Poynting vector — energy flux in EM waves: Worked examples

1.8.36 · D3 · Physics › Electromagnetism › Poynting vector — energy flux in EM waves

Scenario matrix

Is topic ke har problem ka answer inhi boxes mein se kisi ek mein aata hai. Hum har row ke liye kam se kam ek example karte hain.

Cell	Kya vary karta hai	Example
A — measured intensity → amplitude	Tumhe $I$ pata hai, $E_{0}$ , $B_{0}$ chahiye	Ex 1
B — amplitude → intensity & power	Tumhe $E_{0}$ (ya $B_{0}$ ) pata hai, $I$ , $P$ chahiye	Ex 2
C — geometry / $S$ ki direction	$d A$ aur $S$ ke beech angle zero nahi	Ex 3 (figure)
D — degenerate: $E ∥ B$	cross product $\to 0$ , koi energy flow nahi	Ex 4 (figure)
E — limiting / instantaneous vs average	$S (t)$ vs $⟨ S ⟩$ , aur $S = 0$ ke moments	Ex 5
F — real-world word problem	sun, distance, inverse-square	Ex 6
G — radiation pressure twist	absorb vs reflect, area par force	Ex 7
H — exam trap: resistor / static fields	$S$ jahan tum wave expect nahi karte	Ex 8 (figure)

Har example se pehle, Forecast: — ruko aur answer ka size ya sign guess karo. Pehle guess karna hi number ko dimaag mein chipkaata hai.

Constants jo hum baar baar use karte hain (ye teen yaad karo):

Cell A — measured intensity → field amplitude

Worked example Ex 1. Ek radio station ka field tumhare window par

Ek radio transmitter tumhare ghar par intensity $I = 0.020 W/m^{2}$ produce karta hai. Electric-field amplitude $E_{0}$ aur magnetic-field amplitude $B_{0}$ nikalo.

Forecast: $E_{0}$ volts-per-metre mein hoga — kya ye kuch, kuch dozen, ya hazaaron hoga? (Sunlight ke $\sim 1000$ V/m se chhota hoga kyunki $I$ bahut chhota hai.)

Intensity formula se shuru karo $I = \frac{1}{2} ε_{0} c E_{0}^{2}$ . Ye step kyun? Intensity, $S$ ka time average hai; parent note ne derive kiya tha ki $⟨ cos^{2} ⟩ = \frac{1}{2}$ , to ye jo cheez hum measure karte hain ( $I$ ) aur jo chahiye ( $E_{0}$ ) ke beech ka bridge hai.
$E_{0}$ ke liye invert karo: $E_{0} = \frac{2 I}{ε _{0} c}$ . Ye step kyun? $E_{0}$ ek square ke andar daba hua hai, to hum square root se undo karte hain.
Numbers plug karo: $E_{0} = \frac{2 ( 0.020 )}{( 8.85 \times 1 0 ^{- 12} ) ( 3.0 \times 1 0 ^{8} )} \approx 3.88 V/m$
Wave lock $E_{0} = c B_{0}$ se $B_{0}$ nikalo: $B_{0} = E_{0} / c = 3.88/ (3.0 \times 1 0^{8}) \approx 1.29 \times 1 0^{- 8} T$ . Ye step kyun? Ek plane wave mein $E$ aur $B$ independent nahi hote — parent note ne dikhaya tha ki har instant par $E = c B$ hota hai.

Verify: $E_{0}$ se intensity rebuild karo: $\frac{1}{2} (8.85 \times 1 0^{- 12}) (3 \times 1 0^{8}) (3.88)^{2} \approx 0.020 W/m^{2}$ . ✓ Units: $(W/m^{2}) / (F/m \cdot m/s)$ collapse hokar V/m ban jaata hai (algebra par trust karo; check confirm kar deta hai).

Cell B — amplitude → intensity aur power

Worked example Ex 2. Magnetic amplitude se beam power tak

Ek laser beam ka magnetic amplitude $B_{0} = 1.0 \times 1 0^{- 6} T$ hai aur cross-sectional area $A = 3.0 mm^{2}$ hai. Intensity $I$ aur beam power $P$ nikalo.

Forecast: $B_{0} \sim 1 0^{- 6}$ T sunlight ke $B_{0}$ jaisa hai, to $I$ $1 0^{2}$ – $1 0^{3} W/m^{2}$ order ka expect karo, aur milliwatt-scale power kyunki area bahut chhota hai.

Intensity ka $B_{0}$ wala form chuno: $I = \frac{c B _{0}^{2}}{2 μ _{0}}$ . Ye step kyun? Hume $B_{0}$ diya gaya hai, $E_{0}$ nahi. $B$ -version use karne se pehle convert nahi karna padta — galti ki chances kam hain.
Plug karo: $I = \frac{( 3.0 \times 1 0 ^{8} ) ( 1.0 \times 1 0 ^{- 6} ) ^{2}}{2 ( 4 π \times 1 0 ^{- 7} )} \approx 119 W/m^{2}$ .
Area convert karo: $A = 3.0 mm^{2} = 3.0 \times 1 0^{- 6} m^{2}$ . Ye step kyun? $I$ square metre per hai; mm² use karne se answer $1 0^{6}$ guna galat aata.
Power = intensity × area (beam $∥$ area, isliye $S \cdot d A = S d A$ ): $P = I A = 119 \times 3.0 \times 1 0^{- 6} \approx 3.6 \times 1 0^{- 4} W \approx 0.36 mW$ .

Verify: $E_{0} = c B_{0} = 300 V/m$ se cross-check karo, phir $I = \frac{1}{2} ε_{0} c E_{0}^{2} = \frac{1}{2} (8.85 \times 1 0^{- 12}) (3 \times 1 0^{8}) (300)^{2} \approx 119 W/m^{2}$ . Do tareekon se same number. ✓

Cell C — geometry: $S$ surface se tilted

Worked example Ex 3. Ek tilted panel par padti wave

Ek wave $⟨ S ⟩ = 800 W/m^{2}$ carry karti hai horizontally travel karte hue. Area $A = 2.0 m^{2}$ ka ek flat solar panel itna tilt hai ki uska normal beam se $θ = 6 0^{\circ}$ ka angle banata hai. Ye kitni power collect karta hai?

Forecast: face-on se door tilt hai, to ise face-on $800 \times 2 = 1600$ W se kam collect karna chahiye. $6 0^{\circ}$ par lagbhag aadha expect karte hain.

Flux ko dot product ki tarah likho: $P = \oint S \cdot d A = S A cos θ$ . Ye step kyun? Sirf $S$ ka woh component jo surface normal ke saath hai, surface ke through energy deliver karta hai. Dot product exactly wahi component nikalta hai — figure mein red arrow ko green normal par project hote dekho.
Cosine kyun, sine kyun nahi? Kyunki $θ$ normal se measure kiya jaata hai; face-on ( $θ = 0$ ) maximum dena chahiye, aur $cos 0 = 1$ deta hai. Agar sine use karte, to face-on zero deta — galat hota.
Plug karo: $P = 800 \times 2.0 \times cos 6 0^{\circ} = 800 \times 2.0 \times 0.5 = 800 W$ .

Verify: face-on $1600$ W ka exactly aadha, forecast se match karta hai. Limiting checks: $θ = 0^{\circ} \Rightarrow 1600$ W (full), $θ = 9 0^{\circ} \Rightarrow 0$ W (edge-on, beam side se guzar jaati hai). ✓

Cell D — degenerate: parallel fields, zero flow

Worked example Ex 4. Crossed vs parallel fields

Region 1 mein, $E$ $+ \overset{x}{^}$ point karta hai aur $B$ $+ \overset{y}{^}$ point karta hai, dono ka magnitude itna hai ki perpendicular hone par $S = 50 W/m^{2}$ milta. Region 2 mein, koi claim karta hai ki $E$ aur $B$ dono $+ \overset{x}{^}$ point karte hain (parallel). Har region mein $S$ kya hai?

Forecast: region 1 → energy $+ \overset{z}{^}$ ke along flow karti hai. Region 2 → parallel vectors ka cross product zero hoga, isliye bilkul koi energy flow nahi.

Region 1: $\overset{x}{^} \times \overset{y}{^} = \overset{z}{^}$ , to $S$ magnitude $50 W/m^{2}$ ke saath $+ \overset{z}{^}$ point karta hai. Ye step kyun? Right-hand rule: ungliyan $E$ ( $\overset{x}{^}$ ) ke along, $B$ ( $\overset{y}{^}$ ) ki taraf curl, thumb $\overset{z}{^}$ point karta hai — wahi energy direction hai.
Region 2: $E \times B$ dono $\overset{x}{^}$ ke along hain: $\overset{x}{^} \times \overset{x}{^} = 0$ . Ye step kyun? Cross-product ka magnitude $E B sin ϕ$ hai jahan $ϕ$ unke beech ka angle hai; $ϕ = 0 \Rightarrow sin 0 = 0$ . Exactly isliye parent note ne cross product par zor diya tha — parallel fields koi propagating energy carry nahi karte, aur sirf cross product ko ye pata hota hai.
General reminder: $∣ S ∣ = \frac{1}{μ _{0}} E B sin ϕ$ . Ye $ϕ = 9 0^{\circ}$ par max hota hai (ek real wave) aur $ϕ = 0^{\circ}$ ya $18 0^{\circ}$ par zero ho jaata hai.

Verify: region 2 ka magnitude $\propto sin 0 = 0$ . ✓ Dot product $E \cdot B$ yahan $E B cos 0 \neq = 0$ deta — exactly wahi galti hai jiske baare mein parent warn karta hai.

Cell E — instantaneous vs average, aur zero ke moments

Worked example Ex 5. Ek bright wave mein

$S$ kab zero hota hai? Ek wave ka $E (t) = E_{0} cos (ω t)$ hai jahan $E_{0} = 100 V/m$ . Nikalo (a) peak instantaneous $S$ , (b) time-averaged $S$ , aur (c) kis phase par $S = 0$ hoga.

Forecast: average peak ka exactly aadha hona chahiye (famous $\frac{1}{2}$ ). Aur $S$ momentarily har cycle mein do baar zero hit karta hai, jab fields zero cross karte hain.

Instantaneous: $S (t) = ε_{0} c E (t)^{2} = ε_{0} c E_{0}^{2} cos^{2} (ω t)$ . Ye step kyun? Parent ne dikhaya tha ki $S = ε_{0} c E^{2}$ har instant par hold karta hai, kisi averaging se pehle.
Peak ( $cos^{2} = 1$ par): $S_{m a x} = ε_{0} c E_{0}^{2} = (8.85 \times 1 0^{- 12}) (3 \times 1 0^{8}) (100)^{2} \approx 26.6 W/m^{2}$ .
Average: $cos^{2}$ ko uske average $\frac{1}{2}$ se replace karo: $⟨ S ⟩ = \frac{1}{2} ε_{0} c E_{0}^{2} \approx 13.3 W/m^{2}$ . Ye step kyun? Ek full cycle mein $⟨ cos^{2} ⟩ = \frac{1}{2}$ ; intensity wahi hai jo ek slow detector read karta hai, isliye ye average hai, peak nahi.
Zeros: $S = 0$ jab $cos (ω t) = 0$ , yani $ω t = \frac{π}{2}, \frac{3 π}{2}, \dots$ — har period mein do baar. Ye step kyun? $S \propto E^{2} \geq 0$ , isliye ye kabhi negative nahi hota; ye sirf zero par dip karta hai jab bhi field zero se guzarta hai, phir wapas upar aata hai.

Verify: $⟨ S ⟩ / S_{m a x} = 13.3/26.6 = 0.5$ exactly. ✓ Note karo ki $S$ , $2 ω$ par oscillate karta hai (double frequency), kyunki cosine ko square karne se uski frequency double ho jaati hai.

Cell F — real-world word problem (inverse square)

Worked example Ex 6. Sun ki power output

Earth ke orbit par sunlight ki intensity $I = 1360 W/m^{2}$ hai. Earth–Sun distance $r = 1.5 \times 1 0^{11} m$ hai. Sun ki total radiated power (luminosity) kya hai?

Forecast: Sun bahut bada hai — $1 0^{26}$ watts jaisa kuch expect karo.

Power ko poori sphere par spread karo: Sun sabhi directions mein equally radiate karta hai, isliye radius $r$ par same total power $P$ , area $4 π r^{2}$ wali sphere par spread hoti hai. Isliye $I = \frac{P}{4 π r ^{2}}$ . Ye step kyun? $S$ power per area hai; total power ke liye hum intensity ko us area se multiply karte hain jis par wo spread hai — yahan ek sphere, jo inverse-square law $I \propto 1/ r^{2}$ deta hai.
$P$ ke liye solve karo: $P = I (4 π r^{2})$ .
Plug karo: $P = 1360 \times 4 π (1.5 \times 1 0^{11})^{2} \approx 3.8 \times 1 0^{26} W$ .

Verify: accepted solar luminosity $\approx 3.85 \times 1 0^{26}$ W hai. ✓ Dimension check: $(W/m^{2}) (m^{2}) = W$ . Do guna door ( $r \to 2 r$ ) jaane par intensity quarter ho jaayegi — inverse square confirm.

Cell G — radiation-pressure twist

Worked example Ex 7. Ek sail par push: absorb vs reflect

Ek solar sail ka area $A = 100 m^{2}$ hai aur ye wahan hai jahan $I = 1360 W/m^{2}$ hai. Uska force nikalo (a) agar ye light ko perfectly absorb kare, (b) agar ye use perfectly reflect kare.

Forecast: force bahut chhota hoga (light barely push karti hai), micronewton jaisa, aur reflection absorbing case ka double dega.

Absorber par radiation pressure: $P_{rad} = I / c$ . Ye step kyun? Light momentum flux carry karti hai jo uske energy flux ka $c$ se divided hota hai; use absorb karne par wo momentum sail par transfer ho jaata hai. (Dekho Radiation pressure.)
Force = pressure × area: $F_{abs} = \frac{I}{c} A = \frac{1360}{3 \times 1 0 ^{8}} \times 100 \approx 4.5 \times 1 0^{- 4} N$ .
Reflection ise double karta hai: $P_{rad} = 2 I / c$ , kyunki light ka momentum sirf ruka nahi balki reverse bhi hua, isliye sail double kick feel karta hai. $F_{ref} = \frac{2 I}{c} A \approx 9.1 \times 1 0^{- 4} N$ . Ye step kyun? Bounce ke liye momentum change $2 p$ hota hai vs catch ke liye $p$ — same reason hai ki wall se bounce karne wali ball us ball se zyada push karti hai jo wall se chipak jaaye.

Verify: $F_{ref} / F_{abs} = 2$ exactly. ✓ Dono $\sim 1 0^{- 4}$ N hain — 100 m² par force ki ek bahut halki phoonk, forecast se match karta hai.

Cell H — exam trap: resistor mein energy

Worked example Ex 8. Resistor ki heat kahan se aati hai?

Ek cylindrical resistor jiska radius $a$ aur length $L$ hai, current $I$ carry karta hai aur resistance $R$ hai. Uski surface par fields use karke dikhao ki Poynting flux uske andar $I^{2} R$ ke barabar hai.

Forecast: ye familiar Joule heating $I^{2} R$ ke barabar hona chahiye — lekin energy flow ki direction tumhe surprise karegi.

Wire ke along electric field: voltage drop $V = I R$ length $L$ par hai, isliye $E = V / L = I R / L$ , axis ke along point karta hai (current ki direction mein). Ye step kyun? $E$ current drive karta hai; uska magnitude voltage gradient hai.
Wire ke around magnetic field: surface par Ampère's law $B = \frac{μ _{0} I}{2 π a}$ deta hai, wire ke around circle banata hua. Ye step kyun? Ek straight current apne around $B$ wrap karta hai — ye surface ke bahar ka field hai.
Poynting vector ki direction: $S = \frac{1}{μ _{0}} E \times B$ . $E$ axial aur $B$ azimuthal hone par, $E \times B$ radially inward point karta hai — energy wire mein uski curved side se andar flow karti hai (figure mein red arrows). Ye step kyun? Axial se azimuthal tak right-hand rule radial deta hai; sign inward work out hota hai. Yahi trap hai: energy sides se andar aati hai, "wire ke neeche se" nahi.
Magnitude aur integrate karo: $S = \frac{E B}{μ _{0}} = \frac{1}{μ _{0}} \frac{I R}{L} \frac{μ _{0} I}{2 π a} = \frac{I ^{2} R}{2 π a L}$ . Side area $2 π a L$ se multiply karo: $P = S (2 π a L) = \frac{I ^{2} R}{2 π a L} \times 2 π a L = I^{2} R .$

Verify: $2 π a L$ perfectly cancel ho jaata hai, $P = I^{2} R$ bachta hai — exactly Joule heating. ✓ (Poynting's theorem is bookkeeping ki guarantee karta hai.)

Recall Ye problem kis cell ka hai?

Sirf $I$ diya, $E_{0}$ maanga? ::: Cell A — $I = \frac{1}{2} ε_{0} c E_{0}^{2}$ invert karo. Tilted surface se guzarta beam? ::: Cell C — $P = S A cos θ$ use karo jahan $θ$ normal se hai. $E$ aur $B$ parallel? ::: Cell D — $S = 0$ , koi energy flow nahi. $S_{m a x}$ aur $⟨ S ⟩$ mein difference? ::: $⟨ cos^{2} ⟩$ se $\frac{1}{2}$ ka factor aata hai. Reflecting vs absorbing sail? ::: Cell G — reflection double force deta hai ( $2 I / c$ vs $I / c$ ).

Ye bhi dekho: Intensity and amplitude of waves, Energy density of electric and magnetic fields, EM wave equation, Vector calculus identities, Maxwell's equations.