Electromagnetic waves — derivation from Maxwell's equations: Worked examples

1.8.33 · D3 · Physics › Electromagnetism › Electromagnetic waves — derivation from Maxwell's equations

Definition Is page par use hone wale symbols (ek baar define kiye, taaki kuch "assumed" na ho)

$E, B$ — electric aur magnetic field vectors; $E_{0}, B_{0}$ unke peak amplitudes hain (wiggle ki sabse badi value).
$\hat{E}, \hat{B}, \hat{k}$ — unit vectors (length 1) jo $E$ , $B$ , aur travel ki directions batate hain.
$k$ — the wavenumber: ek metre mein wave ke kitne radians fit hote hain, $k = 2 π / λ$ . Bada $k$ = tightly packed crests.
$ω$ — the angular frequency: ek second mein oscillation ke kitne radians hote hain, $ω = 2 π f$ . Bada $ω$ = fast wiggling.
$J$ — the current density: unit area se kitna electric current flow karta hai (vacuum mein $J = 0$ ).
$u_{E} = \frac{1}{2} ε_{0} E^{2}$ — electric energy density (cubic metre mein $E$ dwara stored joules); $u_{B} = \frac{1}{2 μ _{0}} B^{2}$ magnetic version hai.
$S$ — the Poynting vector: ek square metre se per second flow hone wali energy, $S = \frac{1}{μ _{0}} E \times B$ . Iska time-average $⟨ S ⟩$ likha jaata hai.

The scenario matrix

Har EM-wave problem jo tum meet karte ho woh actually inhi cells mein se ek hai. Column "Example" batata hai ki neeche kaunsa worked example us cell ko nail karta hai.

#	Cell class	Tricky kyun hai	Example
C1	Amplitude conversion $B_{0} \to E_{0}$ ya wapas	kaunsa "bada" hai, factor $c$	Ex 1
C2	Frequency ↔ wavelength, tiny/huge scales	$c = f λ$ spectrum mein	Ex 2
C3	Direction / triad geometry ( $\hat{E}, \hat{B}, \hat{k}$ ke saare axis & sign combos)	right-hand rule, kaunsa axis kaunsa	Ex 3 (figure)
C4	Zero / degenerate input ( $E ∥ k$ , $B_{0} = 0$ , $J = 0$ )	kya wave exist bhi karta hai?	Ex 4
C5	Energy density & Poynting flux (real-world power)	equal energies, unit traps	Ex 5
C6	Verify a candidate solution (dispersion $ω = c k$ )	kya yeh actually ek wave hai?	Ex 6
C7	Limiting behaviour ( $c$ agar $μ_{0}$ ya $ε_{0}$ scale ho, medium slow-down)	$c$ constants ke saath kaise respond karta hai	Ex 7
C8	Exam twist (given a full field, extract everything)	ek expression se $k, ω, \hat{k}$ padhna	Ex 8 (figure)

Do constants har jagah repeat hote hain, isliye inhe ek baar pin kar lo: $μ_{0} = 4 π \times 1 0^{- 7} T\cdotm/A = 1.2566 \times 1 0^{- 6}, ε_{0} = 8.854 \times 1 0^{- 12} F/m, c = 2.998 \times 1 0^{8} m/s .$

Ex 1 — Amplitude conversion (cell C1)

Worked example Ek measured

$B$ -field se $E$ -field nikalna Vacuum mein ek radio wave ka magnetic amplitude $B_{0} = 3.0 \times 1 0^{- 9} T$ hai. $E_{0}$ nikalo.

Forecast: abhi guess karo — $E_{0}$ kuch volts/metre ke aas-paas hoga, ya lagbhag ek billion? (Tesla mein itne chhote numbers aksar $c$ se multiply hone par bahut bade ho jaate hain.)

Relation likho. $E_{0} = c B_{0}$ . Yeh step kyun? Faraday's law ko plane wave par apply karne se $k E_{0} = ω B_{0}$ mila, aur $ω / k = c$ , isliye amplitudes ka ratio exactly $c$ hai — parent note ka plane-wave section dekho.
Substitute karo. $E_{0} = (2.998 \times 1 0^{8}) (3.0 \times 1 0^{- 9}) = 0.8994 V/m$ . Yeh step kyun? Pure plug-in; bada $c$ aur chhota $B_{0}$ almost cancel kar dete hain, aur result $1 V/m$ ke paas aata hai.

Verify: wapas divide karo — $E_{0} / B_{0} = 0.8994/3.0 \times 1 0^{- 9} = 2.998 \times 1 0^{8} = c$ . ✓ Units: $(m/s) (T) = (m/s) (V\cdots/m^{2}) = V/m$ . ✓

Ex 2 — Frequency ↔ wavelength across the spectrum (cell C2)

Worked example

Electromagnetic Spectrum ke do ends ek saath (a) Ek FM station $f = 90.0 MHz$ par broadcast karta hai. $λ$ nikalo. (b) Ek gamma ray ka $λ = 1.0 \times 1 0^{- 12} m$ hai. $f$ nikalo.

Forecast: FM wavelength — kisi insaan se bada hoga ya chhota? Gamma frequency — 20 zeros se zyada honge ya kam?

Relation. $c = f λ$ , isliye $λ = c / f$ aur $f = c / λ$ . Yeh step kyun? $v = ω / k$ mein $ω = 2 π f$ , $k = 2 π / λ$ daalne se $v = f λ$ milta hai; vacuum mein $v = c$ .
(a) $λ = \frac{2.998 \times 1 0 ^{8}}{9.00 \times 1 0 ^{7}} = 3.33 m$ . Yeh step kyun? MHz $= 1 0^{6} Hz$ ; kuch-metre ka answer match karta hai is baat se ki FM antennas metre-scale ke kyun hote hain.
(b) $f = \frac{2.998 \times 1 0 ^{8}}{1.0 \times 1 0 ^{- 12}} = 3.0 \times 1 0^{20} Hz$ . Yeh step kyun? Chhoti wavelength ⇒ enormous frequency, exactly yahi "gamma" ka matlab hai.

Verify: wapas multiply karo. (a) $(9.00 \times 1 0^{7}) (3.33) = 2.997 \times 1 0^{8} \approx c$ . ✓ (b) $(3.0 \times 1 0^{20}) (1 0^{- 12}) = 3.0 \times 1 0^{8} = c$ . ✓

Ex 3 — The direction triad, every axis and sign (cell C3)

Figure — Electromagnetic waves — derivation from Maxwell's equations

Worked example

$B$ kis taraf point karta hai — saare axis aur sign combos cover karke? Rule ka use karo ki $\hat{E}, \hat{B}, \hat{k}$ ek right-handed set banate hain. Hum $\hat{k}$ ke roop mein teeno coordinate axes ko walk karte hain, aur $\hat{E}$ ke dono signs ko, taaki har axis/sign combination represent ho: (A) $\hat{k} = + \overset{x}{^}, \hat{E} = + \overset{y}{^}$ ; (B) $\hat{k} = + \overset{x}{^}, \hat{E} = - \overset{y}{^}$ ; (C) $\hat{k} = + \overset{z}{^}, \hat{E} = + \overset{x}{^}$ ; (D) $\hat{k} = - \overset{x}{^}, \hat{E} = + \overset{y}{^}$ ; (E) $\hat{k} = + \overset{y}{^}, \hat{E} = + \overset{z}{^}$ ; (F) $\hat{k} = + \overset{z}{^}, \hat{E} = - \overset{x}{^}$ ; (G) $\hat{k} = - \overset{y}{^}, \hat{E} = + \overset{x}{^}$ .

Forecast: agar sirf travel direction $\hat{k}$ flip karo (case A → case D), toh kya $\hat{B}$ bhi flip hoga?

Rule. Kyunki $\hat{E}, \hat{B}, \hat{k}$ mutually perpendicular unit vectors hain aur $\hat{E} \times \hat{B} = \hat{k}$ hai, same right-hand set se $\hat{B} = \hat{k} \times \hat{E}$ milta hai. Yeh step kyun? Faraday's law $\nabla \times E = - \partial_{t} B$ $B$ ko $E$ aur $k$ dono ke perpendicular lock karta hai; right-hand rule sign fix karta hai.
Case A ( $\hat{k} = \overset{x}{^}, \hat{E} = \overset{y}{^}$ ): $\hat{B} = \overset{x}{^} \times \overset{y}{^} = + \overset{z}{^}$ (figure mein violet arrow). Yeh step kyun? Baseline case — seedha $\hat{B} = \hat{k} \times \hat{E}$ mein plug in karo; do "positive" unit vectors ka cross teesra deta hai.
Case B ( $\hat{k} = \overset{x}{^}, \hat{E} = - \overset{y}{^}$ ): $\hat{B} = \overset{x}{^} \times (- \overset{y}{^}) = - \overset{z}{^}$ . Yeh step kyun? $\hat{E}$ flip karo ⇒ $\hat{B}$ bhi flip hoga, toh wave phir bhi $+ \overset{x}{^}$ travel karta hai — yeh same axis pair ka sign variation hai.
Case C (ek alag axis, $\hat{k} = \overset{z}{^}, \hat{E} = \overset{x}{^}$ ): $\hat{B} = \overset{z}{^} \times \overset{x}{^} = + \overset{y}{^}$ . Yeh step kyun? "Kaunsa axis" wala confusion cover karta hai — rule $\hat{B} = \hat{k} \times \hat{E}$ bilkul same hai chahe koi bhi axis koi bhi role play kare; yahan cyclic order $z \to x \to y$ $+ \overset{y}{^}$ deta hai.
Case D (travel flip, $\hat{k} = - \overset{x}{^}, \hat{E} = \overset{y}{^}$ ): $\hat{B} = (- \overset{x}{^}) \times \overset{y}{^} = - \overset{z}{^}$ . Yeh step kyun? Sirf $\hat{k}$ flip karna ( $\hat{E}$ nahi) $\hat{B}$ ko flip kar deta hai — forecast ka jawab: haan, flip hota hai.
Case E ( $\hat{k} = + \overset{y}{^}, \hat{E} = + \overset{z}{^}$ ): $\hat{B} = \overset{y}{^} \times \overset{z}{^} = + \overset{x}{^}$ . Yeh step kyun? Teesra coordinate axis ( $+ \overset{y}{^}$ ) travel direction ke roop mein cover karta hai, teeno positive axes ko $\hat{k}$ ke roop mein complete karta hai.
Case F ( $\hat{k} = + \overset{z}{^}, \hat{E} = - \overset{x}{^}$ ): $\hat{B} = \overset{z}{^} \times (- \overset{x}{^}) = - \overset{y}{^}$ . Yeh step kyun? Case C jaisa hi $\hat{k}$ hai lekin $\hat{E}$ ka opposite sign — dikhata hai ki sign flip is axis pair par bhi carry through hota hai.
Case G ( $\hat{k} = - \overset{y}{^}, \hat{E} = + \overset{x}{^}$ ): $\hat{B} = (- \overset{y}{^}) \times \overset{x}{^} = + \overset{z}{^}$ . Yeh step kyun? Travel direction ke roop mein ek negative axis aur positive $\hat{E}$ — confirm karta hai ki negative- $\hat{k}$ cases bhi identical rule follow karte hain, aur poora axis/sign census close ho jaata hai.

Verify: har baar $\hat{E} \times \hat{B} = \hat{k}$ check karo. A: $\overset{y}{^} \times \overset{z}{^} = \overset{x}{^}$ . ✓ B: $(- \overset{y}{^}) \times (- \overset{z}{^}) = \overset{x}{^}$ . ✓ C: $\overset{x}{^} \times \overset{y}{^} = \overset{z}{^}$ . ✓ D: $\overset{y}{^} \times (- \overset{z}{^}) = - \overset{x}{^}$ . ✓ E: $\overset{z}{^} \times \overset{x}{^} = \overset{y}{^}$ . ✓ F: $(- \overset{x}{^}) \times (- \overset{y}{^}) = \overset{z}{^}$ . ✓ G: $\overset{x}{^} \times \overset{z}{^} = - \overset{y}{^}$ . ✓ Har coordinate axis (dono signs) $\hat{k}$ ke roop mein, aur $\hat{E}$ ke dono signs, same triad follow karte hain — yeh Polarization ke peeche ki geometry hai.

Ex 4 — Zero aur degenerate inputs (cell C4)

Worked example Teen "kya wave exist bhi karta hai?" traps

Yaad karo $J$ current density hai (current per unit area; vacuum mein zero). Decide karo ki har configuration valid vacuum EM wave hai ya nahi: (a) $E$ travel direction $\hat{k}$ ke along point kar raha ho; (b) ek field jisme $E_{0} \neq = 0$ lekin $B_{0} = 0$ ho; (c) plain Ampère $\nabla \times B = μ_{0} J$ use karo $J = 0$ ke saath (Displacement Current term drop karke).

Forecast: teenon mein se kitne survive karte hain?

(a) $\nabla \cdot E = 0$ se $ik E_{0 k} = 0 \Rightarrow E_{0 k} = 0$ milta hai. Yeh step kyun? Gauss's law vacuum mein $\hat{k}$ ke along koi bhi field component forbid karta hai. Ek longitudinal EM wave impossible hai — yeh degenerate hai, ruled out. Sound longitudinal ho sakti hai; light nahi.
(b) Plane wave ke liye Faraday deta hai $k E_{0} = ω B_{0}$ . Agar $E_{0} \neq = 0$ toh $B_{0} = k E_{0} / ω \neq = 0$ . Yeh step kyun? $E$ wiggle nahi kar sakta bina $B$ ke — ek changing $E$ $B$ ka source hai (Ampère–Maxwell). $B_{0} = 0$ contradictory hai, wave nahi.
(c) Hum parent note ki derivation repeat karte hain lekin crippled Ampère law ke saath. Faraday's law ka curl lete hain: $\nabla \times (\nabla \times E) = \nabla \times (- \frac{\partial B}{\partial t}) = - \frac{\partial}{\partial t} (\nabla \times B) .$ Left side, curl-of-curl identity aur $\nabla \cdot E = 0$ use karke, $- \nabla^{2} E$ hai. Right side ab crippled law $\nabla \times B = μ_{0} J = 0$ use karta hai (kyunki $J = 0$ ), jo deta hai $- \frac{\partial}{\partial t} (0) = 0$ . Dono sides equate karne par: $- \nabla^{2} E = 0$ , yaani $\nabla^{2} E = 0$ — Laplace's equation, koi time term nahi, koi wave nahi. Yeh step kyun? Displacement current $μ_{0} ε_{0} \partial_{t} E$ ke bina, right side mein kuch nahi bachta jo $\partial_{t}^{2} E$ term ban sake. Woh missing term literally wahi hai jo light banata hai.

Verify: teeno degenerate cases fail hote hain, har ek alag Maxwell equation ki wajah se: (a) Gauss, (b) Faraday, (c) Ampère–Maxwell. Ek genuine wave ke liye charo ka saath kaam karna zaroori hai. ✓

Ex 5 — Energy density aur Poynting power (cell C5, real-world)

Worked example Atmosphere ke top par solar power

Yaad karo (symbol box se): $u_{E} = \frac{1}{2} ε_{0} E^{2}$ electric energy density hai (J/m³), $u_{B} = \frac{1}{2 μ _{0}} B^{2}$ magnetic wali, aur $S = \frac{1}{μ _{0}} E \times B$ Poynting vector hai (power per m²), time-average $⟨ S ⟩$ ke saath. Sunlight $E_{0} = 1030 V/m$ ke saath aati hai. (a) $B_{0}$ nikalo. (b) $⟨ S ⟩ = \frac{1}{2} ε_{0} c E_{0}^{2}$ nikalo. (c) Amplitudes use karke $u_{E} = u_{B}$ confirm karo.

Forecast: kya $⟨ S ⟩$ famous "solar constant" lagbhag $1360 W/m^{2}$ ke paas aayega?

(a) $B_{0} = E_{0} / c = 1030/2.998 \times 1 0^{8} = 3.435 \times 1 0^{- 6} T$ . Yeh step kyun? Wahi $E_{0} = c B_{0}$ relation, Ex 1 ki tarah ulta run kiya.
(b) $⟨ S ⟩ = \frac{1}{2} ε_{0} c E_{0}^{2} = \frac{1}{2} (8.854 \times 1 0^{- 12}) (2.998 \times 1 0^{8}) (1030)^{2} = 1408 W/m^{2}$ . Yeh step kyun? $S = E B / μ_{0}$ ko ek cycle par time-average karne se factor $\frac{1}{2}$ aata hai; $B = E / c$ aur $c = 1/ μ_{0} ε_{0}$ substitute karne par $\frac{1}{2} ε_{0} c E_{0}^{2}$ form milta hai. Result real solar constant se match karta hai. ✓
(c) $u_{E} = \frac{1}{2} ε_{0} E_{0}^{2}$ aur $u_{B} = \frac{1}{2 μ _{0}} B_{0}^{2}$ . Dono compute karo. Yeh step kyun? Parent note equal energy claim karta hai; hum usse trust karne ki jagah numerically test karte hain.

Verify: $u_{E} = \frac{1}{2} (8.854 \times 1 0^{- 12}) (1030)^{2} = 4.696 \times 1 0^{- 6} J/m^{3}$ ; $u_{B} = \frac{1}{2 ( 1.2566 \times 1 0 ^{- 6} )} (3.435 \times 1 0^{- 6})^{2} = 4.696 \times 1 0^{- 6} J/m^{3}$ . 4 sig figs tak equal. ✓ $⟨ S ⟩$ ke units: $(F/m) (m/s) (V/m)^{2} = W/m^{2}$ . ✓

Ex 6 — Ek candidate solution verify karna (cell C6)

Worked example Kya yeh expression actually ek wave hai?

Test karo ki kya $E_{y} = E_{0} cos (k x - ω t)$ equation $\frac{\partial ^{2} E}{\partial x ^{2}} = μ_{0} ε_{0} \frac{\partial ^{2} E}{\partial t ^{2}}$ solve karta hai, aur $ω, k$ par condition nikalo. (Yahan $k$ wavenumber hai aur $ω$ angular frequency, symbol box mein define hai.)

Forecast: koi bhi $ω$ aur koi bhi $k$ , ya sirf ek special pair?

Space derivative do baar. $\frac{\partial E}{\partial x} = - k E_{0} sin (k x - ω t)$ , phir $\frac{\partial ^{2} E}{\partial x ^{2}} = - k^{2} E$ . Yeh step kyun? Cosine ke do $x$ -derivatives $(- k^{2})$ neeche kheechtey hain aur cosine wapas return karte hain — shape survive karta hai.
Time derivative do baar. $\frac{\partial E}{\partial t} = + ω E_{0} sin (k x - ω t)$ , phir $\frac{\partial ^{2} E}{\partial t ^{2}} = - ω^{2} E$ . Yeh step kyun? $t$ mein same mechanism, $(- ω^{2})$ deta hai.
Equation impose karo. $- k^{2} E = μ_{0} ε_{0} (- ω^{2}) E \Rightarrow k^{2} = μ_{0} ε_{0} ω^{2} \Rightarrow \frac{ω}{k} = \frac{1}{μ _{0} ε _{0}} = c$ . Yeh step kyun? Cosine dono sides se cancel ho jaata hai; jo bachta hai woh dispersion relation hai — yeh tab hi wave hai jab $ω = c k$ .

Verify: $k = 2 π / λ$ , $ω = 2 π f$ ke saath, condition $ω = c k$ ban jaata hai $f λ = c$ — wahi rule jo Ex 2 mein use hua. Consistent. ✓ Numerically $k = 1$ , $ω = c$ ke liye: $k^{2} = 1$ aur $μ_{0} ε_{0} ω^{2} = μ_{0} ε_{0} c^{2} = 1$ . ✓

Ex 7 — $c$ ka limiting behaviour (cell C7)

Worked example Agar constants change ho jaayein toh?

(a) Ek student galti se $ε_{0}$ double kar deta hai (thought experiment). $c$ kis factor se change hoga? (b) Alag se, $μ_{0}$ double karta hai instead. Ab kis factor se? (c) Glass mein relative permittivity $ε_{r} = 2.25$ (aur $μ_{r} = 1$ ) ke saath, wave speed $v$ aur refractive index $n = c / v$ nikalo.

Forecast: kisi bhi ek constant ko double karna wave ko speed up karta hai ya slow down — aur kya fark padta hai kaun sa double kiya?

(a) $c = \frac{1}{μ _{0} ε _{0}}$ , isliye $ε_{0} \to 2 ε_{0}$ deta hai $c \to \frac{1}{2} c$ . Yeh step kyun? $c$ depends karta hai $ε_{0}^{- 1/2}$ par; $ε_{0}$ scale karne par $c$ inverse square root se scale hota hai. Bada $ε_{0}$ ⇒ slower wave.
(b) Dono constants ke symmetric role ki wajah se, $μ_{0} \to 2 μ_{0}$ bhi deta hai $c \to \frac{1}{2} c$ . Yeh step kyun? Formula $c = 1/ μ_{0} ε_{0}$ mein $μ_{0}$ aur $ε_{0}$ ko identically treat kiya jaata hai — kisi ko bhi double karne ka exact same effect hota hai. Yeh edge case dikhata hai ki koi bhi constant "special" nahi hai.
(c) Medium mein $v = \frac{1}{μ _{0} μ _{r} ε _{0} ε _{r}} = \frac{c}{μ _{r} ε _{r}} = \frac{c}{2.25} = \frac{c}{1.5}$ . Yeh step kyun? Ek dielectric Ampère–Maxwell mein $ε_{0} \to ε_{0} ε_{r}$ replace karta hai; derivation bilkul same hai lekin naye constant ke saath.
Compute karo. $v = 2.998 \times 1 0^{8} /1.5 = 1.999 \times 1 0^{8} m/s$ , $n = c / v = 1.5$ . Yeh step kyun? $n = μ_{r} ε_{r} = 2.25 = 1.5$ — glass mein light slow hoti hai aur bend karti hai.

Verify: (a) aur (b) dono factor $\frac{1}{2} = 0.7071$ dete hain; square karne par $1/ (μ_{0} \cdot 2 ε_{0})$ aur $1/ (2 μ_{0} \cdot ε_{0})$ wapas milta hai — identical. ✓ (c) $n \cdot v = 1.5 \times 1.999 \times 1 0^{8} = 2.998 \times 1 0^{8} = c$ . ✓

Ex 8 — Exam twist: ek expression se sab kuch padhna (cell C8)

Worked example "Saate quantities extract karo" waala question

Given (SI units) $E = (120 V/m) cos [(1.00 \times 1 0^{7} m^{- 1}) z - (3.00 \times 1 0^{15} s^{- 1}) t] \overset{x}{^}$ . Nikalo: (1) $E_{0}$ , (2) $k$ (wavenumber), (3) $ω$ (angular frequency), (4) travel ki direction, (5) wave speed, (6) $B_{0}$ , (7) $B$ ki direction.

Forecast: jo speed milegi — kya woh $c$ niklegi, confirm karte hue ki yeh vacuum wave hai?

Amplitudes aur shape padho. $E_{0} = 120 V/m$ ; $k = 1.00 \times 1 0^{7} m^{- 1}$ ; $ω = 3.00 \times 1 0^{15} s^{- 1}$ . Yeh step kyun? $cos (k z - ω t)$ mein position ko multiply karne wala number $k$ hai, $t$ ko multiply karne wala $ω$ hai (dono symbol box mein define hain).
Travel ki direction. Phase hai $k z - ω t$ (ek $z$ -term minus ek $t$ -term) ⇒ $+ \overset{z}{^}$ ke along travel karta hai. Yeh step kyun? Constant phase ke ek point ke liye $z$ badhna zaroori hai jab $t$ badhta hai, isliye wave $+ z$ ki taraf move karta hai — figure mein orange travel arrow.
Speed. $v = ω / k = 3.00 \times 1 0^{15} /1.00 \times 1 0^{7} = 3.00 \times 1 0^{8} m/s = c$ . Yeh step kyun? $ω / k$ phase velocity hai; $c$ milna vacuum confirm karta hai.
$B_{0}$ . $B_{0} = E_{0} / c = 120/2.998 \times 1 0^{8} = 4.003 \times 1 0^{- 7} T$ . Yeh step kyun? Faraday's law se amplitude relation, exactly Ex 1 ki tarah.
$B$ ki direction. $\hat{E} = \overset{x}{^}$ , $\hat{k} = \overset{z}{^}$ , isliye $\hat{B} = \hat{k} \times \hat{E} = \overset{z}{^} \times \overset{x}{^} = \overset{y}{^}$ . Yeh step kyun? Triad rule (Ex 3); $B$ $\pm \overset{y}{^}$ ke along oscillate karta hai, $E$ ke saath in phase.

Verify: $v = ω / k = 3.00 \times 1 0^{8}$ matches $c$ . ✓ $\hat{E} \times \hat{B} = \overset{x}{^} \times \overset{y}{^} = \overset{z}{^} = \hat{k}$ . ✓ $E_{0} / B_{0} = 120/ (4.003 \times 1 0^{- 7}) = 2.998 \times 1 0^{8} = c$ . ✓

Recall Rapid self-test (jawab dene ke baad reveal karo)

Kaunsa Maxwell equation longitudinal EM wave forbid karta hai? ::: Gauss's law $\nabla \cdot E = 0$ ( $\hat{k}$ ke along component ko khatam karta hai). Agar $B_{0} = 5 \times 1 0^{- 8}$ T ho, toh $E_{0}$ kya hai? ::: $E_{0} = c B_{0} = 15.0 V/m$ . Ek wave mein $\hat{E} = + \overset{y}{^}$ , $\hat{k} = + \overset{z}{^}$ hai; $B$ kis taraf hai? ::: $\hat{B} = \hat{k} \times \hat{E} = \overset{z}{^} \times \overset{y}{^} = - \overset{x}{^}$ . $n = 1.5$ wale glass mein light kitni fast jaati hai? ::: $v = c /1.5 \approx 2.00 \times 1 0^{8} m/s$ . Displacement current hatane se wave equation ka kya hota hai? ::: Woh collapse ho jaata hai $\nabla^{2} E = 0$ mein (Laplace's equation — koi wave nahi).