Maxwell's equations are a set of four coupled partial differential equations that, together with the Lorentz force law, form the foundation of classical electromagnetism, classical optics, and electric circuits. They describe how electric and magnetic fields are generated by charges, currents, and changes of the fields.
James Clerk Maxwell unified electricity and magnetism by introducing the concept of the displacement current, which proved that light itself is an electromagnetic wave.
Physical Meaning: Electric charges act as sources or sinks for the electric field. The net electric flux through any closed surface is proportional to the enclosed electric charge.
Integral Form: $\oint \mathbf{E} \cdot d\mathbf{a} = \frac{Q_{enc}}{\varepsilon_0}$
Differential Form: Using the Divergence Theorem, $\oint \mathbf{E} \cdot d\mathbf{a} = \int (\nabla \cdot \mathbf{E}) d\tau$. Since $Q_{enc} = \int \rho d\tau$, we get:
$$ \nabla \cdot \mathbf{E} = \frac{\rho}{\varepsilon_0} $$Physical Meaning: There are no magnetic monopoles. Magnetic field lines always form continuous, closed loops.
Integral Form: $\oint \mathbf{B} \cdot d\mathbf{a} = 0$
Differential Form: Using the Divergence Theorem similarly gives:
$$ \nabla \cdot \mathbf{B} = 0 $$Physical Meaning: A time-varying magnetic field induces a circulating electric field.
Integral Form: $\oint \mathbf{E} \cdot d\mathbf{l} = -\frac{d}{dt} \int \mathbf{B} \cdot d\mathbf{a}$
Differential Form: Using Stokes' Theorem, $\oint \mathbf{E} \cdot d\mathbf{l} = \int (\nabla \times \mathbf{E}) \cdot d\mathbf{a}$. Equating the integrands gives:
$$ \nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t} $$Physical Meaning: Magnetic fields can be generated in two ways: by electrical current (Ampere's law) and by changing electric fields (Maxwell's addition: displacement current).
Differential Form: Ampere's original law was $\nabla \times \mathbf{B} = \mu_0 \mathbf{J}$. However, the divergence of a curl is always zero ($\nabla \cdot (\nabla \times \mathbf{B}) = 0$), but $\nabla \cdot \mathbf{J} = -\frac{\partial \rho}{\partial t} \neq 0$ for non-steady currents. Maxwell added the displacement current $\mathbf{J}_d = \varepsilon_0 \frac{\partial \mathbf{E}}{\partial t}$ to satisfy the continuity equation:
$$ \nabla \times \mathbf{B} = \mu_0 \mathbf{J} + \mu_0 \varepsilon_0 \frac{\partial \mathbf{E}}{\partial t} $$Use Faraday's Law: $\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}$
Calculate the curl of $\mathbf{E}$:
$$ \nabla \times \mathbf{E} = \begin{vmatrix} \hat{x} & \hat{y} & \hat{z} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ E_0 \cos(kz-\omega t) & 0 & 0 \end{vmatrix} = \hat{y}\frac{\partial}{\partial z}(E_0 \cos(kz-\omega t)) = \hat{y} (-E_0 k \sin(kz-\omega t)) $$So, $-\frac{\partial \mathbf{B}}{\partial t} = -E_0 k \sin(kz-\omega t) \hat{y}$.
Integrating with respect to time $t$:
$$ \mathbf{B} = \int E_0 k \sin(kz-\omega t) \hat{y} dt = E_0 k \frac{-\cos(kz-\omega t)}{-\omega} \hat{y} = \frac{E_0 k}{\omega} \cos(kz-\omega t) \hat{y} $$Since $\frac{k}{\omega} = \frac{1}{c}$, we get $\mathbf{B} = \frac{E_0}{c} \cos(kz-\omega t) \hat{y}$.
If $\mathbf{B}(t)$ is uniform in space but varies with time, then $\nabla \times \mathbf{B} = 0$ (because spatial derivatives of a uniform field are zero).
According to the Ampere-Maxwell law in vacuum: $\nabla \times \mathbf{B} = \mu_0 \varepsilon_0 \frac{\partial \mathbf{E}}{\partial t}$.
This would imply $\frac{\partial \mathbf{E}}{\partial t} = 0$. But if we take the curl of Faraday's law ($\nabla \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}$), we get a contradiction. Specifically, $\nabla \times (\nabla \times \mathbf{E}) = \nabla(\nabla\cdot\mathbf{E}) - \nabla^2\mathbf{E} = 0$, meaning $\nabla \times \mathbf{E}$ must be spatially constant, so $\frac{\partial \mathbf{B}}{\partial t}$ is spatially constant. This alone is fine, but it violates the wave equation derived from combining them unless the field is zero. It implies an infinite propagation speed, which violates special relativity and Maxwell's complete framework.