possible to show that [latex]\mathbf{F} = \mathbf{G}[/latex] by shrinking the area of [latex]S[/latex] to zero by taking a limit.

Therefore, we can let area [latex]D(t)[/latex] shrink to zero by taking a limit and obtain the differential form of Faraday’s law:

[latex]\text{curl }{\bf{E}}=-\frac{\partial{\bf{B}}}{\partial{t}}[/latex].

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.

Calculate the curl of electric field [latex]{\bf{E}}[/latex] if the corresponding magnetic field is constant field [latex]{\bf{B}}(t)=\langle1,-4,2\rangle[/latex].

Show Solution

Since the magnetic field does not change with respect to time, [latex]-\frac{\partial{\bf{B}}}{\partial{t}}=0[/latex]. By Faraday’s law, the curl of the electric field is therefore also zero.

Analysis

A consequence of Faraday’s law is that the curl of the electric field corresponding to a constant magnetic field is always zero.