4B · EM Waves — Energy, Numerics, Verification + 4 Solved PYQs

Poynting vector, refractive-index numerics with full division work, EM wave properties, given E-B pair verification. Easy theory then solved questions.

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

1. Poynting vector + energy density

An EM wave carries energy. The Poynting vector $\vec{S}$ tells how much power flows per unit area and in which direction. Like water flow: amount per second through each square metre, pointing downstream.

Formula 1. $\vec{S} = \dfrac{1}{\mu_0}(\vec{E}\times\vec{B}) = \vec{E}\times\vec{H}$, unit W/m$^2$, points along the direction of propagation.

Symbols: $\vec{E}$ electric field (V/m), $\vec{B}$ magnetic field (T), $\vec{H} = \vec{B}/\mu_0$ in vacuum (A/m), $\mu_0 = 4\pi\times 10^{-7}$ H/m.

Energy density

Energy stored per unit volume: electric part $u_E = \tfrac{1}{2}\varepsilon_0 E^2$, magnetic part $u_B = B^2/(2\mu_0)$. For a plane wave $E = cB$, the two halves are equal, total $u = \varepsilon_0 E^2 = B^2/\mu_0$. Then $S = uc$ (energy density $u$ moving at speed $c$ gives flux $uc$).

Note. For a sine wave, the time-averaged $\langle S\rangle$ is half the peak value.

Unit check

$[E\times B/\mu_0] = (\text{V/m·T})/(\text{N/A}^2) = \text{W/m}^2$, power per area. Correct.

Transverse EM wave with E along k-hat and B along i-hat, propagating along y

Fig 1. $\vec{S} = (\vec{E}\times\vec{B})/\mu_0$ points along $+y$ when $\vec{E}\parallel\hat{k}$ and $\vec{B}\parallel\hat{\imath}$ (because $\hat{k}\times\hat{\imath} = \hat{\jmath}$).

2. Speed and wavelength in a medium

Light slows down inside glass or water. Refractive index $n$ tells by what factor.

Formula 2. $v = c/n$, $\lambda = \lambda_0/n$, where $c = 3\times 10^8$ m/s. Frequency $f$ does not change (wave crests must match at the boundary).

Worked divisions

Water $n = 1.33$: $1.33\times 2.25 = 2.9925 \approx 3.00$, so $v \approx 2.26\times 10^8$ m/s.

Glass $n = 1.658$, $\lambda_0 = 5893$ Å: $1.658\times 3554 = 5892.57 \approx 5893$, so $\lambda \approx 3554$ Å.

Glass speed: $1.658\times 1.81 = 3.00098 \approx 3.00$, so $v \approx 1.81\times 10^8$ m/s.

3. Properties of EM waves + how to verify a given field pair

Six properties: transverse, speed $c$, ratio $E/B = c$, carries energy and momentum, needs no medium, obeys reflection/refraction/interference/diffraction/polarization.

4-point verification test

For a given $\vec{E}$, $\vec{B}$ pair, check:

$\vec{E}\cdot\vec{B} = 0$ (perpendicular).
Find propagation direction from phase $(ky - \omega t)$ — wave travels along $+y$ (constant phase $y = (\omega t + \text{const})/k$ grows with $t$).
$E_0/B_0 = \omega/k = c$ (amplitude ratio).
Faraday $\nabla\times\vec{E} = -\partial\vec{B}/\partial t$ holds component-wise.

Reference. Brij Lal & Subrahmanyam Ch-11 (Poynting, energy density); Ajoy Ghatak Ch-7 (wave properties).

Solved PYQs

S4-Q6

Define Poynting vector (and its unit).

201820192022×3 Recall

$\vec{S} = (\vec{E}\times\vec{B})/\mu_0 = \vec{E}\times\vec{H}$ gives energy flux of the fields. Direction = propagation direction by right-hand rule. Unit W/m$^2$.

Steps

Write $\vec{S} = (\vec{E}\times\vec{B})/\mu_0 = \vec{E}\times\vec{H}$ (reason: energy flux of the fields).
Direction = propagation direction by right-hand rule (reason: cross product gives a perpendicular vector).
Unit derivation: $(\text{V/m})\times\text{T}\,\div\,(\text{N/A}^2) = \text{W/m}^2$ (reason: SI base unit check).

Answer

The Poynting vector is the energy flux density of an EM field. $\boxed{\vec{S} = \dfrac{1}{\mu_0}(\vec{E}\times\vec{B}) = \vec{E}\times\vec{H},\quad [\vec{S}] = \text{W/m}^2}$.

Exam tip

Asked ×3. One-line definition + formula + unit is full marks. Always add "direction = propagation direction" for safety.

S4-Q7

Refractive index & velocity of light in a medium: velocity/wavelength in water ($n = 1.33$) and glass ($n = 1.658$) for $\lambda = 5893$ Å.

202220232024×3 Recall

$v = c/n$, $\lambda = \lambda_0/n$, $c = 3.00\times 10^8$ m/s, $\lambda_0 = 5893$ Å.

Steps

Water: $v = 3.00\times 10^8/1.33 = 2.26\times 10^8$ m/s (reason: $1.33\times 2.25 = 2.9925 \approx 3.00$).
Glass wavelength: $\lambda = 5893/1.658 = 3554.3$ Å (reason: $1.658\times 3554 = 5892.57 \approx 5893$).
Glass speed: $v = 3.00\times 10^8/1.658 = 1.81\times 10^8$ m/s (reason: $1.658\times 1.81 = 3.00098 \approx 3.00$).
Convert if needed: $3554$ Å $= 3.554\times 10^{-7}$ m.

Answer

$\boxed{v_{\text{water}} \approx 2.26\times 10^8\text{ m/s},\quad \lambda_{\text{glass}} \approx 3554\text{ Å},\quad v_{\text{glass}} \approx 1.81\times 10^8\text{ m/s}}$.

Exam tip

Show each multiplication check ($1.33\times 2.25$, $1.658\times 3554$) — examiners give step marks for visible division work.

S4-Q8

Write down the main properties of electromagnetic waves.

2023 Recall

Six properties of EM waves: transverse, $E/B = c$, $v = c/n$, energy flux $\vec{S}$, carries momentum, travels in vacuum, obeys optical phenomena.

Steps

Transverse: $\vec{E}\perp\vec{B}$, both perpendicular to propagation (reason: from curl equations).
Speed $c = 1/\sqrt{\mu_0\varepsilon_0}$ in vacuum, $v = c/n$ in a medium (reason: wave equation).
Amplitude ratio $E/B = c$ (reason: substitute plane wave into Maxwell equations).
Carries energy (Poynting vector $\vec{S}$) and momentum (radiation pressure $p = u$ on absorption) (reason: from energy density argument).
Needs no medium — travels through vacuum (reason: derived from Maxwell equations alone).
Shows reflection, refraction, interference, diffraction, polarization (reason: light is an EM wave).

Answer

Transverse; $\vec{E}\perp\vec{B}$; speed $c = 1/\sqrt{\mu_0\varepsilon_0}$ in vacuum, $v = c/n$ in a medium; $E/B = c$; carries energy (Poynting vector) and momentum (radiation pressure); travels in vacuum; obeys reflection, refraction, interference, diffraction and polarization. $\boxed{\text{Transverse wave},\ E/B = c,\ v = c/n,\ \text{energy flux }\vec{S},\ \text{momentum},\ \text{vacuum travel}}$.

Exam tip

Five to six points with two formulas ($c$ and $E/B$) is a full-marks answer. One mark per correct point.

S4-Q9

Show that $\vec{E} = E_0\cos(ky-\omega t)\,\hat{k}$ and $\vec{B} = B_0\cos(ky-\omega t)\,\hat{\imath}$ represent an electromagnetic field.

2023 Recall

Use the 4-point test: perpendicular, propagation direction, amplitude ratio $E_0/B_0 = \omega/k = c$, and check Faraday's law component-wise.

Steps

$\vec{E}\cdot\vec{B} = E_0B_0\cos^2(ky-\omega t)(\hat{k}\cdot\hat{\imath}) = 0$ (reason: $\hat{k}\cdot\hat{\imath} = 0$, fields perpendicular).
Phase $(ky-\omega t)$: constant phase gives $y = (\omega t + \text{const})/k$, which grows with $t$, so propagation is $+y$ (reason: phase-front moves to larger $y$); also $\vec{S} \propto \hat{k}\times\hat{\imath} = \hat{\jmath}$ confirms $+y$.
Require $E_0/B_0 = \omega/k = c$ (reason: amplitude ratio for a vacuum wave).
Faraday check: $\nabla\times\vec{E} = -E_0k\sin(ky-\omega t)\,\hat{\imath}$ (reason: only $E_z$ varies, only with $y$), and $-\partial\vec{B}/\partial t = -B_0\omega\sin(ky-\omega t)\,\hat{\imath}$; equality needs $E_0k = B_0\omega$, i.e. $E_0/B_0 = \omega/k$. Same check works for Ampere–Maxwell.

Answer

All four tests pass. The given pair is a valid $y$-propagating plane EM wave with $\vec{E}\perp\vec{B}$ and amplitude ratio $E_0/B_0 = \omega/k = c$. $\boxed{\text{Valid plane wave with }\vec{E}\perp\vec{B},\ E_0/B_0 = \omega/k = c,\ \text{propagation }+\hat{\jmath}}$.

Exam tip

Do all four checks in order. Most students skip the curl check (step 4) and lose a mark.

Reference. Brij Lal & Subrahmanyam Ch-11 (Poynting, energy density); Ajoy Ghatak Ch-7 (wave properties).