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# 4 Maxwell equations: everything you need to know!

from the category: Theories

## Definition

Maxwell equations - are four coupled differential equations that summarize the experimental findings of electrodynamics and represent the fundamental equations for the electromagnetic properties of nature.

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:

1. 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.
2. 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.
3. 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.
4. 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.
Any device that exploits electricity or magnetism is fundamentally based on the Maxwell equations.

The goal of this article is not to derive the Maxwell equations theoretically or experimentally, but to present them as simple and understandable as possible. I will also briefly explain the mathematics that occur in the Maxwell equations. However, you should know what a partial derivative is and what an integral is before you continue to read the article. To understand all Maxwell equations you first need to know what an electric and magnetic field is.

## Electric field

Consider a large electrically charged sphere with the charge $$Q$$ and a small sphere with the charge $$q$$. The electric force $$F_{\text e}$$ between these two spheres, which are at a distance $$r$$ from each other, is given by Coulombs law: 1 $F_{\text e} = \frac{1}{4\pi \, \varepsilon_0} \, \frac{Q \, q}{r^2}$ here $$1/4 \ pi \, \varepsilon_0$$ is a constant prefactor with the electric field constant $$\varepsilon_0$$, which provides the right unit of force, namely the unit Newton [N].

Now what if you know the value of the big charge and want to know the value of the force that the big charge exerts on the small charge? 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 Coulombs law. To do so, you simply divide Coulombs law by $$q$$ on both sides: 1.1 $\frac{F_{\text e}}{q} = \frac{1}{4\pi \, \varepsilon_0} \, \frac{Q}{r^2}$

This way, on the right hand side, the small charge 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 $$Q$$: 1.2 $E := \frac{F_{\text e}}{q}$ By calling it source charge we imply that the charge $$Q$$ is the source of the electric field. The electric field $$E$$ thus indicates the electric force which would act on a small charge $$q$$ when placed at the distance $$r$$ to the source charge $$Q$$. The electric field $$E$$ represents the electric force which would act on a small charge $$q$$, if this small charge is placed at a distance $$r$$ to the source charge $$Q$$.

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 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 $$E_{\text z}$$: 1.3 $\boldsymbol{E} = \left(\begin{array}{c} E_{\text x} \\ E_{\text y} \\ E_{\text z} \end{array}\right)$

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.

## Magnetic field (magnetic flux density)

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 at the velocity $$v$$ in an external magnetic field experiences a magnetic force $$F_{\text m}$$, which deflects the particle. 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 speed 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 $$B$$. So the magnetic force is given by: 2 $F_{\text m} = q \, v \, B$ The unit of the quantity $$B$$ must be such that the right hand side of the equation 2 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 $$B$$ must be [kg/As²]. This is what we will call the unit of Tesla [T = kg/As²]. And we call $$B$$ magnetic flux density (or short: magnetic field). The magnetic flux density describes the external magnetic field and thus determines the magnitude of the force on a charged particle.

The equation 2 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: 2.1 $\boldsymbol{F_{\text m}} = \left(\begin{array}{c} F_{\text{mx}} \\ F_{\text{my}} \\ F_{\text{mz}} \end{array}\right), ~~~ \boldsymbol{v} = \left(\begin{array}{c} v_{\text x} \\ v_{\text y} \\ v_{\text z} \end{array}\right), ~~~ \boldsymbol{B} = \left(\begin{array}{c} B_{\text x} \\ B_{\text y} \\ B_{\text z} \end{array}\right)$

Now, the three quantities are not scalars, but three-dimensional vectors with components in the x-, y- and z-direction. Now the question is: How should the velocity vector $$\boldsymbol{v}$$ be vectorially multiplied by the magnetic field vector $$\boldsymbol{B}$$? 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 so-called cross product.

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

Cross product of two vectors 2.2 $\boldsymbol{v} ~\times~ \boldsymbol{B} ~=~ \left(\begin{array}{c} v_yB_z-v_zB_y \\ v_zB_x-v_xB_z \\ v_xB_y-v_yB_x \end{array}\right)$

For the force $$\boldsymbol{F_{\text m}}$$ to be always orthogonal to $$\boldsymbol{v}$$ and $$\boldsymbol{B}$$, we take the cross product of $$\boldsymbol{v}$$ and $$\boldsymbol{B}$$ in our equation 2. So in vector form the magnetic force is generally given by: 2.3 $\boldsymbol{F}_{\text m} = q \, \boldsymbol{v} \times \boldsymbol{B}$

The charge $$q$$ is an ordinary scalar factor. The magnetic flux density $$\boldsymbol{B}$$ 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 $$\boldsymbol{B}$$. Both are so-called 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 $$\boldsymbol{B}(x, ~ y, ~ z)$$ field vector indicating both magnitude and direction of the electric and magnetic fields. Electric and magnetic fields are generally three-dimensional vector fields. $\boldsymbol{E} = \left(\begin{array}{c} E_{\text x} \\ E_{\text y} \\ E_{\text z} \end{array}\right), ~~~ \boldsymbol{B} = \left(\begin{array}{c} B_{\text x} \\ B_{\text y} \\ B_{\text z} \end{array}\right)$

There are a total of four Maxwell equations. These four Maxwell equations can be represented in two different ways. There is the so-called 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. One theorem is called divergence integral theorem and the other one curl integral theorem. If you understand the two theorems, it will be easier for you to understand the Maxwell equations. Let's first look at the divergence integral theorem.

## Divergence integral theorem

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

Divergence integral theorem 3 $\int_{V} \left(\nabla \cdot \boldsymbol{F}\right) \, \text{d}v ~=~ \oint_{A}\boldsymbol{F} \cdot \text{d}\boldsymbol{a}$

First, let's look at the right hand side of the equation 3. 3.1 $\oint_A \boldsymbol{F} \cdot \text{d}\boldsymbol{a}$

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 $$\boldsymbol{B}$$ when considering the Maxwell equations. So it is a vector with three components. $$\text{d} \boldsymbol{a}$$ 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 so-called 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 - Definition 3.2 $\boldsymbol{F} \cdot \text{d}\boldsymbol{a} = F_{\text x} \, \text{d}a_{\text x} ~+~ F_{\text y} \, \text{d}a_{\text y} ~+~ F_{\text z} \, \text{d}a_{\text z}$

As you can see from the definition 3.2, 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 so-called 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 $$\boldsymbol{F}_{\perp}$$: 3.3 $\boldsymbol{F} = \boldsymbol{F}_{||} + \boldsymbol{F}_{\perp}$

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 3.4 $\boldsymbol{F}_{\perp} \cdot \text{d}\boldsymbol{a} = 0$ however, the scalar product between the part $$\boldsymbol{F}_{||}$$ and the $$\text{d} \boldsymbol{a}$$ element is generally not zero.

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 3.1, the scalar products for all locations of the considered surface $$A$$ are added up. That is the task of the integral. The right-hand side of the divergence integral theorem 3 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 $$A$$: 3.5 $\Phi = \oint_A \boldsymbol{F} \cdot \text{d}\boldsymbol{a}$

This description is based on what this surface integral means: It measures how much of the vector field $$\boldsymbol{F}$$ flows out or flows into a considered surface $$A$$. An area integral measures how much of the vector field $$\boldsymbol{F}$$ enters the surface $$A$$ or exits the surface $$A$$. The surface integral measures the flux of the vector field through this area.

Electric and magnetic flux

If the vector field $$\boldsymbol{F}$$ in the surface integral 3.5 is an electric field $$\boldsymbol{E}$$, then this surface integral is called electric flux $$\Phi_{\text e }$$ through the surface $$A$$: 3.6 $\Phi_{\text e} = \int_A \boldsymbol{E} \cdot \text{d}\boldsymbol{a}$

And if the vector field $$\boldsymbol{F}$$ is a magnetic field $$\boldsymbol{B}$$, the surface integral is called magnetic flux $$\Phi_{\text m}$$ through the surface $$A$$: 3.7 $\Phi_{\text m} = \int_A \boldsymbol{B} \cdot \text{d}\boldsymbol{a}$

Now let's look at the left hand side of the theorem 3: 3.8 $\int_{V} \left(\nabla \cdot \boldsymbol{F}\right) \, \text{d}v$

$$V$$ 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:

Nabla operator 3.9 $\nabla ~=~ \left(\begin{array}{c} \frac{\partial}{\partial x} \\ \frac{\partial}{\partial y} \\ \frac{\partial}{\partial z} \end{array}\right)$

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 also happens in the integral 3.8. The Nabla operator is applied to the vector field $$\boldsymbol{F}$$ by taking the scalar product between the Nabla operator and the vector field: 3.10 $\nabla ~\cdot~ \boldsymbol{F} ~=~ \frac{\partial F_x}{\partial x} + \frac{\partial F_y}{\partial y} + \frac{\partial F_z}{\partial z}$

As you can see in 3.10, 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 3.8, 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.

So let's summarize the statement of the divergence integral theorem 3: On the left side is the sum of the sources and sinks of the vector field within a volume and on the right side is the total flow of the vector field through the surface of that volume. And the two sides should be the same. The divergence integral theorem thus states that the sum of the sources and sinks of a vector field $$\boldsymbol{F}$$ within a volume $$V$$ is the same as the flow $$\Phi$$ of the vector field through the surface $$A$$ of that volume.

## Curl integral theorem

Now consider the second important theorem necessary for understanding the Maxwell equations, namely the curl integral theorem.

Curl integral theorem 4 $\int_{A} (\nabla \times \boldsymbol{F}) \cdot \text{d}\boldsymbol{a} ~=~ \oint_{L} \boldsymbol{F} \cdot \text{d}\boldsymbol{l}$

If you understand 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 4. The $$L$$ is 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. 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.

Then the scalar products (in 4) 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.

Now you know what happens on the right hand side of the curl integral theorem 4: 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 $$L$$.

Electric and magnetic 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): 4.2 $U_{\text e} = \int_L \boldsymbol{E} \cdot \text{d}\boldsymbol{l}$

On the other hand, when the vector field $$\boldsymbol{F}$$ is a magnetic field $$\boldsymbol{B}$$, the line integral is called magnetic voltage $$U_{\text m}$$ along the line $$L$$: 4.3 $U_{\text m} = \int_L \boldsymbol{B} \cdot \text{d}\boldsymbol{l}$

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 4.2 of the electric field, that is the 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 4. Let's look at the left hand side of 4 now: 4.4 $\int_{A} (\nabla \times \boldsymbol{F}) \cdot \text{d}\boldsymbol{a}$

Here the area $$A$$ occurs again. This area, unlike 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 piece of the surface $$A$$ and 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 2.3. 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:

Curl 4.5 $\nabla \times \boldsymbol{F} ~=~ \left(\begin{array}{c} \frac{\partial F_{\text z}}{\partial y} - \frac{\partial F_{\text y}}{\partial z} \\ \frac{\partial F_{\text x}}{\partial z} - \frac{\partial F_{\text z}}{\partial x} \\ \frac{\partial F_{\text y}}{\partial x} - \frac{\partial F_{\text x}}{\partial y} \end{array}\right)$
Vector $$\nabla \times \boldsymbol{F}$$ expresses how much of the field $$\boldsymbol{F}$$ rotates around a point within the surface $$A$$.

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 4.4) within the surface $$A$$ are summed up by means of the integral.

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 4. The curl integral theorem thus states that the total curl of a vector field $$\boldsymbol{F}$$ within a surface $$A$$ corresponds to the rotation of the vector field $$\boldsymbol{F}$$ along the edge $$L$$ of that surface.

Well, it is somehow clear that the rotation of the vector field inside of the surface cancels in the summation and only the rotation of the vector field along the edge $$L$$ remains.

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

## The first Maxwell equation

This is the first Maxwell equation in integral form:

1. Maxwell equation in integral form 5 $\oint_A \boldsymbol{E} ~\cdot~ \text{d}\boldsymbol{a} ~=~ \frac{Q}{\varepsilon_0}$

The left hand side of the Maxwell equation 5 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 $$A$$: 5.1 $\Phi_{\text e} = \oint_A \boldsymbol{E} ~\cdot~ \text{d}\boldsymbol{a}$

On the right hand side in 5 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. 5.2 $\Phi_{\text e} = \frac{Q}{\varepsilon_0}$ The first Maxwell equation states that 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 3, which combines a volume integral with the surface integral, the surface integral on the left hand side of the first Maxwell equation 5 can be replaced by a volume integral of the divergence of the electric field: 5.3 $\int_{V} \left(\nabla \cdot \boldsymbol{F}\right) \, \text{d}v ~=~ \frac{Q}{\varepsilon_0}$

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: 5.4 $\int_{V} \left(\nabla \cdot \boldsymbol{F}\right) \, \text{d}v ~=~ \frac{1}{\varepsilon_0} \int_{V} \rho \, \text{d}v$

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:

First Maxwell equation in differential form 5.5 $\nabla ~\cdot~ \boldsymbol{E} ~=~ \frac{1}{\varepsilon_0} \, \rho$

On the left hand side of the differential form 5.5 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.
The first Maxwell equation in differential form states that the electric charges are the sources and sinks of the electric field. Charges generate the electric field!
Good to know! Coulomb’s law 1.1 is a special case of the first Maxwell equation.

## The second Maxwell equation

Second Maxwell equation in integral form 6 $\oint_{A} \boldsymbol{B} ~\cdot~ \text{d}\boldsymbol{a} ~=~ 0$

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 $$\boldsymbol{B}$$. According to the equation the magnetic flux $$\Phi_{\text m}$$ through the closed surface $$A$$ is always zero: 6.1 $\Phi_{\text m} = \oint_A \boldsymbol{B} ~\cdot~ \text{d}\boldsymbol{a} = 0$ The second Maxwell equation states that 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 3, the surface integral in 6 can be transformed into a volume integral; this way, the divergence $$\nabla \cdot \boldsymbol{B}$$ of the magnetic field comes into play: 6.2 $\oint_{V} (\nabla \cdot \boldsymbol{B}) \, \text{d}v ~=~ 0$

This integral in 6.2 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:

Second Maxwell equation in differential form 6.3 $\nabla ~\cdot~ \boldsymbol{B} ~=~ 0$

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. The second Maxwell equation in differential form states that 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 6 have to be modified. Then the Maxwell equations would look even more symmetrical, more beautiful!

## The third Maxwell equation

This is what the third Maxwell equation looks like in an integral form:

Third Maxwell equation in integral form 7 $\oint_{L} \boldsymbol{E} ~\cdot~ \text{d}\boldsymbol{l} ~=~ - \frac{\partial}{\partial t} \int_{A} \boldsymbol{B} ~\cdot~ \text{d}\boldsymbol{a}$

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

On the left hand side in 7 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 $$L$$: 7.1 $U_{\text e} = \oint_{L} \boldsymbol{E} ~\cdot~ \text{d}\boldsymbol{l}$

On the right hand side in 7 is a surface integral of the magnetic field $$\boldsymbol{B}$$ over an arbitrary surface $$A$$. This integral corresponds to the magnetic flux $$\Phi_{\text m}$$ through the surface $$A$$: 7.2 $\Phi_{\text m} = \int_{A} \boldsymbol{B} ~\cdot~ \text{d}\boldsymbol{a}$

This magnetic flux is differentiated with respect to time $$t$$: 7.3 $\frac{\partial \Phi_{\text m}}{\partial t} = \frac{\partial }{\partial t}\int_{A} \boldsymbol{B} ~\cdot~ \text{d}\boldsymbol{a}$

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 7.1. The minus sign in 7 takes into account the direction of the rotation: 7.4 $U_{\text e} = - \frac{\partial }{\partial t}\int_{A} \boldsymbol{B} ~\cdot~ \text{d}\boldsymbol{a}$

• 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 7, 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! The third Maxwell equation states that 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. In other words, a change in the magnetic flux through the surface $$A$$ creates an electric voltage $$U_{\text e}$$ along the edge of $$A$$.

Example: 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 7 will be eliminated: 7.5 $\oint_{L} \boldsymbol{E} ~\cdot~ \text{d}\boldsymbol{l} = 0$

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 4 you can transform the integral form 7 into the differential form. This theorem connects a line integral with a surface integral. To do this, simply replace the line integral in 7 with the surface integral in 4: 7.6 $\int_{A} \left(\nabla \times \boldsymbol{E} \right) ~\cdot~ \text{d}\boldsymbol{a} ~=~ - \frac{\partial}{\partial t} \int_{A} \boldsymbol{B} ~\cdot~ \text{d}\boldsymbol{a}$

This brings $$\nabla \times \boldsymbol{E}$$ into play. On the other side in 7.6 you may pull the time derivative inside the integral. 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:

Third Mawell equation in differential form 7.7 $\nabla \times \boldsymbol{E} ~=~ -\frac{\partial \boldsymbol{B}}{\partial t}$
The third Maxwell equation in differential form states that a changing magnetic field $$\boldsymbol{B}$$ causes a rotating electric field $$\boldsymbol{E}$$ and vice versa in such a way that the energy conservation is fulfilled.

## The fourth Maxwell equation

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

Fourth Maxwell equation in integral form 8 $\oint_{L} \boldsymbol{B} ~\cdot~ \text{d}\boldsymbol{l} ~=~ \mu_0 \, I + \mu_0 \, \varepsilon_0 \, \frac{\partial }{\partial t} \int_{A} \boldsymbol{E} ~\cdot~ \text{d}\boldsymbol{a}$

What kind of an integral is on the left side in 8? Exactly, a line integral of the magnetic field $$\boldsymbol{B}$$ along the closed line $$L$$. This is the definition of the magnetic voltage $$U_{\text m}$$. On the right side in 8 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 8 is the same. In addition, something new occurs here, namely the electric current $$I$$. When electric charges flow along a conductor, they generate a current $$I$$.

Also there is another summand in 8. You know the surface integral of the electric field. This is the electric flux $$\Phi_{\text e}$$ through the surface $$A$$. In addition, a time derivative is ahead of the electric flux. So the whole thing is the temporal change of the electric flux: 8.1 $U_{\text m} ~=~ \mu_0 \, I + \mu_0 \, \varepsilon_0 \, \frac{\partial \Phi_{\text e}}{\partial t}$

In summary: On the right side in 8.1 there are two summands: One contribution by the current and one contribution by the change of the electric flux. The fourth Maxwell equation states that the rotating magnetic field is generated first by an electric current through the surface $$A$$ and second by the changing electric field.

Ampere's law (magnetostatics)

An important special case arises, if the electric flux does not change over time, then the second summand in 8 is zero: 8.2 $\oint_{L} \boldsymbol{B} ~\cdot~ \text{d}\boldsymbol{l} ~=~ \mu_0 \, I$

Let us now derive the differential form of 8. With the curl integral theorem 4, you can transform the line integral into a surface integral, thus bringing the curl of the magnetic field $$\boldsymbol{B}$$ into play: 8.3 $\int_{A} (\nabla \times \boldsymbol{B}) ~\cdot~ \text{d}\boldsymbol{a} ~=~ \mu_0 \, I + \mu_0 \, \varepsilon_0 \, \frac{\partial }{\partial t} \int_{A} \boldsymbol{E} ~\cdot~ \text{d}\boldsymbol{a}$

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 $$A$$: 8.4 $\int_{A} (\nabla \times \boldsymbol{B}) ~\cdot~ \text{d}\boldsymbol{a} ~=~ \mu_0 \, \int_{A} \boldsymbol{j} \cdot \text{d}\boldsymbol{a} ~+~ \mu_0 \, \varepsilon_0 \, \frac{\partial }{\partial t} \int_{A} \boldsymbol{E} ~\cdot~ \text{d}\boldsymbol{a}$

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 $$A$$: 8.5 $\int_{A} (\nabla \times \boldsymbol{B}) ~\cdot~ \text{d}\boldsymbol{a} ~=~ \mu_0 \, \int_{A} \left(\boldsymbol{j} + \mu_0 \, \varepsilon_0 \, \frac{\partial }{\partial t} \int_{A} \boldsymbol{E} \right) ~\cdot~ \text{d}\boldsymbol{a}$

For the equation 8.5 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:

Fourth Maxwell equation in differential form 8.6 $\nabla ~\times~ \boldsymbol{B} ~=~ \mu_0 \left( \boldsymbol{j} ~+~ \varepsilon_0 \frac{\partial \boldsymbol{E}}{\partial t} \right)$
The differential form thus states that the curl of the magnetic field $$\boldsymbol{B}$$ at a point in space is caused in two ways: by the current density $$\boldsymbol{j}$$ and by an electric field $$\boldsymbol{E}(t)$$ changing at this point in space.
It is not over yet... 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! But that's a topic for another article.
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