What are Maxwell's Equations?

Maxwell’s equations are a set of four differential equations that describe the electromagnetic field due to charges and currents. By the early 19th century, the phenomena of electricity and magnetism were clear. Coulomb's law of electrostatics was given in 1780, followed by Ampere's law in 1825. Later, Michael Faraday discovered electromagnetic induction. According to Faraday, there is always an electric field associated with an electric charge density and a magnetic field associated with a current density. Through these fields, the electric and magnetic fields communicate. James Clerk Maxwell followed Faraday's idea of electric and magnetic fields and wrote a comprehensive theory of electric fields/forces and magnetic fields/forces. In 1865, Maxwell summarized the mathematical foundation of his work known as Maxwell's equation of electromagnetism.

Maxwell's Equations

Let's start with the four Maxwell equations.

$$\begin{gathered} \nabla ·\overrightarrow{E}=\frac{\rho }{{\epsilon }_0} (Gauss\text{'} Law) \left(1\right) \\ \nabla \times \overrightarrow{E}=-\frac{\partial \overrightarrow{B}}{\partial t} (Faraday\text{'}s law ) \left(2\right) \\ \nabla ·\overrightarrow{B}=0 (magnetic monopole) \left(3\right) \\ \nabla \times \overrightarrow{B}=-{\mu }_0\overrightarrow{J}+{\mu }_0{\epsilon }_0\frac{\partial \overrightarrow{E}}{\partial t} (Ampere-Maxwell Law) \left(4\right) \end{gathered}$$

where ρ and $\overrightarrow{J}$ are the charge density and current density, respectively.

Each equation describes a feature of the electric field and the magnetic field. The terms on the left-hand side of each equation quantify the nature of the field, whereas, on the right-hand side, it has the source distribution.

Gauss law

The first equation, also knowns as the Gauss Law in differential form says that the divergence of electric field due to an electric charge is proportional to the magnitude of the source charge.

Gauss law for magnetism

According to this law, an electric charge has no magnetic analogs or magnetic monopoles. On the other hand, the net outflow of the magnetic field through a closed surface is zero because the magnetic field of material is attributed to a dipole. Such magnetic dipoles are inseparable pairs of equal and opposite 'magnetic charges' represented as loops of current. Hence, the total magnetic flux through a Gaussian surface is zero.

Faraday's law

The second equation is also known as Faraday's law of induction. It points out that if there is a changing magnetic field in some region of space, then there is a corresponding electric field in that region. According to it, the work done per unit charge to move around a closed loop is equal to the rate of change of the magnetic flux through the enclosed surface.

Magnetic monopole

The third equation implies that the divergence of the magnetic field is always zero. It contrasts with equation (1), according to which divergence of the electric field is proportional to the magnitude of the charge. This essentially means that the magnetic charge (also known as monopole) cannot exist in nature.

Ampere-Maxwell law

Finally, the fourth equation, Ampere-Maxwell law, says that the curl of the magnetic field in a region is proportional to the source current density. According to Ampere-Maxwell, the induced magnetic field around any closed loop is proportional to the electric current along with the displacement current through the enclosed surface. The displacement current is proportional to the rate of change of electric flux.

The constancy of the speed of light

One of the main achievements of Maxwell's equations was that they predicted light as a phenomenon of electromagnetic waves and its speed being constant in every frame of reference. To see this remarkable result, let's analyze Maxwell's equation in the absence of any sources, i.e.

$$\begin{gathered} \nabla ·\overrightarrow{E}=0 (Gauss\text{'} Law) (1*) \\ \nabla \times \overrightarrow{E}=-\frac{\partial \overrightarrow{B}}{\partial t} (Faraday\text{'}s law ) (2*) \\ \nabla ·\overrightarrow{B}=0 (magnetic monopole) (3*) \\ \nabla \times \overrightarrow{B}={\mu }_0{\epsilon }_0\frac{\partial \overrightarrow{E}}{\partial t} (Ampere-Maxwell Law) (4*) \end{gathered}$$

Taking the curl of Faraday's law

$\nabla \times \nabla \times \overrightarrow{E}=\nabla \times \left(-\frac{\partial \overrightarrow{B}}{\partial t}\right)$ 

Using the identity,

$\nabla \times \nabla \times \overrightarrow{A}=\nabla \left(\nabla ·\overrightarrow{A}\right)-{\nabla }^2\overrightarrow{A}$

Changing the order of derivative on the right-hand side because space and time derivative are independent, thus

$\nabla (\nabla ·\overrightarrow{E})-{\nabla }^{2}E=-\frac{\partial \left(\nabla \times B\right)}{\partial t}$

Plugging source free Ampere-Maxwell law (3*) in equation (4) and noting that source free Gauss' Law reads

$\nabla ·\overrightarrow{E}=0$

Now, equation (4) becomes

$$\begin{gathered} -{\nabla }^2\overrightarrow{E}=-\frac{\partial \left(\nabla \times \overrightarrow{B}\right)}{\partial t} \\ {\nabla }^2\overrightarrow{E}=\frac{\partial }{\partial t}\left({\mu }_0{\epsilon }_0\frac{\partial \overrightarrow{E}}{\partial t}\right) \\ {\nabla }^2\overrightarrow{E}-{\mu }_0{\epsilon }_0\frac{{\partial }^2\overrightarrow{E}}{\partial t^2}=0 \left(5\right) \\ \left({\nabla }^2-{\mu }_0{\epsilon }_0\frac{{\partial }^2}{\partial t^2}\right)\overrightarrow{E}=0 \left(6\right) \end{gathered}$$

A similar wave equation can also be derived for the magnetic field.

Equations (5) and (6) are in the form of the wave equation.

${\nabla }^{2}f(x,t)-\frac{{\partial }^{2}f(x,t)}{\partial t^2}=0 \left(7\right)$

where v is the speed of the wave.

Comparing (7) and (5) gives the speed of electromagnetic waves as

$v_{EM}=\frac{1}{{\mu }_0{\epsilon }_0} \left(8\right)$

This is also incidentally equal to the speed of light c.

Notice that the quantities appearing in expression (8) are constants and property of the medium in which the electromagnetic waves or light waves are traveling. So, we conclude that the speed of light cannot depend upon the choice of the reference frame. Hence, it must be a universal constant. This idea further helped Einstein in developing the Theory of Special Relativity.

$c=\frac{1}{{\mu }_0{\epsilon }_0}$                   (9)

Energy in an electromagnetic wave

Electromagnetic waves (EM) possess energy in the form of electromagnetic radiation. We can assign energy density as the sum of electric and magnetic field energy density. Assume that this energy density travels with the speed of light in a vacuum, we can use this idea to calculate the amount of energy passing through a unit area perpendicular to the direction of propagation of the electromagnetic wave. Since equation (5) is a wave equation, we use

$$\begin{gathered} U=\frac{1}{2}\left({\epsilon }_0{E}^2+\frac{B^2}{{\mu }_0}\right) \left(10\right) \\ \text{This expression can also be written as} \\ U=\frac{{\epsilon }_0}{2}\left(E^2+c^2{B}^2\right) \left(11\right) \end{gathered}$$

So at t=0, the sinusoidal solution to an electric field and magnetic field must vary as

$$\begin{gathered} E^2={E_0}^2{\sin }^{2}y \left(12\right) \\ {B}^2={E_0}^2{c}^2{\sin }^{2}y \left(13\right) \end{gathered}$$

Plugging (12) and (13) in (11) gives,

$U={\epsilon }_0{E_0}^2{\sin }^{2}y (14)$

The mean value of sin²y is 1/2, so it can be said that the mean energy density in the field is

$Energy=\frac{{\epsilon }_0{E_0}^2}{2} \left(15\right)$

and the rate of flow of energy 'u' in the direction perpendicular to y (propagation direction in this case) is

$u=\frac{c{\epsilon }_0{E_0}^2}{2} \left(16\right)$

We can generalize this idea to any continuous and repetitive wave, and posit that the rate of energy transfer in the direction perpendicular to the wave is proportional to

$S={\epsilon }_{0}c{E}^2 \left(17\right)$

The unit of quantity S is Joules per second per square meter.

This calculation of S is based on Maxwell's equations.

In free space, the constant in the equation (17) has a value

${\epsilon }_{0}c\approx \frac{1}{377}$ 

So,

$S\left(watts/m^2\right)=\frac{E^2\left(volt m\right)}{377 ohms} \left(18\right)$

This has the same units as the power dissipation in resistive load derived using Ohm's law.

$P=\frac{V^2}{R}$

The electromagnetic wave induces a current with the contained energy on encountering conductors.

Using Maxwell's equation, a more general form of quantity S in (18) can be derived, which is

$\frac{\partial U}{\partial t}={\epsilon }_0\frac{\partial \overrightarrow{E}}{\partial t}·\overrightarrow{E}+\frac{1}{{\mu }_0}\frac{\partial \overrightarrow{B}}{\partial t}· \overrightarrow{B} \left(19\right)$

$\frac{\partial U}{\partial t}=-\frac{1}{{\mu }_0}\nabla ·\left(\overrightarrow{E}\times \overrightarrow{B}\right) \left(20\right)$

The bracketed term on the right-hand side of equation (20) is known as the Poynting vector.

$\overrightarrow{S}=\frac{\overrightarrow{E}\times \overrightarrow{B}}{{\mu }_0} \left(21\right)$

To know more about the Poynting vector, notice a resemblance with the continuity equation.

$\frac{\partial \rho }{\partial t}=-\nabla ·\overrightarrow{J}$

It can be inferred that the Poynting vector plays the role of some kind of current density. In the case of electromagnetic waves, its power density, i.e., the energy is flowing out per unit time per unit area. In other words, the equation of continuity is a statement of conservation of charge.

Context and Applications

This topic finds significant roles in:

• Masters in Science (Physics)
• Bachelors in Science (Physics)
• Bachelors in Technology (Mechanical Engineering)

Practice Problems

1. Suppose a point charge q=1 nC lies at the center of a cube of side 1m. What is the total flux out of that cube?

1. $1.13\frac{N{m}^2}{C}$
2. $2.26\frac{N{m}^2}{C}$
3. $5\frac{N{m}^2}{C}$
4. $3.56\frac{N{m}^2}{C}$

Answer: Option a

Explanation: According to Gauss' law, the flux out of a volume is proportional to the charge contained in it. Its formula is $\Phi =\frac{q}{{\epsilon }_0}$. Plugging in the given values, we get the answer $1.13\frac{N{m}^2}{C}$

2. The electromagnetic field in some regions has an amplitude of 3.1 V/m and a frequency of 1.2 MHz. What is the energy density in the EM wave?

1. $6.45\times {10}^{-11}\frac{J}{m^3}$
2. $8.5\times {10}^{-11}\frac{J}{m^3}$
3. $5.6\times {10}^{-11}\frac{J}{m^3}$
4. $4.25\times {10}^{-11}\frac{J}{m^3}$

Answer: Option d

Explanation: The formula for energy density in EM waves is $\frac{{\epsilon }_0{E}^2}{2}$. Plugging in the given values, we get the answer $4.25\times {10}^{-11}\frac{J}{m^3}$ 

3. A charged particle moving with velocity v=6.27×10⁶ m/s is perpendicular to the magnetic field and an electric field with strength E=3×10⁶ N/C. What is the value of magnetic field strength?

1. $1 T$
2. $0.36 T$
3. $0.478 T$
4. $2.56 T$

Answer: Option c

Explanation: The electric and magnetic fields for a transverse wave are related as B=Ec. Plugging in the given values, we get the answer $0.478 T$ 

4. Which of the following laws does not follow from Maxwell's equation?

1. Gauss' law
2. Faraday's law
3. Planck's law
4. Ampere's law

Answer: Option c

Explanation: Planck's law is related to energy quantization, which has nothing to do with electromagnetism. Hence, the correct option is Planck's law.

5. Which of the following implies the conservation of charge?

1. Continuity equation
2. Gauss's law
3. Poynting vector
4. Faraday's law

Answer: Option a

Explanation: The continuity equation implies that the rate of accumulation of charge in a volume is equal to the current flowing through it, which means the charge remains conserved. So, the correct option is the continuity equation

Common Mistake

One thing to keep in mind while expressing the rate of flow of energy density in terms of the Poynting vector is that it has units of power density and not energy density. So, the Poynting vector describes energy flowing out per unit time per unit volume, not just per unit volume.

Related Concepts

• Fresnel equation
• Coulomb's Law
• Differential Gauss' Law
• Gauge Freedom
• Retarded Radiation
• Special Relativity