Maxwell's Equations

Derivation of Displacement Current and Inconsistency of Ampere's Law

Conditions: Time-varying electric fields, charging capacitor circuit

• Consider Ampere's circuital law in differential form for steady currents: \nabla \times $B = \mu_0 J$.
• Take the divergence of both sides: \nabla \cdot (\nabla \times B) = \mu_0 (\nabla \cdot J). Since the divergence of a curl is always zero, \nabla \cdot J = 0.
• However, the equation of continuity for charge conservation is \nabla \cdot J + \frac{\partial \rho}{\partial t} = 0, which implies \nabla \cdot J \neq 0 for time-varying fields.
• Substitute Gauss's law $\rho = \nabla \cdot D (or \rho = \epsilon_0 \nabla \cdot E) into the continuity equation: \nabla \cdot J + \frac{\partial}{\partial t}(\nabla \cdot (\epsilon_0 E)) = 0$.
• Rearrange to write \nabla \cdot \left( J + \epsilon_0 \frac{\partial E}{\partial t} \right) = 0.
• Define the displacement current density as $J_d = \epsilon_0 \frac{\partial E}{\partial t}$, leading to the modified Ampere-Maxwell law: \nabla \times $B = \mu_0 J + \mu_0 \epsilon_0 \frac{\partial E}{\partial t}$.

Result: \nabla \times $B = \mu_0 J + \mu_0 \epsilon_0 \frac{\partial E}{\partial t}$

Derivation of the Electromagnetic Wave Equation in Free Space

Conditions: Free space (\rho = 0, J = 0)

• Start with Maxwell's curl equations for free space: \nabla \times $E = -\frac{\partial B}{\partial t}$ and \nabla \times $B = \mu_0 \epsilon_0 \frac{\partial E}{\partial t}$.
• Take the curl of Faraday's equation: \nabla \times (\nabla \times E) = \nabla \times \left(-\frac{\partial B}{\partial t}\right) = -\frac{\partial}{\partial t}(\nabla \times B).
• Use the vector identity \nabla \times (\nabla \times E) = \nabla(\nabla \cdot E) - \nabla^2 E. In free space, \nabla \cdot E = 0, so \nabla \times (\nabla \times E) = -\nabla^2 E.
• Substitute the Ampere-Maxwell law into the right side: -\nabla^2 $E = -\frac{\partial}{\partial t} (\mu_0 \epsilon_0 \frac{\partial E}{\partial t}) = -\mu_0 \epsilon_0 \frac{\partial^2 E}{\partial t^2}$.
• Cancel negative signs to obtain the wave equation for electric field: \nabla^2 $E = \mu_0 \epsilon_0 \frac{\partial^2 E}{\partial t^2}$.
• Identify the propagation speed as $c = \frac{1}{\sqrt{\mu_0 \epsilon_0}}$.

Result: \nabla^2 $E = \mu_0 \epsilon_0 \frac{\partial^2 E}{\partial t^2} \text{}$ and } \nabla^2 $B = \mu_0 \epsilon_0 \frac{\partial^2 B}{\partial t^2}$

Plane Electromagnetic Wave Propagation

Draw: Orthogonal sinusoidal waves representing E vector along y-axis and B vector along z-axis, propagating along the x-axis, with arrows indicating cross-product direction E \times B.

Interpret: Demonstrates transverse nature of electromagnetic waves where electric and magnetic fields oscillate perpendicular to each other and to the direction of propagation.

Exam use: Frequently asked in conceptual questions regarding field orientations and wave vector direction.

Charging Capacitor and Displacement Current Loop

Draw: A parallel plate capacitor connected to a battery, showing conduction current in wires and a dashed contour loop between capacitor plates passing through the electric field.

Interpret: Proves that Ampere's law fails without the displacement current term because the contour can be bounded by different surfaces with the same edge.

Exam use: Tested via theoretical reasoning questions on consistency of Maxwell's equations.

Hertz's Spark Gap Experiment

Draw: An induction coil connected to a spark gap (transmitter) and a distant resonant wire loop with a micro spark gap (receiver).

Interpret: First experimental confirmation of the production and detection of high-frequency electromagnetic waves predicted by Maxwell.

Exam use: Historical and factual multiple-choice questions in PGT exams.

If the electric field in a region is given by $E = E_0 \sin(kx - \omega t)\hat{j}$, find the corresponding magnetic field B assuming propagation in free space.

• Use Faraday's law in differential form: \nabla \times $E = -\frac{\partial B}{\partial t}$.
• Calculate the curl of E: \nabla \times E = \left| \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & 0 & 0 \\ 0 & E_0 \sin(kx - \omega t) & 0 \end{matrix} \right| = \hat{k} \frac{\partial}{\partial x} [E_0 \sin(kx - \omega t)] = k E_0 \cos(kx - \omega t) \hat{k}.
• Integrate with respect to time t: $B = - \int (k E_0 \cos(kx - \omega t) \hat{k}) dt = \frac{k E_0}{\omega} \sin(kx - \omega t) \hat{k}$.
• Since \frac{\omega}{k} = c and \frac{1}{c} = \sqrt{\mu_0 \epsilon_0}, write B in terms of amplitude $B_0 = \frac{E_0}{c}$.

Answer: $B = \frac{E_0}{c} \sin(kx - \omega t)\hat{k}$

A parallel plate capacitor with circular plates of radius R = 0.1 m is being charged with a constant charging current I = 2 A. Calculate the displacement current density through the plates.

• The area of the circular plates is $A = \pi R^2 = \pi (0.1)^2 = 0.01\pi \text{ m}^2$.
• The total displacement current in the capacitor equals the conduction current in the connecting wires: $I_d = I = 2 \text{ A}$.
• Displacement current density J_d is given by $J_d = \frac{I_d}{A}$.
• Substitute the values: $J_d = \frac{2}{0.01\pi} = \frac{200}{\pi} \approx 63.66 \text{ A/m}^2$.

Answer: 63.66 A/m^2

In free space, the magnetic field of an electromagnetic wave is given by $B = 10^{-7} \sin(10^7 x - \omega t) \hat{k} \text{ Tesla}$. Determine the angular frequency \omega and the electric field amplitude E_0.

• From the argument of the sine function, the wave number is $k = 10^7 \text{ m}^{-1}$.
• The relation between angular frequency and wave number for light in free space is \omega = c k.
• Substitute $c = 3 \times 10^8 \text{ m/s}$ and $k = 10^7 \text{ m}^{-1}: \omega = (3 \times 10^8) \times (10^7) = 3 \times 10^{15} \text{ rad/s}$.
• The relation between electric and magnetic field amplitudes is $E_0 = c B_0$.
• Substitute $B_0 = 10^{-7} \text{ T}: E_0 = (3 \times 10^8) \times (10^{-7}) = 30 \text{ V/m}$.

Answer: $\omega = 3 \times 10^{15} \text{ rad/s}$ and $E_0 = 30 \text{ V/m}$