The 4 Maxwell's Equations: Important Basics You Need to Know

by Alexander FufaeV

Table of contents

Practical applications of Maxwell's equations Here you will learn the technical applications of the Maxwell's equations.
1st ingredient: The electric E-field Here you will learn the E-field that occurs in two of the four Maxwell's equations.
2st ingredient: The magnetic B-field Here you will learn the magnetic field that occurs in two of four Maxwell's equations and what it has to do with a cross product.
Maths inside Maxwell equations: Divergence integral theorem One of the two mathematical theorems necessary for a deeper understanding of Maxwell 's equations. You will also learn about the divergence of a vector field and the electric and magnetic flux.
Maths inside Maxwell equations: Curl integral theorem The second mathematical theorem necessary for a deeper understanding of Maxwell 's equations. Here you will also learn the electric and magnetic voltage.
The 1st Maxwell equation Here you learn how the electric field is connected with the electric charge.
The 2nd Maxwell equation Here you learn why there are no magnetic monopoles.
The 3rd Maxwell equation Here you learn how electric vortex fields are related to time-varying magnetic fields.
The 4th Maxwell equation Here you learn how a time-varying electric field and an electric current are related to the magnetic vortex field.

Are you ready to explore Maxwell's equations? In this lesson, you won't be bored with complicated theoretical derivations. Instead, the lesson presents these equations in the simplest, most understandable way possible, making sure you're able to follow along every step of the way. And don't worry, any math that pops up in these equations will be explained with vivid examples and analogies. But before we dive in, make sure you're familiar with partial derivatives and integrals, because they're key to understanding the magic of Maxwell's equations. Are you ready to unlock the electromagnetic secrets of the universe? Let's go!

Practical applications of Maxwell's equations

The four Maxwell equations together with the Lorentz force contain all the knowledge of electrodynamics. There are so many applications of it that I can’t list them all in this video, but some of them are for example:

Electronic devices such as computers and smart phones. These devices contain electrical capacitors, coils, and entire circuits that exploit (take advantage of) the Maxwell equations.
Power generation - whether from nuclear, wind or hydroelectric power plants, the energy released must first be converted into electrical energy so that people can use it. That happens with electric generators. These in turn are based on the Maxwell equations.
Power supply - in order to transport electric power into households with as little energy losses as possible, alternating voltages and transformers based on the Maxwell equations are needed.
And many more - electric welding for assembling car bodies, engines for electric cars, magnetic resonance imaging in medicine, electric kettles in the kitchen, the charger of your smart phone, radio, wifi and so on.

Imagine this!

Any device that exploits electricity or magnetism is fundamentally based on the Maxwell equations.

To understand all four Maxwell equations you first need to know what electric and magnetic fields are, because they are contained in the equations!

1st ingredient: The electric E-field

Consider an electrically charged sphere with the large source charge $Q$ and a sphere with the small test charge $q$. The test charge is at distance $ r $ from the source charge at a given time. The source charge exerts an electric force $ F_{\text e} $ on the test charge, which is given by Coulomb's law:

$$ \begin{align} F_{\text e} ~=~ \frac{1}{4\pi \, \varepsilon_0} \, \frac{Q \, q}{r^2} \end{align} $$

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Eine große Quellladung übt elektrische Kraft auf eine kleine Probeladung aus.

Here $1/4 \pi \, \varepsilon_0 $ is a constant prefactor with the electric field constant $ \varepsilon_0 $, which provides the right unit of force $F_{\text e}$, namely the unit Newton (N).

Now what if you know the value of the big charge $ Q $ and want to know the value of the force that the big charge exerts on the small charge $q$? However, you do not know the exact value of this little charge, or you intentionally leave that value open and only want to look at the electric force exerted by the big charge. So you have to somehow eliminate the small charge $q$ from Coulomb's law. To do so, you simply divide Coulomb's law by $q$ on both sides:

$$ \begin{align} \frac{F_{\text e}}{q} ~=~ \frac{1}{4\pi \, \varepsilon_0} \, \frac{Q}{r^2} \end{align} $$

This way, on the right hand side, the small charge $ q $ drops out and instead lands on the left hand side of the equation. The quotient on the left hand side is defined as the electric field $E$ of the source charge $Q$:

Definition of the electric field

$$ \begin{align} E ~=~ \frac{F_{\text e}}{q} \end{align} $$

What is an electric E-field?

The electric field $E$ indicates the electric force which would act on a small charge $q$ when placed at the distance $r$ to the source charge $Q$. Electric field $E$ is force per charge!

The electric field has the unit "force per charge" N/C or with a more common unit "voltage per distance" V/m. Übrigens: We have called the large charge the source charge to indicate that it is the source of the electric field. And we have chosen a small test charge $q$ so that it does not influence the electric field of the source charge too much.

So far, only the magnitude, that is, the strength of the electric field has been considered, without taking into account the exact direction of the electric field. However, the Maxwell's equations are general and also include the direction of the electric field. So we have to turn the electric field into a vector $\boldsymbol{E}$. Vectors are shown in boldface. Handwritten mostly with a little arrow above the letter to distinguish it from scalar (pure) numbers. I omit the arrows because they make the equations unnecessarily ugly.

The electric field $\boldsymbol{E}$ as a vector in three-dimensional space has three components $E_{\text x}$, $E_{\text y}$ and $ E_{\text z}$:

E-field vector has three components

$$ \begin{align} \boldsymbol{E} ~=~ \begin{bmatrix} E_{\text x} \\ E_{\text y} \\ E_{\text z} \end{bmatrix} \end{align} $$

Let’s look at the first component. The first component $E_{\text x}(x, y, z)$ depends on the space coordinates ($x, ~ y, ~ z $) and is the magnitude of the electric field in $x$-direction. That is, depending on which concrete location is used for ($x, ~ y, ~ z $), the value $E_{\text x}$ is different. The same applies to the other two components $E_{\text y}$ which indicate the magnitude in the $y$-direction and $E_{\text z}$ which indicates the magnitude in the $z$-direction. The components of the electric field thus indicate which electric force would act on a test charge at a specific location in the first, second or third spatial direction.

2st ingredient: The magnetic B-field

Another important physical quantity found in the Maxwell equations is the magnetic field. Experimentally, it is found that a particle with the electric charge $q$ moving with the velocity $v$ in an external magnetic field experiences a magnetic force $F_{\text m}$, which deflects the particle.

Hover the image!

Circular motion due to magnetic Lorentz force, which is perpendicular to the velocity of the electron and to the external magnetic field.

The force on the particle increases in proportion to its charge ($F_{\text m} \sim q $) and its velocity ($F_{\text m} \sim v $), that is doubling the charge or the velocity doubles the force on the particle. But not only that! The force also increases in proportion to the applied magnetic field. To describe this last proportionality of the force and the magnetic field, we introduce the physical quantity $ \class{violet}{B} $. So the magnetic force is given by:

$$ \begin{align} F_{\text m} ~=~ q \, v \, \class{violet}{B} \end{align} $$

The unit of the quantity $\class{violet}{B}$ must be such that the right hand side of the equation 5 gives the unit of force, that is, Newton (N) or equivalently kg m / s². By a simple transformation, you will find that the unit of $\class{violet}{B}$ must be kg/As². This is what we will call the unit of Tesla (T = kg/As²).

We call $\class{violet}{B}$ magnetic flux density (or short: magnetic field or B-field). The magnetic flux density describes the external magnetic field and thus determines the magnitude of the force on a charged particle.

The equation 5 for the magnetic force on a charged particle represents only the magnitude of the force. In order to formulate the magnetic force with vectors - analogous to the electric force - the force, the velocity and the magnetic field are expressed in vector form:

$$ \begin{align} \boldsymbol{F_{\text m}} = \begin{bmatrix} F_{\text{mx}} \\ F_{\text{my}} \\ F_{\text{mz}} \end{bmatrix}, ~~~ \boldsymbol{v} = \begin{bmatrix} v_{\text x} \\ v_{\text y} \\ v_{\text z} \end{bmatrix}, ~~~ \class{violet}{\boldsymbol{B}} = \begin{bmatrix} \class{violet}{B_{\text x}} \\ \class{violet}{B_{\text y}} \\ \class{violet}{B_{\text z}} \end{bmatrix} \end{align} $$

Now, the three quantities are not scalars, but three-dimensional vectors with components in the $x$-, $y$- and $z$-direction. The important question is:

How should the velocity vector $\boldsymbol{v}$ be vectorially multiplied by the magnetic field vector $\class{violet}{\boldsymbol{B}}$?

If you look more closely at the deflection of the charge in the magnetic field, you will notice that the magnetic force always points in the direction orthogonal to the velocity AND to the magnetic field lines. This orthogonality can be easily taken into account with the commonly called cross product.

Cross product between the velocity and the magnetic field vector.

The cross product of two vectors $\boldsymbol{v}$ and $\class{violet}{\boldsymbol{B}}$ is defined so that the result of the cross product $\boldsymbol{v} \times \class{violet}{\boldsymbol{B}}$, which is a vector, is always orthogonal to the two vectors $\boldsymbol{v}$ and $\class{violet}{\boldsymbol{B}}$:

Cross product of the velocity with the magnetic field

$$ \begin{align} \boldsymbol{v} ~\times~ \class{violet}{\boldsymbol{B}} ~=~ \begin{bmatrix} v_{\text y}\,\class{violet}{B_{\text z}}-v_{\text z}\,\class{violet}{B_{\text y}} \\ v_{\text z}\,\class{violet}{B_{\text x}}-v_{\text x}\,\class{violet}{B_{\text z}} \\ v_{\text x}\,\class{violet}{B_{\text y}}-v_{\text y}\,\class{violet}{B_{\text x}} \end{bmatrix} \end{align} $$

For the force $\boldsymbol{F_{\text m}}$ to be always orthogonal to $\boldsymbol{v}$ and $\class{violet}{\boldsymbol{B}}$, we take the cross product of $\boldsymbol{v}$ and $\class{violet}{\boldsymbol{B}}$ in our equation 5. So in vector form the magnetic force is generally given by:

$$ \begin{align} \boldsymbol{F}_{\text m} ~=~ q \, \boldsymbol{v} \times \class{violet}{\boldsymbol{B}} \end{align} $$

What is a magnetic B-field?

The magnetic flux density $\class{violet}{\boldsymbol{B}}$ describes the strength of the magnetic field and thus the magnitude of the magnetic force on a charged particle.

Now you have learned the two important physical ingredients found in the Maxwell equations, namely the electric field $\boldsymbol{E} $ and the magnetic field $ \class{violet}{\boldsymbol{B}} $. They are vector fields. This means that to each location ($x, ~ y, ~ z $) in space you can assign an electric $\boldsymbol{E}(x, ~ y, ~ z) $ and a magnetic $\class{violet}{\boldsymbol{B}}(x, ~ y, ~ z) $ field vector indicating both magnitude and direction of the electric and magnetic fields.

Electric and magnetic fields as three-dimensional vector fields.

$$ \begin{align} \boldsymbol{E} ~=~ \begin{bmatrix} E_{\text x} \\ E_{\text y} \\ E_{\text z} \end{bmatrix}, ~~~ \class{violet}{\boldsymbol{B}} ~=~ \begin{bmatrix} \class{violet}{B_{\text x}} \\ \class{violet}{B_{\text y}} \\ \class{violet}{B_{\text z}} \end{bmatrix} \end{align} $$

These four Maxwell equations can be represented in two different ways:

In integral form, which expresses the Maxwell equations with integrals,
and the differential form, which expresses the Maxwell equations with derivatives.

While the differential form of Maxwell's equations is useful for calculating the magnetic and electric fields at a single point in space, the integral form is there to compute the fields over an entire region in space. The integral form is well suited for the calculation of symmetric problems, such as the calculation of the electric field of a charged sphere, a charged cylinder or a charged plane. The differential form is more suitable for the calculation of complicated numerical problems using computers or, for example, for the derivation of the electromagnetic waves. In addition, the differential form looks much more compact than the integral form. Both forms are useful and can be transformed into each other using two mathematical theorems:

Divergence integral theorem
Curl integral theorem

If you understand the two theorems, it will be easier for you to understand the Maxwell's equations. Let's first look at the divergence integral theorem.

Maths inside Maxwell equations: Divergence integral theorem

This is what the divergence integral theorem (Gauss integral theorem) looks like in all its splendor:

Divergence integral theorem as equation

$$ \begin{align} \int_{V} \left(\nabla \cdot \boldsymbol{F}\right) \, \text{d}v ~=~ \oint_{A}\boldsymbol{F} \cdot \text{d}\boldsymbol{a} \end{align} $$

First, let's look at the right hand side of the equation 10:

$$ \begin{align} \oint_A \boldsymbol{F} \cdot \text{d}\boldsymbol{a} \end{align} $$

The $A$ represents a surface enclosing any volume, for example the surface of a cube, a sphere, or the surface of any three-dimensional body you can think of.
The small circle around the integral indicates that this surface must satisfy a condition: the surface must be closed, in other words it must not contain any holes, so that the equality is met mathematically. The surface $A$ is thus a closed surface.
The $\boldsymbol{F}$ is a vector field and represents either the electric field $\boldsymbol{E}$ or the magnetic field $\class{violet}{\boldsymbol{B}}$ when considering the Maxwell equations. So it is a vector with three components.
The $\text{d} \boldsymbol{a} $ is an infinitesimal surface element, that is, an infinitely small surface element of the considered surface $A$. As you may have already noticed, the $\boldsymbol{a} $ in the $\text{d} \boldsymbol{a} $ element is shown in boldface, so it is a vector, with a magnitude and a direction. The magnitude $\text{d}a $ of the $\text{d} \boldsymbol{a} $ element indicates the area of a small piece of the surface. The $\text{d} \boldsymbol{a} $ element is orthogonal to this piece of the surface and, by definition, points out of the surface.
The dot $\cdot$ between the vector field $\boldsymbol{F} $ and the $\text{d} \boldsymbol{a} $ element represents the scalar product. The scalar product is a way to multiply two vectors. So here, the scalar product between the vector field $\boldsymbol{F} $ and the $\text{d} \boldsymbol{a} $ element is formed.

The scalar product is defined as follows:

Scalar product between vector field and surface element

$$ \begin{align} \boldsymbol{F} ~\cdot~ \text{d}\boldsymbol{a} ~=~ F_{\text x} \, \text{d}a_{\text x} ~+~ F_{\text y} \, \text{d}a_{\text y} ~+~ F_{\text z} \, \text{d}a_{\text z} \end{align} $$

As you can see from the definition 12, the first, second and third components of the two vectors are multiplied and then added up. The result of the scalar product is no longer a vector but an ordinary number, a scalar. To understand what this number means, you must first know that any vector $\boldsymbol{F} $ can be written as the sum of two other vectors:

One vector that is parallel to the $\text {d} \boldsymbol{a} $ element, let’s call it $\boldsymbol{F}_{||} $
and another vector that is orthogonal to the $\text {d} \boldsymbol{a} $ element, let’s call it $\boldsymbol{F}_{\perp} $.

Only the component of the vector field parallel to the surface element contributes to the scalar product in a surface integral.

Vector field split into perpendicular and parallel component

$$ \begin{align} \boldsymbol{F} ~=~ \boldsymbol{F}_{||} ~+~ \boldsymbol{F}_{\perp} \end{align} $$

Another mathematical fact is that the scalar product of two orthogonal vectors always yields zero, which means that in our case scalar product between the part $\boldsymbol{F}_{\perp} $ and the $\text{d} \boldsymbol{a} $ element is zero:

$$ \begin{align} \boldsymbol{F}_{\perp} ~\cdot~ \text{d}\boldsymbol{a} ~=~ 0 \end{align} $$

However, the scalar product between the part $\boldsymbol{F}_{||} $ and the $\text{d} \boldsymbol{a} $ element is generally not zero!

Divergence integral theorem visualized.

So now you can see what the scalar product on the right hand side of the divergence integral theorem 3 does: It just picks out the part of the vector field $\boldsymbol{F} $ that is exactly parallel to the $\text{d} \boldsymbol{a} $ element. The remaining part of the vector field that points in orthogonal direction is eliminated by the scalar product.

Subsequently, in 11, the scalar products for all locations of the considered surface $A$ are added up. That is the task of the integral sign.

The right-hand side of the divergence integral theorem 10 thus sums up all the components of the vector field $\boldsymbol{F} $ that flow into or flow out of the surface $A$. Such an integral, in which small pieces of a surface are summed up, is called surface integral.

If, as in this case, the integrand is a vector field, this surface integral is called the flux $\Phi $ of the vector field $\boldsymbol{F} $ through the surface $A $:

Flux of a vector field through a surface

$$ \begin{align} \Phi ~=~ \oint_A \boldsymbol{F} \cdot \text{d}\boldsymbol{a} \end{align} $$

What does the surface integral inside divergence theorem tell us?

The surface integral on the right-hand side is a number $ \Phi $ that measures how much of the vector field $\boldsymbol{F} $ enters the surface or exits the surface $ A $. It measures the flux $ \Phi $ of the vector field through this surface.

Electric and magnetic flux

Electric flux as the scalar product of the surface orthogonal vector and E-field vector.

If the vector field $\boldsymbol{F} $ in the surface integral 15 is an electric field $\boldsymbol{E} $, then this surface integral is called electric flux $\Phi_{\text e } $ through the surface $A$:

$$ \begin{align} \Phi_{\text e} ~=~ \int_A \boldsymbol{E} \cdot \text{d}\boldsymbol{a} \end{align} $$

And if the vector field $\boldsymbol{F} $ is a magnetic field $ \class{violet}{\boldsymbol{B}} $, the surface integral is called magnetic flux $\Phi_{\text m} $ through the surface $A$:

$$ \begin{align} \Phi_{\text m} ~=~ \int_A \class{violet}{\boldsymbol{B}} \cdot \text{d}\boldsymbol{a} \end{align} $$

Now let's look at the left hand side of the divergence theorem 10:

$$ \begin{align} \int_{V} \left(\nabla \cdot \boldsymbol{F}\right) \, \text{d}v \end{align} $$

$V$ is a volume, but not any volume - it is the volume enclosed by the surface $A$.
$\text{d} v $ is an infinitesimal volume element, in other words an infinitely small volume piece of the considered volume $V$.
The upside down triangle $\nabla $ is called Nabla operator and it has three components like a vector. Its components, however, are not numbers, but derivatives corresponding to the space coordinates.

$$ \begin{align} \nabla ~=~ \begin{bmatrix} \frac{\partial}{\partial x} \\ \frac{\partial}{\partial y} \\ \frac{\partial}{\partial z} \end{bmatrix} \end{align} $$

The first component is the derivative with respect to $x$. The second component is the derivative with respect to $y$. And the third component is the derivative with respect to $z$.

Such an operator, like the Nabla operator, only takes effect when applied to a field. And that happens in the left-hand side 18. The Nabla operator is applied to the vector field $\boldsymbol{F} $ by taking the scalar product between the Nabla operator and the vector field:

Scalar product between Nabla operator and vector field

$$ \begin{align} \nabla ~\cdot~ \boldsymbol{F} ~=~ \frac{\partial F_x}{\partial x} + \frac{\partial F_y}{\partial y} + \frac{\partial F_z}{\partial z} \end{align} $$

As you can see in 20, it is the sum of the derivatives of the vector field with respect to the space coordinates $x$, $y$ and $z$. Such a scalar product between the Nabla operator and a vector field is called the divergence of the vector field $\boldsymbol{F}$. The result at the location ($x, ~ y, ~ z $) is no longer a vector, but a scalar, which can be either positive, negative or zero:

If the divergence at location $(x, ~ y, ~ z) $ is positive: $\nabla \cdot \boldsymbol{F} > 0 $, then there is a source of vector field $\boldsymbol{F}$ at this location. If this location is enclosed by a surface, then the flux $\Phi $ through the surface is also positive. The vector field so to speak 'flows out' of the surface.
If the divergence at location $(x, ~ y, ~ z) $ is negative: $\nabla \cdot \boldsymbol{F} < 0 $, then there is a sink of vector field $\boldsymbol{F}$ at this location. If this location is enclosed by a surface, then the flux $\Phi $ through the surface is also negative. The vector field 'flows into' the surface.
If the divergence at location $(x, ~ y, ~ z) $ disappears: $\nabla \cdot \boldsymbol{F} = 0 $, then that location is neither a sink nor a source of the vector field $\boldsymbol{F}$. The vector field does not flow out or into, or it flows in as much as out, so the two amounts cancel each other out.

Subsequently, the divergence $\nabla \cdot \boldsymbol{F} $ in 18, that is the sources and sinks of the vector field, is summed up at each location within the volume $V$ using the integral. Such an integral, where small pieces $\text{d}v$ of volume $V$ are summed up, is called volume integral.

What does the volume integral inside divergence theorem tell us?

Volume integral on the left-hand side is the sum of the sources and sinks of the vector field $ \boldsymbol{F} $ within a volume $V$.

So let's summarize the statement of the divergence integral theorem 10: On the left-hand side is the sum of the sources and sinks of the vector field within a volume and on the right-hand side is the total flux of the vector field through the surface of that volume. And the two sides should be equal.

What does the divergence theorem tell us?

The sum of the sources and sinks of a vector field $\boldsymbol{F} $ within a volume $V$ is the same as the flux $\Phi $ of the vector field through the surface $A$ of that volume.

Maths inside Maxwell equations: Curl integral theorem

Now consider the second important theorem necessary for understanding the Maxwell's equations:

Curl integral theorem as equation

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

If you understood the divergence integral theorem, then the curl integral theorem should not be totally cryptic to you. You already know the vector field $ \boldsymbol{F} $. The scalar product, the Nabla operator and the $\text{d} \boldsymbol{a} $ element should also be familiar to you now.

First, let's look at the right hand side of the equation 21:

The $L $ represents a line in space.
The circle around the integral sign indicates that this line must be closed, that is it should form a loop whose beginning and end are connected.
The $\text{d} \boldsymbol{l} $ is an infinitesimal line element of the loop, so an infinitely small piece of the line. Again, you should notice that the $\text{d} \boldsymbol{l} $ element is shown in boldface, it's a vector with a magnitude and a direction. The magnitude of the $\text{d} \boldsymbol{l} $ element indicates the length $\text{d} l $ of this small line and its direction points along the line $L$.

Now the scalar product between the vector field $\boldsymbol{F} $ and the line element $\text{d} \boldsymbol{l} $ is formed. You already know what the task of the scalar product is. First, split up the vector field into two parts:

Into $\boldsymbol{F}_{||} $ which is parallel to the $\text{d} \boldsymbol{l} $ element
and into $\boldsymbol{F}_{\perp} $ which is orthogonal to the $\text{d} \boldsymbol{l} $ element.

Line integral of a vector field along a closed line. Only the component of the vector field parallel to the line element contributes to the scalar product.

$$ \begin{align} \boldsymbol{F} ~=~ \boldsymbol{F}_{||} ~+~ \boldsymbol{F}_{\perp} \end{align} $$

The scalar product with the $\text{d} \boldsymbol{l} $ element eliminates the orthogonal component without touching the part of the vector field parallel to the $\text{d} \boldsymbol{l} $ element. Since at each location the $\text{d} \boldsymbol{l} $ element points along the line, only the part of the vector field that runs along the line $L$ is considered in the scalar product; the other part of the vector field drops out:

$$ \begin{align} \boldsymbol{F}_{\perp} ~\cdot~ \text{d}\boldsymbol{a} ~=~ 0 \end{align} $$

Then the scalar products (in 21) for each location on the loop are summed up using the integral. Such an integral, in which small line elements $\text{d} \boldsymbol{l} $ are summed up, is called line integral.

What does the line integral in the curl integral theorem tell us?

The line integral measures how much of the vector field $\boldsymbol{F} $ runs along the line $L$. Because the line is closed, this scalar product returns to the same point where the summation started. The closed line integral thus indicates how much of the vector field $\boldsymbol{F} $ rotates along the loop $L$.

If you take the line integral, as in this case, of a vector field $\boldsymbol{F}$, then the result of this line integral is called voltage $U$:

Voltage definition

$$ \begin{align} U ~=~ \int_L \boldsymbol{F} \cdot \text{d}\boldsymbol{l} \end{align} $$

Electric and magnetic voltage

The line integral of the electric field is the definition of the electric voltage.

If the vector field $\boldsymbol{F} $ in this line integral is an electric field $\boldsymbol{E} $, then this line integral is referred to as electric voltage $U_{\text e }$ along the line $L$ (this is the general definition of electric voltage):

$$ \begin{align} U_{\text e} ~=~ \int_L \boldsymbol{E} \cdot \text{d}\boldsymbol{l} \end{align} $$

On the other hand, when the vector field $\boldsymbol{F} $ is a magnetic field $ \class{violet}{\boldsymbol{B}} $, the line integral is called magnetic voltage $U_{\text m} $ along the line $L$:

$$ \begin{align} U_{\text m} ~=~ \int_L \class{violet}{\boldsymbol{B}} \cdot \text{d}\boldsymbol{l} \end{align} $$

The voltage in the case of an electric field is proportional to the energy that a positively charged particle gains as it passes the line $L$. A negatively charged particle, on the other hand, loses this energy as it passes the line $L$. The line integral 23 of the electric field, that is the electric voltage, measures the energy gain or energy loss of charged particles as they pass through the line $L$ under consideration.

Now you should have understood the right hand side of the curl integral theorem 21. Let's look at the left hand side now:

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

Here the surface $A$ occurs again. This surface, unlike in the divergence integral theorem, must not be a closed surface, but it is simply the surface enclosed by the line $L$.
$\text{d} \boldsymbol{a} $ is again an infinitesimal surface element of the surface $A$. At any location it is orthogonal on that surface.
In addition, here comes the cross product, which you have already met when we discussed the magnetic force 8.

Here, the cross product is formed between the Nabla operator $\nabla $ and the vector field $\boldsymbol{F} $. In addition to the scalar product, it is the second way to multiply two vectors. This cross product between the Nabla operator and the vector field $\boldsymbol{F} $ is called the curl of the vector field $\boldsymbol{F} $. The result, in contrast to the scalar product, is again a vector field:

Cross product between vector field and surface element

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

What does the cross product with the nabla operator mean?

Vector $\nabla \times \boldsymbol{F}$ expresses how much of the field $\boldsymbol{F} $ rotates around a point within the surface $A$.

Curl integral theorem visualized.

Then the scalar product is formed between the new vector field $\nabla \times \boldsymbol{F} $ with the infinitesimal surface element $\text{d} \boldsymbol{a} $. Thus, as you already know, only the part $(\nabla \times \boldsymbol{F})_{||} $ of $\nabla \times \boldsymbol{F} $ is picked out, which runs parallel to the surface element $\text{d} \boldsymbol{a} $. Since the surface element $\text{d} \boldsymbol{a} $ is orthogonal to the surface $A $, the scalar product picks out only the part of the vector field $\nabla \times \boldsymbol{F} $ which is also orthogonal on the surface $A $, so $(\nabla \times \boldsymbol{F})_{||} $.

Subsequently, all scalar products (in 24) within the surface $A $ are summed up. This is the task of the integral sign.

Let's summarize the statement of the curl integral theorem: On the right hand side, the vector field $\boldsymbol{F} $ is summed up along a line $L $. Thus, the rotation of the vector field around the enclosed surface is considered. On the left hand side, the curl $\nabla \times \boldsymbol{F} $ of the vector field $\boldsymbol{F} $ is summed up at each individual point within the surface. Both sides should be equal according to the curl integral theorem 21.

What does the curl integral theorem tell us?

The total curl of a vector field $\boldsymbol{F} $ within a surface $A $ is equal to the rotation of the vector field $\boldsymbol{F} $ along the edge $L$ of that surface.

With the acquired knowledge you are now ready to fully understand the Maxwell equations. Here we go!

The 1st Maxwell equation

This is the first Maxwell's equation expressed with an integral:

1st Maxwell equation in integral form

$$ \begin{align} \oint_A \class{blue}{\boldsymbol{E}} ~\cdot~ \text{d}\boldsymbol{a} ~=~ \frac{\class{red}{Q}}{\varepsilon_0} \end{align} $$

The left hand side of the Maxwell equation 27 should be familiar to you. It is a surface integral in which the electric field $\boldsymbol{E} $ occurs. This integral measures how much of the electric field comes out of or enters the surface A. The integral thus represents the electric flux $\Phi_{\text e} $ through the surface $A$:

$$ \begin{align} \Phi_{\text e} ~=~ \oint_A \boldsymbol{E} ~\cdot~ \text{d}\boldsymbol{a} \end{align} $$

On the right hand side in 27 is the total charge $Q $, which is enclosed by the surface $A$; divided by the electric field constant $\varepsilon_0 $, which provides the correct unit:

$$ \begin{align} \Phi_{\text e} ~=~ \frac{Q}{\varepsilon_0} \end{align} $$

What does the first Maxwell's equation in integral form mean?

The electric flux $\Phi_{\text e} $ through a closed surface $A $ corresponds to the electric charge $Q$ enclosed by this surface.

With the previously learned divergence integral theorem 10, which combines a volume integral with the surface integral, the surface integral on the left hand side of the first Maxwell equation 27 can be replaced by a volume integral of the divergence of the electric field:

$$ \begin{align} \int_{V} \left(\nabla \cdot \boldsymbol{E}\right) \, \text{d}v ~=~ \frac{Q}{\varepsilon_0} \end{align} $$

The enclosed charge $Q $ can also be expressed with a volume integral. By definition, charge density $\rho $ is charge per volume $V$. Bring the volume to the other side, then you have $Q = \rho \, V$. The volume $V$ can generally be written in the form of a volume integral. That is, the volume integral of the charge density $\rho $ over a volume $V$ is the charge enclosed in that volume. This turns the right hand side of the Maxwell equation into a volume integral:

$$ \begin{align} \int_{V} \left(\nabla \cdot \boldsymbol{E}\right) \, \text{d}v ~=~ \frac{1}{\varepsilon_0} \int_{V} \rho \, \text{d}v \end{align} $$

As you can see, we integrate over the same volume $V$ on both sides. For this equation to be satisfied for an arbitrarily chosen volume $V$, the integrands on both sides must be equal (the right integrand being multiplied by the constant $1/\varepsilon_0$). And now you have discovered the differential form of the first Maxwell equation:

1st Maxwell equation in differential form

$$ \begin{align} \nabla ~\cdot~ \class{blue}{\boldsymbol{E}} ~=~ \frac{\class{red}{\rho}}{\varepsilon_0} \end{align} $$

On the left hand side of the differential form 32 you can see the divergence $\nabla ~\cdot~ \boldsymbol{E}$ of the electric field $\boldsymbol{E}$. You know that at a specific point in space it can be positive, negative or zero. The sign of the divergence determines the type of the charge at the considered point in space:

If the divergence is positive: $\nabla ~\cdot~ \boldsymbol{E}(x,y,z) $ > $ 0 $, then the charge density $\rho(x,y,z)$ at this point in space is positive and thus also the charge $Q(x,y,z)$. In this point of space there is therefore a positive charge, which is a source of the electric field.
If the divergence is negative:$\nabla ~\cdot~ \boldsymbol{E} $ < $0 $, then the charge density $\rho(x,y,z)$ is negative and thus also the charge $Q(x,y,z)$. At this point of space there is therefore a negative charge, which is a sink of the electric field.
If the divergence is zero: $\nabla ~\cdot~ \boldsymbol{E} = 0$, then the charge density $\rho(x,y,z)$ is zero as well. At this point in space, there is either no charge, or there is just as much positive charge as negative, so the total charge at this point is cancelled out. At this point in space an ideal electric dipole could be located.

What does the first Maxwell's equation in differential form mean?

The electric charges are the sources and sinks of the electric field. Charges generate electric field!

The 2nd Maxwell equation

2nd Maxwell equation in integral form

$$ \begin{align} \oint_A \class{violet}{\boldsymbol{B}} ~\cdot~ \text{d}\boldsymbol{a} ~=~ 0 \end{align} $$

There is nothing unfamiliar in this equation. Everything should look familiar to you now. On the left hand side you see a surface integral over $A$. Now not an integral of an electric field, as in the first Maxwell equation, but an integral of a magnetic field $\class{violet}{\boldsymbol{B}}$. According to the equation the magnetic flux $\Phi_{\text m}$ through the closed surface $A$ is always zero:

$$ \begin{align} \Phi_{\text m} &~=~ \oint_A \class{violet}{\boldsymbol{B}} ~\cdot~ \text{d}\boldsymbol{a} \\\\
&~=~ 0 \end{align} $$

Magnetix flux through a closed surface A is always zero (2nd Maxwell equation, integral form)

What does the second Maxwell's equation in integral form mean?

There are always just as many magnetic field vectors coming out of a surface $A$ as there are vectors entering the surface.

With the divergence integral theorem 10, the surface integral in 33 can be transformed into a volume integral; this way, the divergence $\nabla \cdot \class{violet}{\boldsymbol{B}}$ of the magnetic field comes into play:

$$ \begin{align} \oint_{V} (\nabla \cdot \class{violet}{\boldsymbol{B}}) \, \text{d}v ~=~ 0 \end{align} $$

This integral in 35 shall be zero. The integral for any volume $V$ is only always zero if the integrand is zero. In this way, the second Maxwell equation emerges in its differential form:

2nd Maxwell equation in differential form

$$ \begin{align} \nabla \cdot \class{violet}{\boldsymbol{B}} ~=~ 0 \end{align} $$

If the divergence is zero, this means that at EACH point $(x,y,z)$ in space there is either no magnetic charge (also called a magnetic monopole) or there is just as much positive magnetic charge as negative, so the total charge at that point cancels out, such as in an ideal magnetic dipole, which always has both a north and a south pole. The north pole corresponds to a positive magnetic charge and the south pole corresponds to a negative magnetic charge. Since there are no magnetic monopoles, there are no separated sources and sinks of the magnetic field.

What does the second Maxwell's equation in differential form mean?

There are no magnetic monopoles that generate a magnetic field. Only magnetic dipoles can exist.

The second Maxwell equation, like the other Maxwell equations, is an experimental result. That is, if some day a magnetic charge should be found, for example, a single north pole without a corresponding south pole, then the second Maxwell equation have to be modified. Then the Maxwell equations would look even more symmetrical, more beautiful!

The 3rd Maxwell equation

You probably already know the third Maxwell equation under the name of Faraday's law of induction. The following third Maxwell equation is the most general form of the law of induction.

3rd Maxwell equation in integral form

$$ \begin{align} \oint_{L} \class{blue}{\boldsymbol{E}} ~\cdot~ \text{d}\boldsymbol{l} ~=~ -\int_{A} \frac{\partial \class{violet}{\boldsymbol{B}} }{\partial t} ~\cdot~ \text{d}\boldsymbol{a} \end{align} $$

Time changing magnetic field generates a rotating electric field and vice versa (3rd Maxwell equation, integral form).

On the left hand side in 37 is a line integral of the electric field $\boldsymbol{E}$ over a closed line $L$, which borders the surface $A$. This integral sums up all the parts of the electric field that run along the line $L$, which means, it sums up how much of the electric field rotates along the line. The integral corresponds to the electric voltage $U_{\text e}$ along the line $L$:

$$ \begin{align} U_{\text e} ~=~ \oint_{L} \boldsymbol{E} ~\cdot~ \text{d}\boldsymbol{l} \end{align} $$

On the right-hand side in 37 is a surface integral of the magnetic field $\class{violet}{\boldsymbol{B}}$ over an arbitrary surface $A$. We may pull out the time derivative out of the integral if the integration limits are time-independent. Physically said: If the area $A$, which the magnetic field penetrates, does not change. Then the integral of the magnetic field corresponds to the magnetic flux $\Phi_{\text m}$ through the surface $A$:

$$ \begin{align} \Phi_{\text m} ~=~ \int_{A} \class{violet}{\boldsymbol{B}} ~\cdot~ \text{d}\boldsymbol{a} \end{align} $$

According to Eq. 37 this magnetic flux is differentiated with respect to time $t$:

$$ \begin{align} \frac{\partial \Phi_{\text m}}{\partial t} ~=~ \frac{\partial }{\partial t}\int_{A} \class{violet}{\boldsymbol{B}} ~\cdot~ \text{d}\boldsymbol{a} \end{align} $$

The time derivative of the magnetic flux indicates how much the magnetic flux changes as time passes. So it's the temporal change of the magnetic flux. The larger the change of the magnetic flux, the greater the rotating electric field $\boldsymbol{E}$ in 38. The minus sign in 37 takes into account the direction of the rotation:

$$ \begin{align} U_{\text e} ~=~ - \frac{\partial }{\partial t}\int_{A} \class{violet}{\boldsymbol{B}} ~\cdot~ \text{d}\boldsymbol{a} \end{align} $$

If the change in the magnetic flux is positive: $ \frac{\partial \Phi_{\text m}}{\partial t} > 0 $, the electric voltage is negative: $U_{\text e} < 0 $.
If the change in the magnetic flux is negative: $ \frac{\partial \Phi_{\text m}}{\partial t} < 0 $, the electric voltage is positive: $U_{\text e} > 0 $.

The electric voltage and the change of the magnetic flux thus behave opposite to each other. The minus sign ensures energy conservation. Maybe you know that by the name: Lenz’s law. As you can see, according to the third Maxwell equation 37, rotating electric field produces time-varying magnetic field and vice versa. The Lenz’s law now states that the magnetic flux, which is generated by the rotating electric field, counteracts its cause. Because if it was not the case, the rotating electric field would amplify itself and thus generate energy out of nowhere. That is impossible!

What does the third Maxwell's equation in integral form mean?

The electric voltage $U_{\text e}$ along a closed line $L$ corresponds to the change in magnetic field through the surface $A$ bordered by that line. A change in the magnetic flux through the surface $A$ creates a voltage $U_{\text e}$ along the edge of $A$.

Special case: Time-independent magnetic field

Consider another important special case. If the magnetic field does NOT change in time, the right side of the Maxwell equation 37 will be eliminated:

$$ \begin{align} \oint_{L} \boldsymbol{E} ~\cdot~ \text{d}\boldsymbol{l} ~=~ 0 \end{align} $$

Then the equation states that the electric voltage along a closed line is always zero. So there is no rotating electric field, as long as the magnetic field doesn’t change over time.

If an electron passes the closed line $L$, it would not change its energy because, as you learned, electric voltage indicates how much energy a charge gains or loses when it passes a line. In this case, the voltage is zero. Therefore no energy gain.

With the curl integral theorem 21 you can transform the integral form 37 into the differential form. This theorem connects a line integral with a surface integral. To do this, simply replace the line integral in 37 with the surface integral in 21:

$$ \begin{align} \int_{A} \left(\nabla \times \boldsymbol{E} \right) ~\cdot~ \text{d}\boldsymbol{a} ~=~ - \int_{A} \frac{\partial \class{violet}{\boldsymbol{B}}}{\partial t} ~\cdot~ \text{d}\boldsymbol{a} \end{align} $$

This brings $\nabla \times \boldsymbol{E}$ into play. Since the equation applies to any surface $A$, the integrands on both sides must be equal. And just now, you have discovered the differential form of the third Maxwell equation:

3rd Maxwell equation in differential form

$$ \begin{align} \nabla \times \class{blue}{\boldsymbol{E}} ~=~ -\frac{\partial \class{violet}{\boldsymbol{B}}}{\partial t} \end{align} $$

The result of the cross product with $\boldsymbol{E}$ gives again a vector field, namely the vector field $\nabla \times \boldsymbol{E}$, which according to the properties of the cross product, is orthogonal to $\boldsymbol{E}$. Since the result of the cross product should be the change of the magnetic field, the vector $\frac{\partial \class{violet}{\boldsymbol{B}}}{\partial t}$ of the magnetic field change is orthogonal to the electric field $\boldsymbol{E}$.

What does the third Maxwell's equation in differential form mean?

A changing magnetic field $\class{violet}{\boldsymbol{B}}$ causes a rotating electric field $\boldsymbol{E}$ and vice versa in such a way that the energy conservation is fulfilled.

If the magnetic field does not change, i.e. is static, the right-hand side of equation 44 is zero and Maxwell's third equation simplifies to the electrostatic case:

Electrostatic 3rd Maxwell equation in differential form

$$ \begin{align} \nabla ~\times~ \boldsymbol{E} ~=~ 0 \end{align} $$

This equation again means: As long as there are no changing magnetic fields, the electric field is always vortex-free. If the rotation of a vector field vanishes, then the field is called conservative (energy conserving). Thus, the electrostatic field $\boldsymbol{E}$ is conservative with respect to 45. By "electrostatic" is meant that the E-field is time independent. It is generated by unmoved sources of the field.

The 4th Maxwell equation

Let's move on to the fourth, the last Maxwell's equation:

4th Maxwell equation in integral form

$$ \begin{align} \oint_{L} \class{violet}{\boldsymbol{B}} ~\cdot~ \text{d}\boldsymbol{l} ~=~ \mu_0 \, \class{red}{I} ~+~ \mu_0 \, \varepsilon_0 \, \int_{A} \frac{\partial \class{blue}{\boldsymbol{E}}}{\partial t} ~\cdot~ \text{d}\boldsymbol{a} \end{align} $$

Electric current and a time-dependent electric field generate a rotating magnetic field and vice versa.

What kind of an integral is on the left-hand side in 46? Exactly, a line integral of the magnetic field $\class{violet}{\boldsymbol{B}}$ along the closed line $L$. This is the definition of the magnetic voltage $U_{\text m}$. On the right side in 46 occurs the electric field constant $\varepsilon_0$ and the magnetic field constant $\mu_0$. They ensure that the unit on both sides of the Maxwell equation 46 is the same. In addition, something new occurs here, namely the electric current $\class{red}{I}$. When electric charges flow along a conductor, they generate a current $I$.

To the current contribution there is added a field contribution in eq. 46, in which the electric field $ \boldsymbol{E} $ occurs. If the surface $A$ does not change in time, then we may pull the time derivative in front of the integral. Then the surface integral of the electric field corresponds to the electric flux $\Phi_{\text e}$ through the surface $A$. In addition, the time derivative is in front of the electrical flux:

$$ \begin{align} U_{\text m} ~=~ \mu_0 \, \class{red}{I} ~+~ \mu_0 \, \varepsilon_0 \, \frac{\partial \Phi_{\text e}}{\partial t} \end{align} $$

On the right side of Eq. 47 are two contributions that can produce a vortex magnetic field: One contribution due to the current and one contribution due to the time changing electric flux.

What does the fourth Maxwell's equation in integral form mean?

The rotating magnetic field $ \class{violet}{\boldsymbol{B}}(t) $ is generated by an electric current $ I $ through the surface $A$ and by the changing electric field $ \boldsymbol{E}(t) $.

An electric current generates a rotating magnetic field.

An important special case arises, if the electric field $\boldsymbol{E}$ does not change over time, then the second summand in 46 is zero:

Ampere's law (magnetostatics)

$$ \begin{align} \oint_{L} \class{violet}{\boldsymbol{B}} ~\cdot~ \text{d}\boldsymbol{l} ~=~ \mu_0 \, \class{red}{I} \end{align} $$

Let us now derive the differential form of 46. With the curl integral theorem 21, you can transform the line integral into a surface integral, thus bringing the curl of the magnetic field $\class{violet}{\boldsymbol{B}}$ into play:

$$ \begin{align} \int_{A} (\nabla \times \class{violet}{\boldsymbol{B}}) ~\cdot~ \text{d}\boldsymbol{a} ~=~ \mu_0 \, \class{red}{I} + \mu_0 \, \varepsilon_0 \int_{A} \frac{\partial \boldsymbol{E}}{\partial t} ~\cdot~ \text{d}\boldsymbol{a} \end{align} $$

Now we have to express the current $I$ with a surface integral so that we get a single integrand on the right. We can do that simply by using the current density $j$. It indicates the current $I$ per area $A$ through which the current flows. Consequently, the current can also be written as the surface integral of the current density $\boldsymbol{j}$ over the surface $A$:

$$ \begin{align} \int_{A} (\nabla \times \class{violet}{\boldsymbol{B}}) ~\cdot~ \text{d}\boldsymbol{a} ~=~ \mu_0 \, \int_{A} \class{red}{\boldsymbol{j}} \cdot \text{d}\boldsymbol{a} ~+~ \mu_0 \, \varepsilon_0 \int_{A} \frac{\partial \boldsymbol{E}}{\partial t} ~\cdot~ \text{d}\boldsymbol{a} \end{align} $$

Note that here in the integral the scalar product of the current density $\boldsymbol{j}$ is taken with the $\text{d}\boldsymbol{a}$ element. So we pick only the part $\boldsymbol{j}_{||}$ of the current density vector that runs parallel to the $\text{d}\boldsymbol{a}$ element. Only this parallel part of the current density contributes to the current $I$ through the surface $A$.

You can pull the time derivative inside the integral. The two surface integrals can now be combined into one because we integrate over the same surface $A$:

$$ \begin{align} \int_{A} (\nabla \times \class{violet}{\boldsymbol{B}}) ~\cdot~ \text{d}\boldsymbol{a} ~=~ \mu_0 \, \int_{A} \left(\class{red}{\boldsymbol{j}} + \varepsilon_0 \, \frac{\partial \boldsymbol{E}}{\partial t} \right) ~\cdot~ \text{d}\boldsymbol{a} \end{align} $$

For the equation 51 to be satisfied for any surface $A$, the integrands on both sides must be equal. And you already have the differential form of the fourth Maxwell equation that you were looking for:

4th Maxwell equation in differential form

$$ \begin{align} \nabla ~\times~ \class{violet}{\boldsymbol{B}} ~=~ \mu_0 \left( \class{red}{\boldsymbol{j}} ~+~ \varepsilon_0 \frac{\partial \boldsymbol{E}}{\partial t} \right) \end{align} $$

The magnetic field $\class{violet}{\boldsymbol{B}}$ is orthogonal to the current density $\boldsymbol{j}$ and to the electric field change $\frac{\partial \boldsymbol{E}}{\partial t}$ because of the cross product.

What does the fourth Maxwell's equation in differential form mean?

The curl of the magnetic field $\class{violet}{\boldsymbol{B}}$ at a point in space is caused in two ways: 1) by the current density $\boldsymbol{j}$ and 2) by an electric field $\boldsymbol{E}(t)$ changing at this point in space.

Now you finally learned the foundation of all electrodynamics. Isn't it amazing how much knowledge and how many technical applications are contained in these four equations? By the way Maxwells equations still hide something interesting that can be revealed by a few steps of transforming the equations: Electromagnetic waves, light!