Exercises — Electromagnetic waves — derivation from Maxwell's equations

HUM KYA DHUNDH RAHE HAIN: woh term jo tab bhi bachti hai jab koi charges aur currents nahi hote. Chaar vacuum equations hain: $\nabla \cdot \vec{E}=0$, $\nabla \cdot \vec{B}=0$, $\nabla \times \vec{E}=-{\partial }_t\vec{B}$, aur $\nabla \times \vec{B}={\mu }_0{\epsilon }_0{\partial }_t\vec{E}$. $\vec{J}=0$ set karo: ordinary Ampère term ${\mu }_0\vec{J}$ gayab ho jaata hai. Chauthe equation ki right side par sirf ${\mu }_0{\epsilon }_0{\partial }_t\vec{E}$ bachta hai — yahi displacement current term hai. Jawab: Ampère–Maxwell law, apne displacement-current term ${\mu }_0{\epsilon }_0{\partial }_t\vec{E}$ ke zariye. Iske bina, vacuum mein $\nabla \times \vec{B}=0$ ho jaata hai aur koi wave equation nahi banta.

HUM KYA KARTE HAIN: dono equations mein ${\partial }_t^2$ ke coefficient ko compare karo. Hamari equation mein coefficient ${\mu }_0{\epsilon }_0$ hai; template mein $1/v^2$ hai. YEH KYUN EQUAL HONE CHAHIYE: ek hi field ko describe karne wali identical form ki do equations ke coefficients identical hone chahiye, isliye $\frac{1}{v^2}={\mu }_0{\epsilon }_0⟹v=\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}=c.$ Jawab: $v=\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}$.

HUM KYA JAANTE HAIN: triad $(\hat{E},\hat{B},\hat{k})$ right-handed hai, matlab $\hat{E}\times \hat{B}=\hat{k}$. Yahan $\hat{E}=\hat{y}$ aur $\hat{k}=\hat{x}$ hai. Humein $\hat{B}$ chahiye jiske saath $\hat{y}\times \hat{B}=\hat{x}$ ho. $\hat{B}=\hat{z}$ test karo: $\hat{y}\times \hat{z}=\hat{x}$. ✓ Jawab: $\vec{B}$ $+\hat{z}$ direction mein point karta hai. Neeche diye figure mein dekho — $\vec{E}$ upar, $\vec{B}$ page-plane se bahar, wave daayein chalti hui.

$c=f\lambda$ KYUN: $v=\omega /k$ se, jahan $\omega =2\pi f$ aur $k=2\pi /\lambda$, $2\pi$ cancel ho jaata hai aur milta hai $c=f\lambda$. $\lambda =\frac{c}{f}=\frac{3.00\times 10^8}{2.45\times 10^9}=0.1224\text{m}\approx 12.2\text{cm}.$ Jawab: $\lambda \approx 0.122\text{m}$ (lagbhag 12 cm — isliye microwave cavities haath ke size ki hoti hain).

KYUN: ek plane wave ke liye Faraday's law deta hai $k{E}_0=\omega B_0$, yaani $E_0/B_0=\omega /k=c$. $B_0=\frac{E_0}{c}=\frac{90}{3.00\times 10^8}=3.0\times 10^{-7}\text{T}.$ Jawab: $B_0=3.0\times 10^{-7}\text{T}$. Dhyan do ki $B_0$ numerically bahut chhota hai — yeh units ka effect hai, koi "weak" field nahi.

HUM KYA KARTE HAIN: multiply karo, reciprocal square root lo. ${\mu }_0{\epsilon }_0=(1.2566\times 10^{-6})(8.854\times 10^{-12})=1.1127\times 10^{-17}{\text{s}}^2/{\text{m}}^2.$ $\frac{1}{\sqrt{1.1127\times 10^{-17}}}=2.998\times 10^8\text{m/s}.$ Jawab: $c\approx 3.00\times 10^8\text{m/s}$. Do lab constants, dono ka light se koi seedha rishta nahi, phir bhi speed of light de dete hain.

$f=\frac{c}{\lambda }=\frac{3.00\times 10^8}{633\times 10^{-9}}=4.74\times 10^{14}\text{Hz}.$ $\omega =2\pi f=2\pi (4.74\times 10^{14})=2.98\times 10^{15}\text{rad/s}.$ Jawab: $f\approx 4.74\times 10^{14}\text{Hz}$, $\omega \approx 2.98\times 10^{15}\text{rad/s}$.

IDENTITY KYA DETA HAI: $\nabla \times (\nabla \times \vec{E})=\nabla (\nabla \cdot \vec{E})-{\nabla }^2\vec{E}$. Assumption yeh hai: Maxwell ki pehli vacuum equation hai $\nabla \cdot \vec{E}=0$ (Gauss's law $\rho =0$ ke saath). Yeh $\nabla (\nabla \cdot \vec{E})=\nabla (0)=0$ bana deta hai. Toh left side sirf $-{\nabla }^2\vec{E}$ reh jaata hai, aur clean wave equation milti hai. Agar $\rho \ne 0$ ho: Gauss's law ban jaata hai $\nabla \cdot \vec{E}=\rho /{\epsilon }_0$, toh surviving term hai $\nabla (\rho /{\epsilon }_0)$. Equation ban jaata hai ${\nabla }^2\vec{E}-{\mu }_0{\epsilon }_0{\partial }_t^2\vec{E}=\nabla \left(\frac{\rho }{{\epsilon }_0}\right),$ ek driven (inhomogeneous) wave equation. Charges source term ki tarah kaam karte hain aur pristine free wave disturb ho jaati hai.

HUM KYA COMPUTE KARTE HAIN: $\nabla \cdot \vec{E}={\partial }_x{E}_x+{\partial }_y{E}_y+{\partial }_z{E}_z$. $e^{i(kx-\omega t)}$ form ki wave ke liye, sirf $x$ par spatial dependence hai, aur ${\partial }_x$ ek factor $ik$ neeche laata hai. Toh $\nabla \cdot \vec{E}=ik{E}_x$ (sirf $x$-component ka apna $x$-derivative nonzero ho sakta hai yahan). Zero set karo: $ik{E}_{0x}=0$. Kyunki $k\ne 0$, humein chahiye $E_{0x}=0$. Matlab: $\vec{E}$ ka woh component jo travel direction $\hat{x}$ ke saath hai, woh zero hona chahiye. Field sirf $y$–$z$ plane mein rehti hai, $\vec{k}$ ke perpendicular — wave transverse hai. $\vec{B}$ par bhi yahi argument ($\nabla \cdot \vec{B}=0$ use karke) $\vec{B}$ ko bhi transverse bana deta hai.

Space mein do baar differentiate karo: ${\partial }_{x}E=E_{0}k\cos (kx-\omega t)$, ${\partial }_x^{2}E=-E_0{k}^2\sin (kx-\omega t)=-k^{2}E$. Time mein do baar differentiate karo: ${\partial }_{t}E=E_0\omega \cos (kx-\omega t)\cdot (-1)\cdot (-1)$... dhyan se: ${\partial }_t(kx-\omega t)=-\omega$, toh ${\partial }_{t}E=-E_0\omega \cos (kx-\omega t)$, aur ${\partial }_t^{2}E=-E_0{\omega }^2\sin (kx-\omega t)=-{\omega }^{2}E$. Substitute karo: $-k^{2}E={\mu }_0{\epsilon }_0(-{\omega }^{2}E)$. $-E$ cancel karo: $k^2={\mu }_0{\epsilon }_0{\omega }^2⟹\frac{\omega }{k}=\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}=c.$ Jawab: yeh solution hai iff $\omega =ck$. Yeh dispersion relation hai — vacuum mein yeh ek straight line hai ($\omega$, $k$ ke proportional), matlab har frequency ek hi speed $c$ par travel karti hai (koi spreading nahi).

(a) $B_0=E_0/c=200/(3.00\times 10^8)=6.667\times 10^{-7}\text{T}$. (b) $u_{\max }=\frac{1}{2}{\epsilon }_0{E}_0^2=\frac{1}{2}(8.854\times 10^{-12})(200{)}^2=1.771\times 10^{-7}{\text{J/m}}^3$. (c) $\frac{1}{2{\mu }_0}{B}_0^2=\frac{(6.667\times 10^{-7}{)}^2}{2(1.2566\times 10^{-6})}=1.768\times 10^{-7}{\text{J/m}}^3$. Jawab: dono energy densities agree karti hain (rounding ke saath, $\approx 1.77\times 10^{-7}{\text{J/m}}^3$), confirm karta hai ki electric aur magnetic parts equal energy store karte hain — L2-trap numbers se resolve hua. Dekho Poynting Vector and EM Energy.

$E_0$ solve karo: $E_0=\sqrt{\frac{2I}{c{\epsilon }_0}}$. $E_0=\sqrt{\frac{2(1000)}{(3.00\times 10^8)(8.854\times 10^{-12})}}=\sqrt{7.53\times 10^5}=868\text{V/m}.$ Phir $B_0$: $B_0=E_0/c=868/(3.00\times 10^8)=2.9\times 10^{-6}\text{T}$. Jawab: $E_0\approx 8.7\times 10^2\text{V/m}$, $B_0\approx 2.9\times 10^{-6}\text{T}$. Sunlight ka magnetic field Earth ke apne field ($\sim 5\times 10^{-5}$ T) se comparable hai lekin $\sim 10^{14}$ Hz par oscillate karta hai.

Key fact: vacuum mein dono $c$ par travel karte hain, toh $f=c/\lambda$ aur ratio sirf wavelengths par depend karta hai. $\frac{f_2}{f_1}=\frac{c/{\lambda }_2}{c/{\lambda }_1}=\frac{{\lambda }_1}{{\lambda }_2}=\frac{700}{400}=\mathrm{1.75.}$ Jawab: violet ki frequency red ki $1.75\times$ hai. Yeh kahan baithte hain uske liye dekho Electromagnetic Spectrum.

Step 1 — Ampère–Maxwell law ka curl lo. $\nabla \times \vec{B}={\mu }_0{\epsilon }_0{\partial }_t\vec{E}$ se shuru karo aur dono sides ka curl lo: $\nabla \times (\nabla \times \vec{B})={\mu }_0{\epsilon }_0\nabla \times ({\partial }_t\vec{E}).$ Kyun: curl lene se right side par $\nabla \times \vec{E}$ aa jaata hai, aur Faraday bilkul usi ke liye formula deta hai. Step 2 — identity se left side. $\nabla \times (\nabla \times \vec{B})=\nabla (\nabla \cdot \vec{B})-{\nabla }^2\vec{B}$. Kyunki $\nabla \cdot \vec{B}=0$, pehla term khatam: $\nabla \times (\nabla \times \vec{B})=-{\nabla }^2\vec{B}.$ Step 3 — right side, order swap karo aur Faraday daalo. Smooth fields ke liye space-curl aur time-derivative commute karte hain: ${\mu }_0{\epsilon }_0\nabla \times ({\partial }_t\vec{E})={\mu }_0{\epsilon }_0{\partial }_t(\nabla \times \vec{E})={\mu }_0{\epsilon }_0{\partial }_t\left(-{\partial }_t\vec{B}\right)=-{\mu }_0{\epsilon }_0{\partial }_t^2\vec{B}.$ Step 4 — equate karo. $-{\nabla }^2\vec{B}=-{\mu }_0{\epsilon }_0{\partial }_t^2\vec{B}$; minus signs cancel: $\boxed{{\nabla }^2\vec{B}={\mu }_0{\epsilon }_0{\partial }_t^2\vec{B}}$ Same form, same speed $c=1/\sqrt{{\mu }_0{\epsilon }_0}$. $\vec{E}$ aur $\vec{B}$ identical wave equations follow karte hain — unhe karna hi chahiye, kyunki woh ek hi wave ke do chehere hain.

(a) $c^{\prime }=1/\sqrt{2{\mu }_0{\epsilon }_0}=\frac{1}{\sqrt{2}}\cdot \frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}=c/\sqrt{2}$. Light $1/\sqrt{2}\approx 0.707$ factor se slow hogi, yaani $c^{\prime }\approx 2.12\times 10^8\text{m/s}$. (b) Fixed $f$ par, $\lambda =c/f$, toh ${\lambda }^{\prime }=c^{\prime }/f=(c/\sqrt{2})/f=\lambda /\sqrt{2}$. Wavelength usi factor $\approx 0.707$ se shrink ho jaata hai. Jawab: speed aur wavelength dono $1/\sqrt{2}$ se girte hain; frequency (source se set hoti hai) unchanged rehti hai. Yeh bilkul waise hi hai jaise ek dielectric medium behave karta hai — zyada effective $\epsilon$ light slow karta hai aur $\lambda$ chhota karta hai.

${\partial }_t\vec{E}=0$ set karo: right side zero ho jaata hai, bacha ${\nabla }^2\vec{E}=0$ — Laplace's equation. Interpretation: yeh static fields ki equation hai (electrostatics), waves ki nahi. Koi propagation nahi kyunki kuch bhi time mein change nahi ho raha — seesaw hil nahi raha, toh koi push transmit nahi hoti. Seekh: ek wave ke liye time variation zaroori hai. Displacement-current coupling tabhi act karta hai jab ${\partial }_t\vec{E}\ne 0$ ho. Zero time-derivative woh degenerate limit hai jahan wave equation gracefully static case mein reduce ho jaata hai — koi contradiction nahi, bas koi wave nahi.

(a) $\vec{B}$ ki direction: chahiye $\hat{E}\times \hat{B}=\hat{k}$, yaani $\hat{y}\times \hat{B}=\hat{x}$, deta hai $\hat{B}=\hat{z}$ (kyunki $\hat{y}\times \hat{z}=\hat{x}$). Amplitude $B_0=E_0/c$. $\vec{E}$ ke saath phase mein: $\vec{B}=\frac{E_0}{c}\hat{z}\cos (kx-\omega t).$ (b) Poynting magnitude maano $E_0=100\text{V/m}$ ke saath: $B_0=100/(3.00\times 10^8)=3.333\times 10^{-7}\text{T}$. $S=\frac{E_0{B}_0}{{\mu }_0}=\frac{(100)(3.333\times 10^{-7})}{1.2566\times 10^{-6}}=26.5{\text{W/m}}^2.$ Direction: $\vec{S}\propto \vec{E}\times \vec{B}=\hat{y}\times \hat{z}=\hat{x}$ — energy propagation direction mein flow karti hai, jaise hona chahiye. Dekho Poynting Vector and EM Energy aur Polarization. Jawab: $\vec{B}=\frac{E_0}{c}\hat{z}\cos (kx-\omega t)$; peak $S\approx 26.5{\text{W/m}}^2$ $+\hat{x}$ ke saath.

Kya tumhe mila: 2.1 ≈ 0.122 m ::: haan / 2.2 = $3.0\times 10^{-7}$ T ::: haan / 2.4 $f\approx 4.74\times 10^{14}$ Hz ::: haan / 4.2 $E_0\approx 868$ V/m ::: haan / 5.2 factor $1/\sqrt{2}$ ::: haan / 5.4 $S\approx 26.5$ W/m² ::: haan