4A · Maxwell Equations + Wave Equation — Theory + 5 Solved PYQs

This page covers the equations side of Maxwell’s theory. It has four short theory sections, one figure, and five solved PYQs.

1. Conduction current vs displacement current

In a wire charges move. That is conduction current. In the gap of a charging capacitor no charges move, but the electric field between the plates changes with time. Maxwell said this changing field acts like a current. That is displacement current.

Capacitor-gap argument

Take a capacitor charging with current $I_c$. Draw an Amperian loop around the wire. Surface 1 cuts the wire: current through it is $I_c$. Surface 2 passes through the gap between the plates: conduction current through it is zero, but the electric flux $\Phi_E = EA$ grows with time. So $I_d$ in the gap continues the conduction current in the wire.

$$I_d = \varepsilon_0\frac{d\Phi_E}{dt} = \frac{dQ}{dt} = I_c$$

Unit check

$[\varepsilon_0,\partial E/\partial t] = (\text{C}^2/\text{N·m}^2)(\text{N/C·s}) = \text{C/s·m}^2 = \text{A/m}^2$, same as $J_c$. Correct.

2. Ampere law for non-steady currents + Maxwell correction

Old Ampere law $\oint\vec{B}\cdot d\vec{l} = \mu_0 I$ works for steady currents only. For a charging capacitor the same loop gives two answers, so it must be corrected.

Derivation of the correction

Take divergence of $\nabla\times\vec{B} = \mu_0\vec{J}$. The left side is zero (div of a curl is zero). So $\nabla\cdot\vec{J} = 0$. But continuity says

$$\nabla\cdot\vec{J} + \frac{\partial\rho}{\partial t} = 0$$

For non-steady currents $\partial\rho/\partial t \ne 0$, so Ampere law contradicts continuity. Fix: use Gauss law $\nabla\cdot\vec{E} = \rho/\varepsilon_0$ to replace $\rho$. Then

$$\nabla\times\vec{B} = \mu_0\vec{J} + \mu_0\varepsilon_0\frac{\partial\vec{E}}{\partial t}$$

This is the Ampere–Maxwell law. The integral form is $\oint\vec{B}\cdot d\vec{l} = \mu_0(I_c + I_d)$.

Unit check

$[\mu_0\varepsilon_0,\partial E/\partial t] = \text{T/m}$, same as $\nabla\times\vec{B}$. Correct.

3. All four Maxwell equations

Four laws that contain all of electricity and magnetism.

In a material medium, replace $\varepsilon_0 \to \varepsilon = K\varepsilon_0$ and $\mu_0 \to \mu$, with $\vec{D} = \varepsilon\vec{E}$ and $\vec{H} = \vec{B}/\mu$: $\nabla\cdot\vec{D} = \rho_f$, $\nabla\cdot\vec{B} = 0$, $\nabla\times\vec{E} = -\partial\vec{B}/\partial t$, $\nabla\times\vec{H} = \vec{J}_f + \partial\vec{D}/\partial t$.

4. EM wave equation in free space + speed of light

In free space ($\rho = 0$, $\vec{J} = 0$) the two curl equations combine to give a wave whose speed is the speed of light.

Fig 1. Free-space EM wave: $\vec{E}\perp\vec{B}$, both perpendicular to propagation. Speed $c = 1/\sqrt{\mu_0\varepsilon_0}$.

Quick derivation steps

In free space Faraday and Ampere–Maxwell give

$$\nabla\times\vec{E} = -\frac{\partial\vec{B}}{\partial t}, \qquad \nabla\times\vec{B} = \mu_0\varepsilon_0\frac{\partial\vec{E}}{\partial t}$$

Take curl of Faraday (to remove $\vec{B}$): $\nabla\times(\nabla\times\vec{E}) = -\partial(\nabla\times\vec{B})/\partial t$. Left side by identity $\nabla\times(\nabla\times\vec{E}) = \nabla(\nabla\cdot\vec{E}) - \nabla^2\vec{E}$. In free space $\nabla\cdot\vec{E} = 0$, so left side $= -\nabla^2\vec{E}$. Right side using Ampere–Maxwell $= -\mu_0\varepsilon_0,\partial^2\vec{E}/\partial t^2$. Equate:

$$\nabla^2\vec{E} = \mu_0\varepsilon_0\frac{\partial^2\vec{E}}{\partial t^2}$$

Compare with $\nabla^2 f = (1/v^2)\partial^2 f/\partial t^2$. So $1/v^2 = \mu_0\varepsilon_0$ and $c = 1/\sqrt{\mu_0\varepsilon_0}$. Putting $\mu_0 = 4\pi\times 10^{-7}$ H/m, $\varepsilon_0 = 8.85\times 10^{-12}$ F/m gives $c \approx 3.00\times 10^8$ m/s.

Solved PYQs

Reference. Brij Lal & Subrahmanyam Ch-11 (Maxwell equations, displacement current); Ajoy Ghatak Ch-7 (wave equation).