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Stokes' Curl Theorem Simply Explained

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In addition to the Gauss Divergence Theorem, we will also need the Stokes Curl Theorem in order to understand Maxwell's equations in depth, for example. The Stokes integral theorem states that the rotation of a vector field within a surface is equal to the rotation of the vector field along the edge of this surface. Expressed mathematically, this integral theorem looks like this:

$$ \begin{align} \int_{A}\left(\nabla\times\boldsymbol{F}\right) \cdot \mathrm{d}\boldsymbol{a} ~=~ \oint_{L}\boldsymbol{F} \cdot \mathrm{d}\boldsymbol{l} \end{align} $$

If you have understood the Gauss' divergence theorem, the Stokes' curl theorem should no longer seem totally cryptic to you. You are already familiar with the vector field $\boldsymbol{F}$. It depends on three spatial coordinates: $\boldsymbol{F} = \boldsymbol{F}(x, y, z)$, and as a vector, it has three components. The scalar product in $\boldsymbol{F} \cdot d\boldsymbol{l}$, as well as the Nabla operator $\nabla$ and the infinitesimal $d\boldsymbol{a}$ surface element, should be familiar to you if you have read the lesson on Gauss' divergence theorem.

The Line Integral in the Stokes Curl Theorem

Let's first consider the line integral on the right side of Stokes' Curl Theorem 1. The symbol $ L $ represents a line in three-dimensional space. The circle around the integral sign indicates that this line must be closed, meaning its start point and end point are connected. We briefly refer to such a closed line as a loop.

The $ \mathrm{d}\boldsymbol{l} $ is an infinitesimal line element of the line, essentially an infinitely small piece of the line. Also, you should notice that the $ \mathrm{d}\boldsymbol{l} $ line segment is in boldface, indicating that it is a vector with three components: $ \mathrm{d}l_x $, $ \mathrm{d}l_y $, and $ \mathrm{d}l_z $. The magnitude of the $ \mathrm{d}\boldsymbol{l} $ line segment represents the length of this line segment, while its direction points along the line.

A line element $ d\boldsymbol{l} $ on the loop $L $ at an example position $(x,y,z)$.

Then the scalar product $ \boldsymbol{F} \cdot \mathrm{d}\boldsymbol{l} $ is formed on the right-hand side between a vector field $ \boldsymbol{F} $ and the line element $ \mathrm{d}\boldsymbol{l} $:

$$ \begin{align} \boldsymbol{F} \cdot \mathrm{d}\boldsymbol{l} ~=~ F_x \, {\mathrm{d}l}_x ~+~F_y \, {\mathrm{d}l}_y ~+~ F_z \, {\mathrm{d}l}_z \end{align} $$

You have already learned what the purpose of this scalar product is in the Gauss divergence theorem. Let's recap! First, divide the vector field $ \boldsymbol{F}= \class{blue}{\boldsymbol{F}_{||}} +\boldsymbol{F}_{\perp} $ into two components:

Into the component $ \class{blue}{\boldsymbol{F}_{||}} $, which points parallel to the $\mathrm{d}\boldsymbol{l}$ line element.
Into the component $ \boldsymbol{F}_{\perp} $ that points perpendicular to the $\mathrm{d}\boldsymbol{l}$ line element.

The scalar product $ \boldsymbol{F} \cdot \mathrm{d}\boldsymbol{l} $ eliminates the perpendicular component $ \boldsymbol{F}_{\perp} $ of the vector field $ \boldsymbol{F} $ and retains only the component parallel to the $ \mathrm{d}\boldsymbol{l} $ line element, denoted as $ \class{blue}{\boldsymbol{F}_{||}} $. Why again is the perpendicular component zero? Because the scalar product of two vectors perpendicular to each other, $ \boldsymbol{F}_{\perp} $ and $ \mathrm{d}\boldsymbol{l} $, is mathematically zero:

$$ \begin{align} \boldsymbol{F} \cdot \mathrm{d}\boldsymbol{l} &~=~ \left(\class{blue}{\boldsymbol{F}_{||}} + \boldsymbol{F}_{\perp}\right) \cdot \mathrm{d}\boldsymbol{l} \\\\
&~=~ \class{blue}{\boldsymbol{F}_{||}} \cdot \mathrm{d}\boldsymbol{l} ~+~ \underbrace{\boldsymbol{F}_{\perp} \cdot \mathrm{d}\boldsymbol{l}}_{0}
~=~ \class{blue}{\boldsymbol{F}_{||}} \cdot \mathrm{d}\boldsymbol{l} \end{align} $$

Since the $ \mathrm{d}\boldsymbol{l} $-element runs parallel to the line at every point of the line, only the parallel component $ \class{blue}{\boldsymbol{F}_{||}} $ of the vector field remains in the scalar product 3, which of course also runs along the line $L$; all other components of the vector field are eliminated.

Subsequently, on the right side of Stokes' theorem 1, the scalar products $\boldsymbol{F} \cdot \mathrm{d}\boldsymbol{l}\left(x, y, z\right)$ for each postion $\left(x, y, z\right)$ on the line $L$ are summed up using the integral in 1. Let's denote the right-hand side briefly as $U$.

$$ \begin{align} U ~:=~ \oint_{L}\boldsymbol{F} \cdot \mathrm{d}\boldsymbol{l} \end{align} $$

The line integral, therefore, results in a number $U$, which is a measure of how much of the vector field $\boldsymbol{F}$ runs along the line $L$. Because the line $L$ is closed, the summation returns to the same point $\left(x, y, z\right)$ where the summation began. The closed line integral $U$ thus intuitively indicates how much of the vector field $\boldsymbol{F}$ rotates along the loop $L$.

The Surface Integral in the Stokes Curl Theorem

Let us now consider the surface integral on the left-hand side of the Stokes Curl Theorem 1:

$$ \begin{align} \int_{A}\left(\nabla\times\boldsymbol{F}\right) \cdot \mathrm{d}\boldsymbol{a} \end{align} $$

In the surface integral, the area $A$ is involved. In contrast to the surface integral with a circle around the integral sign, as in Gauss' divergence theorem, here we are considering an open surface. It does not enclose any volume; it's simply a surface enclosed by the loop $L$.

A surface element $ d\boldsymbol{a} $ within the area $ A $ at an example position $(x,y,z)$.

The vector $\mathrm{d}\boldsymbol{a} = ({\mathrm{d}a}_x,~{\mathrm{d}a}_y, ~{\mathrm{d}a}_z) $ represents an infinitely small area of the surface $A$ and this vector is perpendicular to each positon $\left(x,y,z\right)$ on this surface.

In the surface integral, the cross product $ \mathrm{\nabla}\times\boldsymbol{F} $ between the Nabla operator $ \mathrm{\nabla} $ and the vector field $ \boldsymbol{F} $ also appears. The intuitive meaning of the cross product should be familiar to you from Maths for Physics I. The cross product, along with the dot product, is the second way to multiply vectors. The vector $ \mathrm{\nabla}\times\boldsymbol{F} $ corresponds to the curl of the vector field $ \boldsymbol{F} $. Unlike the dot product, the result of the cross product is again a vector field that is perpendicular to $ \boldsymbol{F} $. Why perpendicular? Because that's a characteristic property of the cross product! If we explicitly write out the cross product, the resulting vector $ \mathrm{\nabla}\times\boldsymbol{F} $ looks like this.

$$ \begin{align} \nabla \times \boldsymbol{F} ~=~ \begin{bmatrix} \partial_{\text y} \, F_{\text z} - \partial_{\text z} \, F_{\text y} \\ \partial_{\text z} \, F_{\text x} - \partial_{\text x} \, F_{\text z} \\ \partial_{\text x} \, F_{\text y} - \partial_{\text y} \, F_{\text x} \end{bmatrix} \end{align} $$

What does the intuitive meaning of the curl $\nabla \times \boldsymbol{F}$ imply? The vector $\nabla \times \boldsymbol{F}\left(x, y, z\right)$ indicates how much the vector field $ \boldsymbol{F} $ rotates at a particular position $\left(x, y, z\right)$ within the surface $A$.

Then, in the surface integral 5, the scalar product $\left(\nabla\times\boldsymbol{F}\right) \cdot \mathrm{d}\boldsymbol{a}$ between the curl vector field $ \nabla\times\boldsymbol{F} $ and the infinitesimal surface element $\mathrm{d}\boldsymbol{a}$ is formed. As we already know, the scalar product picks out only the component $\class{blue}{\left(\nabla\times\boldsymbol{F} \right)_{||}} $ of the curl vector field $ \nabla\times\boldsymbol{F} $ that runs parallel to the surface element:

$$ \begin{align} \left(\nabla\times\boldsymbol{F}\right) \cdot \mathrm{d}\boldsymbol{a} &~=~ \left[ \class{blue}{\left(\nabla\times\boldsymbol{F} \right)_{||}} + \left(\nabla\times\boldsymbol{F} \right)_{\perp} \right] \cdot \mathrm{d}\boldsymbol{a} \\\\
&~=~ \class{blue}{\left(\nabla\times\boldsymbol{F} \right)_{||}} \cdot \mathrm{d}\boldsymbol{a} ~+~ \underbrace{\left(\nabla\times\boldsymbol{F} \right)_{\perp} \cdot \mathrm{d}\boldsymbol{a}}_0 \\
&~=~ \class{blue}{\left(\nabla\times\boldsymbol{F} \right)_{||}} \cdot \mathrm{d}\boldsymbol{a} \end{align} $$

Since the surface element vector $\mathrm{d}\boldsymbol{a}$ is perpendicular to the respective surface element at a specific point, the scalar product 7 picks out only the component of the vector field $ \boldsymbol{F} $ that is also perpendicular to the surface element. Therefore, only the component $\class{blue}{\left(\nabla\times\boldsymbol{F} \right)_{||}}$ remains in the surface integral.

Subsequently, on the left side of Stokes' theorem, the scalar products $\left(\nabla\times\boldsymbol{F}\right) \cdot \mathrm{d}\boldsymbol{a}$ at each point $(x, y, z)$ are summed up using the integral within the surface $A$.

The vector field components $ \class{blue}{(\nabla \times \boldsymbol{F})_{||}} $ protruding perpendicularly from the surface elements are summed up at each point $(x, y, z)$ within the surface $A$.

Let us now summarize the statements of the surface integral (right-hand side) and line integral (left-hand side) of the Stokes Curl Theorem 1:

On the left-hand side, the curl $\mathrm{\nabla}\times\boldsymbol{F}$ of the vector field $\boldsymbol{F}$ is summed up at each individual point within the area $A$.
On the right-hand side, the vector field $ \boldsymbol{F} $ is summed up along the edge $L$ of the area $A$. The right-hand side therefore corresponds to a number that measures the rotation of the vector field on the edge.

Both integrals (i.e. both sides) should be equal according to the Stokes integral theorem 1. The equality therefore clearly states: The total circulation of a vector field $\boldsymbol{F}$ within the area $A$ corresponds exactly to the circulation of the vector field along the edge $L$ of this area.

Exercises with Solutions

Use this formula eBook if you have problems with physics problems.

Exercise #1: $ \class{violet}{\boldsymbol{B}} $-Field Inside and Outside a Wire

Consider an infinitely extended straight wire with radius $R$ through which a constant current $ \class{red}{I} $ flows. Assume that the current is homogeneous across the cross-sectional area $ A $ of the wire.

Calculate the magnetic field $ \class{violet}{\boldsymbol{B}} $ outside the wire.
Calculate the magnetic field $ \class{violet}{\boldsymbol{B}} $ inside the wire.

Hint: Use the Ampère's law from electrostatics: $$ \oint_L \class{violet}{\boldsymbol{B}} \cdot \text{d}\boldsymbol{l} = \mu_0 \, \class{red}{I} $$

And consider the current density $ \class{red}{\boldsymbol{j}} $ for inside the wire: $$ \oint_A \class{red}{\boldsymbol{j}} \cdot \text{d}\boldsymbol{a} = \class{red}{I} $$ which allows you to determine the enclosed current $ \class{red}{I} $.

Solution to Exercise #1.1

To determine the magnetic field outside a long, straight wire, you use Ampère's law by creating an Ampère loop $L$ around the wire. The radius $r$ of the Ampère loop - to fully enclose the wire - is greater than the radius $R$ of the wire. The wire carrying current should penetrate the area enclosed by this Ampère loop!

Ampère's law, which relates the magnetic field to the current, states: \[ \oint_{L}\class{violet}{\boldsymbol{B}} \cdot \text{d}\boldsymbol{l} = \mu_0 \, \class{red}{I} \]

Since the imagined Ampère loop encloses the entire conductor, the current enclosed by it is the total current $I$ flowing through the wire (this total current is known).

This is a cylindrical symmetric problem. Therefore, use cylindrical coordinates {$r_{\perp}$, $ \varphi $, $z$}. Here, $ r_{\perp} := r $ is set for easier notation. The infinitesimal $ \text{d}\boldsymbol{s} $ element is summed along the Ampère loop $L$ from 0 to $ 2\pi $, and it lies at a distance $ r $ from the origin and along the $ \varphi $ coordinate. The infinitesimal element is then: \[ \text{d}\boldsymbol{l} ~=~ r \, \text{d}\varphi \, \boldsymbol{\hat{\varphi}} \]

The $ \boldsymbol{\hat{\varphi}} $ here is the unit vector in the $ \varphi $ direction.

The magnetic field - applying the right-hand rule - circulates around the wire along the $ \varphi $ coordinate: $ \class{violet}{\boldsymbol{B}} = \class{violet}{B} \, \boldsymbol{\hat{\varphi}}$. Therefore, when you substitute and simplify everything, you get: \[ \oint_0^{2\pi} \class{violet}{B} \, r \, \text{d}\varphi = \mu_0 \, \class{red}{I} \]

Due to cylindrical symmetry, the $ \class{violet}{B} $-field is independent of $ \varphi $, so you can pull out $ \varphi $ as well as $ r $ and integrate: \[ 2\pi \, \class{violet}{B} \, r = \mu_0 \, \class{red}{I} \]

As you can see, the $ \class{violet}{B} $-field, with constant current, depends only on the distance $r$ from the wire. Just rearrange for $ \class{violet}{B} $ and consider the $ \boldsymbol{\hat{\varphi}} $ direction of the $ \class{violet}{B} $-field.

Magnetic field lines of a thin current-carrying wire.

The magnetic field (outside) of a current-carrying wire with constant current and in the region $R$ ≤ $r$ is given by: \[ \class{violet}{\boldsymbol{B}(}r\class{violet}{)} = \frac{ \mu_0 \, \class{red}{I} }{ 2\pi } \, \frac{1}{r} \, \boldsymbol{\hat{\varphi}} \]

Solution to Exercise #1.2

To determine the magnetic field inside a long, straight wire, you use - as in Exercise #1.1 - Ampère's law by creating an Ampère loop $L$ inside the wire. Its radius $ r $ - to be inside the wire - is smaller than the radius $R$ of the wire.

When you calculate the line integral in Ampère's law, you get the same result as in Exercise #1.1; with the difference that now the radius $ r $ of the imagined Ampère loop is smaller - not larger - than the radius $R$ of the wire. So, you have until now: \[ 2\pi \, r \, \class{violet}{B} = \mu_0 \, \class{red}{I_{\text{in}}} \]

Now you have to be careful because now the current enclosed by the loop, let's call it $ \class{red}{I_{\text{in}}} $, is no longer the total current $ \class{red}{I} $, but only a part of it. How do you get the enclosed current $ \class{red}{I_{\text{in}}} $ now? First, by considering the current passing through the cross-sectional area $ A $: \[ \int_A \class{red}{\boldsymbol{j}} \cdot \text{d}\boldsymbol{a} = \class{red}{I} \]

This cross-sectional area $A$ is the cross-sectional area of the wire and corresponds to a circle: $ A = \pi \, R^2 $.

What happens now if the cross-sectional area $A_{\text L}$ of the Ampère loop $ L $ is smaller than the entire cross-sectional area $A$ of the wire? We must cut off a part of the current! According to the exercise, you know that the current density is homogeneous - no matter which cross-sectional area you consider: $ \class{red}{\boldsymbol{j}} = \class{red}{j} \, \boldsymbol{\hat{z}} $. This makes the integral in Eq. 8 to: \[ \int_A \class{red}{\boldsymbol{j}} \cdot \text{d}\boldsymbol{a} = \class{red}{j} \, A \]

So the current density is: \[ \class{red}{j} ~=~ \frac{\class{red}{I}}{A} ~=~ \frac{\class{red}{I}}{\pi\,R^2} \]

Thus, you also have the enclosed current, which is the product of current density and the area enclosed by the Ampère loop $ A_{\text L} $: \[ \class{red}{I_{\text{in}}} ~=~ \class{red}{j} \, A_{\text L} ~=~ \frac{\class{red}{I}}{\pi \, R^2} \, \pi \, r^2 \]

Cancel and substitute $ \class{red}{I_{\text{in}}} $ into Ampère's law, then you have: \[ 2\pi \, r \, \class{violet}{B} ~=~ \mu_0 \, \frac{\class{red}{I}}{R^2} \, r^2 \]

If you now just rearrange for $ \class{violet}{B} $ and consider the magnetic field direction $ \boldsymbol{\hat{\varphi}} $, then you have determined the magnetic field.

The magnetic field inside a current-carrying conductor with constant current and in the range $r$ ≤ $R$ is given by: \[ \class{violet}{\boldsymbol{B}(}r\class{violet}{)} = \frac{ \mu_0 \, \class{red}{I}}{2\pi \, R^2} \, r ~ \boldsymbol{\hat{\varphi}} \]

Diagram for the magnetic field $\class{violet}{B}$ as a function of distance $r$ inside and outside the cylindrical conductor.

If you calculate the B-field at the position $R$ (edge of the wire), you will see that the two formulas (for inside and outside) yield the same magnetic field. $ \class{violet}{\boldsymbol{B}} $ is thus continuous everywhere.

The Line Integral in the Stokes Curl Theorem

The Surface Integral in the Stokes Curl Theorem

Exercises with Solutions

Exercise #1: \( \class{violet}{\boldsymbol{B}} \)-Field Inside and Outside a Wire

Solution to Exercise #1.1

Solution to Exercise #1.2