Exercises — Poynting vector — energy flux in EM waves

Galat. $\vec{S}$ ek energy flux hai: energy per area per time.

• ${\text{J/m}}^3$ energy density $u$ ki unit hai — energy jo ek volume mein baithti hai.
• $\vec{S}$ energy ko ek surface ke paar le jaata hai, isliye yeh $\frac{\text{energy}}{\text{area}\cdot \text{time}}=\frac{\text{J}}{{\text{m}}^2\cdot \text{s}}=\frac{\text{W}}{{\text{m}}^2}$ hai.

Answer: ${\text{W/m}}^2$ (watts per square metre).

$\vec{S}=\frac{1}{{\mu }_0}\vec{E}\times \vec{B}$ ko propagation direction $+x$ ki taraf point karna chahiye. Ungliyaan $\vec{E}$ ($+y$) ke along point karo, $\vec{B}$ ki taraf curl karo, thumb $+x$ pe land karna chahiye. Humein $\hat{y}\times \hat{B}=+\hat{x}$ chahiye. Kyunki $\hat{y}\times \hat{z}=\hat{x}$, isliye $\vec{B}$ $==+z==$ ke along hona chahiye.

Check: $\hat{y}\times \hat{z}=\hat{x}$ ✓ — to $\vec{E}\times \vec{B}$ $+x$ direction mein point karta hai, yaani jis direction mein wave jaati hai.

Neeche figure (s01): teen arrows ek mutually perpendicular triad banate hain — kala $\vec{E}$ upar point karta hai ($+y$), kala $\vec{B}$ "page ke andar/upar-daayein" ($+z$), aur laal $\vec{S}$ daayein ($+x$). Apni right hand ki ungliyaan $\vec{E}$ se $\vec{B}$ ki taraf curl karo; thumb laal arrow pe land karega. Woh laal arrow hi akela cheez hai jis par energy transport dhyan deta hai.

$I=\frac{1}{2}{\epsilon }_{0}c{E}_0^2$ Instantaneous flux hai $S={\epsilon }_{0}c{E}^2={\epsilon }_{0}c{E}_0^2{\cos }^2(kx-\omega t)$. Time pe average karne par, $⟨{\cos }^2⟩=\frac{1}{2}$, jo ==$\frac{1}{2}$== le aata hai. Intensity woh hai jo ek detector read karta hai, aur detectors average report karte hain.

(a) $I=\frac{1}{2}{\epsilon }_{0}c{E}_0^2$ ko invert karo (hum $I$ measure karte hain, $E_0$ chahiye): $E_0=\sqrt{\frac{2I}{{\epsilon }_{0}c}}=\sqrt{\frac{2(1360)}{(8.85\times 10^{-12})(3\times 10^8)}}\approx 1.01\times 10^3\text{V/m}.$ (b) Ek wave mein $E=cB$, isliye $B_0=\frac{E_0}{c}=\frac{1.01\times 10^3}{3\times 10^8}\approx 3.4\times 10^{-6}\text{T}.$ Bahut chhota — isliye light ka magnetic push normally negligible hota hai.

$\vec{S}$ power per area hai; beam seedha aperture se guzarti hai, isliye $\vec{S}\parallel d\vec{A}$ aur $P=⟨S⟩A$. Area convert karo: $A=2.0{\text{mm}}^2=2.0\times 10^{-6}{\text{m}}^2$. $P=500\times 2.0\times 10^{-6}=1.0\times 10^{-3}\text{W}=1.0\text{mW}.$ (Yahan $P$ power hai, conventions box ke hisaab se.)

Kyun ek wave mein $u_E=u_B$ hota hai. Dono energy densities hain $u_E=\frac{1}{2}{\epsilon }_0{E}^2$ aur $u_B=\frac{1}{2{\mu }_0}{B}^2$. Ek plane wave mein $E=cB$, isliye $B=E/c$ aur $c^2=1/({\mu }_0{\epsilon }_0)$ substitute karo: $u_B=\frac{1}{2{\mu }_0}{\left(\frac{E}{c}\right)}^2=\frac{E^2}{2{\mu }_0{c}^2}=\frac{E^2}{2{\mu }_0}({\mu }_0{\epsilon }_0)=\frac{1}{2}{\epsilon }_0{E}^2=u_E.$ To magnetic aur electric "halves" har instant pe exactly equal hain, isliye hum likh sakte hain $u=u_E+u_B=2\cdot \frac{1}{2}{\epsilon }_0{E}^2={\epsilon }_0{E}^2.$ Number: $u=(8.85\times 10^{-12})(300{)}^2=7.97\times 10^{-7}{\text{J/m}}^3.$ Flux: $S=uc=7.97\times 10^{-7}\times 3\times 10^8=239{\text{W/m}}^2$. Cross-check: ${\epsilon }_{0}c{E}^2=(8.85\times 10^{-12})(3\times 10^8)(300{)}^2=239{\text{W/m}}^2$ ✓.

Reflection ke liye light momentum reverse karti hai, isliye pressure absorption ke comparison mein double ho jaata hai: $P_{\text{pres}}=\frac{2I}{c}=\frac{2(1360)}{3\times 10^8}=9.07\times 10^{-6}\text{Pa}.$ Force $=P_{\text{pres}}\cdot A=9.07\times 10^{-6}\times 100=9.07\times 10^{-4}\text{N}$ ≈ $0.91\text{mN}$. (Note $P_{\text{pres}}$ = pressure hai, power $P$ se alag; conventions box dekho.) Radiation pressure dekho — momentum flux $S/c$ hai, aur reflection use wapas karta hai, jisse factor 2 milta hai.

Wave ke perpendicular ek face ka area $A$ consider karo. Time $dt$ mein, face ke peeche thickness $cdt$ ki slab ke andar saari energy isse cross karti hai (wave $c$ pe move karti hai).

• Slab mein energy $=u\times (\text{volume})=u(Acdt)$.
• Flux $S=\frac{\text{energy crossed}}{Adt}=\frac{uAcdt}{Adt}=uc.$ ✓

Neeche figure (s02): laal slab woh chunk hai jo face ke theek peeche baitha hai. Iska thickness $cdt$ hai (woh distance jo light time $dt$ mein travel karti hai), isliye us time mein poori laal slab $A$ area ke kale face ke paar sweep ho jaati hai. Iski energy $u(Acdt)$ ko $Adt$ se divide karne par $uc$ milta hai. Yeh parent note se "river current = water density × flow speed" wali picture hai.

(a) Peak flux tab hota hai jab ${\cos }^2=1$: $S_{\max }={\epsilon }_{0}c{E}_0^2=(8.85\times 10^{-12})(3\times 10^8)(200{)}^2=106.2{\text{W/m}}^2.$ (b) Intensity average hai: $I=\frac{1}{2}{\epsilon }_{0}c{E}_0^2=53.1{\text{W/m}}^2.$ Ratio $I/S_{\max }=\frac{1}{2}$, jaisa $⟨{\cos }^2⟩=\frac{1}{2}$ se expect tha.

(a) Wire ke across voltage $V=IR$ hai, length $L$ pe drop hota hai, isliye axial field hai $E=\frac{V}{L}=\frac{IR}{L}(\text{wire ke along}).$ Radius $a$ ki loop pe Ampère's law: $B(2\pi a)={\mu }_{0}I\Rightarrow B=\frac{{\mu }_{0}I}{2\pi a}$ (wire ke around ghoomta hai).

(b) $\vec{E}$ (axial) $\times \vec{B}$ (azimuthal) radially inward point karta hai — energy surrounding field se wire ke andar flow karti hai. Magnitude: $S=\frac{EB}{{\mu }_0}=\frac{1}{{\mu }_0}\cdot \frac{IR}{L}\cdot \frac{{\mu }_{0}I}{2\pi a}=\frac{I^{2}R}{2\pi aL}.$

(c) Side pe outward area element $d\vec{A}$ radially bahar point karta hai, jabki $\vec{S}$ radially andar point karta hai, isliye $\vec{S}\cdot d\vec{A}=-SdA$ — ek negative flux, matlab energy andar aa rahi hai (sign convention box ke consistent). Side area $A_{side}=2\pi aL$ pe magnitudes leke: $|\oint \vec{S}\cdot d\vec{A}|=S\cdot A_{side}=\frac{I^{2}R}{2\pi aL}\cdot 2\pi aL=I^{2}R.$ Dissipated heat sides se andar aati hai, fields ke through — Poynting's theorem exactly Joule heating balance karta hai.

Neeche figure (s03): grey bar resistor hai; kala axial arrow current $I$ ke along $\vec{E}$ hai; laal arrows $\vec{S}$ hain jo wire ke andar seedha point karte hain chaaron taraf se. Woh energy jo resistor ko heat karti hai, woh sideways aati hai field se, "copper ke andar se" nahi.

(a) $P=IA=1360\times 2=2720\text{W}$ (power). (b) Kyun momentum flux $I/c$ hai. Ek EM wave momentum carry karti hai: field energy $U$ ka ek packet momentum $p=U/c$ carry karta hai (electromagnetism ka ek standard result, aur massless energy ke liye relativity se bhi yahi milta hai). Ab dekho ek second mein ek square metre pe kya land karta hai: aanewali energy $I$ hai (yahi intensity ka matlab hai), isliye aanewala momentum $I/c$ hai per area per second. Full absorption ke liye woh momentum ek baar deliver hota hai, pressure $I/c$ deta hai aur force $F=\frac{I}{c}A=\frac{1360}{3\times 10^8}\times 2=9.07\times 10^{-6}\text{N}.$ (c) Yeh bhi $F=\frac{P}{c}=\frac{2720}{3\times 10^8}=9.07\times 10^{-6}\text{N}$ ✓. Kyunki $p=U/c$, time mein differentiate karne par $\frac{dp}{dt}=\frac{1}{c}\frac{dU}{dt}=\frac{P}{c}$ milta hai, yaani force $=$ power$/c$.

Start: $I=\frac{1}{2}{\epsilon }_{0}c{E}_0^2$. Form 2: ek factor $E_0=c{B}_0$ substitute karo: $I=\frac{1}{2}{\epsilon }_{0}c{E}_0(c{B}_0)=\frac{1}{2}{\epsilon }_0{c}^2{E}_0{B}_0.$ $c^2=\frac{1}{{\mu }_0{\epsilon }_0}\Rightarrow {\epsilon }_0{c}^2=\frac{1}{{\mu }_0}$ use karo: $I=\frac{E_0{B}_0}{2{\mu }_0}.✓$ Form 3: ab Form 2 mein $E_0=c{B}_0$ replace karo: $I=\frac{(c{B}_0)B_0}{2{\mu }_0}=\frac{c{B}_0^2}{2{\mu }_0}.✓$ Agar sirf $B_0$ pata hai, seedha $I=\frac{c{B}_0^2}{2{\mu }_0}$ use karo — pehle $E_0$ compute karne ki zaroorat nahi.

Kyun $E$ aur $B$ exactly in phase hain. Is wave ke liye Faraday's law, $\frac{\partial E}{\partial x}=-\frac{\partial B}{\partial t}$, dono fields ko link karta hai. $E=E_0\cos (kx-\omega t)$ ke saath, left side $-k{E}_0\sin (kx-\omega t)$ hai, isliye $\frac{\partial B}{\partial t}=k{E}_0\sin (kx-\omega t)\Rightarrow B=\frac{k{E}_0}{\omega }\cos (kx-\omega t)=\frac{E_0}{c}\cos (kx-\omega t),$ $\omega /k=c$ use karke. To $B$ ka $E$ jaisa hi same $\cos (kx-\omega t)$ hai: dono saath upar-neeche hote hain (in phase), aur unke amplitudes $B_0=E_0/c$ obey karte hain. Isliye $S={\epsilon }_{0}cEB={\epsilon }_{0}c{E}_0^2{\cos }^2(kx-\omega t).$

• Peaks (${\cos }^2=1$): $S={\epsilon }_{0}c{E}_0^2$ — max forward flux, jab field sabse bada hota hai (chahe $+E_0$ ho ya $-E_0$).
• Zeros ($\cos =0$): $S=0$ — woh instant jab field zero se guzarta hai; tab koi energy cross nahi ho rahi.
• Kabhi negative nahi: ek squared quantity negative nahi ho sakti, isliye wave hamesha $+x$ mein energy carry karti hai; yeh kabhi backward nahi bahti. Isliye ek single travelling wave ka flux strictly non-negative hota hai.
• Average: $⟨S⟩=\frac{1}{2}{\epsilon }_{0}c{E}_0^2=I$.

(Contrast: ek standing wave mein $E$ aur $B$ 90° out of phase hain, isliye $⟨S⟩=0$ — energy slosh karti hai par propagate nahi karti.)

Incoming momentum flux (per area) $=I/c$. Absorbed fraction $(1-r)$ apna momentum ek baar deliver karta hai; reflected fraction $r$ double deliver karta hai (aata hai aur ulta jaata hai): $P_{\text{pres}}=\frac{I}{c}[(1-r)+2r]=\frac{I}{c}(1+r).$

• $r=0$ (black): $P_{\text{pres}}=I/c$ ✓.
• $r=1$ (mirror): $P_{\text{pres}}=2I/c$ ✓. $I=1360{\text{W/m}}^2$ ke liye, ek $r=0.5$ grey surface deta hai $P_{\text{pres}}=\frac{1360}{3\times 10^8}(1.5)=6.80\times 10^{-6}\text{Pa}$.

Ise quick quiz ki tarah use karo: question padho, awaaz mein jawab do, phir check karo.

• $\vec{S}$ ki units? — ${\text{W/m}}^2$.
• $E_0$ se intensity? — $I=\frac{1}{2}{\epsilon }_{0}c{E}_0^2$.
• Radiation pressure, reflecting? — $2I/c$; partial: $\frac{I}{c}(1+r)$.
• Kyun ek single travelling wave ka flux kabhi negative nahi hota? — $S\propto {\cos }^2\ge 0$ kyunki $E$ aur $B$ in phase hain.
• Resistor ki heat kahan se enter karti hai? — Sides se radially, $\vec{S}$ ke through; surface integral $I^{2}R$ deta hai.