Poynting vector — energy flux in EM waves: Question bank

1.8.36 · D5 · Physics › Electromagnetism › Poynting vector — energy flux in EM waves

Quick vocabulary refresher, taaki neeche koi symbol surprise na kare:

$S$ — Poynting vector, energy flux: watts per square metre, ek vector jo batata hai energy kahan flow kar rahi hai.
$u$ — energy density: joules per cubic metre, ek point par fields mein stored energy.
$I = ⟨ ∣ S ∣ ⟩$ — intensity: $S$ ke magnitude ka time-average. Plane wave mein $S$ kabhi direction nahi palatata, isliye yeh $∣ ⟨ S ⟩ ∣$ ke barabar hota hai; hum $I$ ko poore time ek scalar (power per area) maante hain.
$E, B$ — electric aur magnetic field vectors; plane wave mein yeh perpendicular hote hain aur $E = c B$ se locked hain.
$J$ — current density: unit area se per second cross karne wala charge (A/m²), moving charges ka flow. Yeh Poynting's theorem mein $- J \cdot E$ term ke through aata hai, jo woh rate hai jis par fields un charges par kaam karti hain.

Geometry hamesha samne rakhne ke liye, teen pictures is page ko anchor karti hain: $E \times B$ ke liye right-hand rule, standing-wave nodes jahan $S = 0$ hai, aur ek resistor mein radial energy inflow ki surprising picture.

Figure — Poynting vector — energy flux in EM waves

Figure dekho: ungliyan blue $E$ ke along point karti hain, yellow $B$ ki taraf curl karti hain, aur green thumb $S$ dono ke perpendicular bahar nikalti hai — isliye $S$ ek cross product hai aur isliye yeh propagation direction ke along land karta hai.

True or false — justify

True or false:

S

us direction mein point karta hai jis mein wave travel karti hai.

True, vacuum mein plane wave ke liye —

E \times B

propagation direction ke along land karta hai. Lekin

S

generally wahan point karta hai jahan energy flow karti hai, jo wave direction nahi bhi ho sakti (jaise resistor ke paas yeh radially inward point karta hai).

True or false: Poynting vector non-zero ho sakta hai jab koi wave propagate nahi kar rahi.

True. Static crossed fields (ek charged capacitor jo ek magnet ke andar rakha ho)

E \times B \neq = 0

dete hain, isliye

S \neq = 0

— energy bina kisi wave ke steady fields mein circulate kar sakti hai.

True or false: Agar

E

aur

B

parallel hain, toh

S = 0

True. Parallel vectors ka cross product zero hota hai, isliye

∣ S ∣ = \frac{1}{μ _{0}} E B sin 0 = 0

— us direction mein koi net energy transport nahi.

True or false:

u = S c

, yaani energy density, flux times speed ke barabar hai.

False — baat ulti hai:

S = u c

. Flux zyada hota hai

c

ke factor se, kyunki time

d t

mein

c d t

length ke slab mein stored energy face cross karti hai, isliye flux

= u \times c

True or false: Plane wave mein

u

mein electric aur magnetic contributions equal hoti hain.

True.

B = E / c

aur

c^{2} = 1/ (μ_{0} ε_{0})

use karke,

\frac{B ^{2}}{2 μ _{0}} = \frac{E ^{2} / c ^{2}}{2 μ _{0}} = \frac{1}{2} ε_{0} E^{2}

—

u

ke dono halve har instant par exactly match karte hain.

True or false: Wave ki intensity

ε_{0} c E_{0}^{2}

hoti hai.

False — yeh do guna zyada hai. Missing

\frac{1}{2}

time-averaging

cos^{2}

se aata hai:

I = \frac{1}{2} ε_{0} c E_{0}^{2}

. Dekho Intensity and amplitude of waves.

True or false: Amplitude

E_{0}

double karne se intensity double ho jaati hai.

False. Intensity

\propto E_{0}^{2}

hai, isliye

E_{0}

double karne se

I

chaar guna ho jaata hai.

True or false: Poynting's theorem energy conservation ka ek statement hai.

True.

\frac{\partial u}{\partial t} + \nabla \cdot S = - J \cdot E

kehta hai: fields se jo energy nikli woh ya toh bahar flow hoti hai (

\nabla \cdot S

) ya moving charges par kaam karne mein kharach hoti hai (

J \cdot E

, current density dotted with field). Kuch bhi gayab nahi hota. Yeh seedha Maxwell's equations se aata hai.

Spot the error

Spot the error: "

S = \frac{1}{μ _{0}} E \cdot B

Dot product ek scalar deta hai jis mein koi direction nahi hoti — lekin energy kahi na kahi flow karti hai, isliye

S

ek vector hona chahiye. Sahi operation cross product

E \times B

hai.

Spot the error: "Wave mein

B_{0} = E_{0}

hota hai, woh equal partners hain."

Woh in-phase oscillate karte hain, lekin amplitudes

E_{0} = c B_{0}

follow karte hain, isliye

B_{0}

c \approx 3 \times 1 0^{8}

se chhota hota hai. "Equal partners" barabar energy density ke baare mein hai, barabar numerical value ke baare mein nahi.

Spot the error: "

S

ki units J/m³ hain, kyunki yeh field energy measure karta hai."

Yeh energy density

u

ki unit hai.

S

energy per area per time hai, W/m² = J/(m²·s) — ek flux hai, density nahi.

Spot the error: "Perfect absorber par radiation pressure ke liye,

P_{r a d} = 2 I / c

Absorption se

P_{r a d} = I / c

milta hai; factor

2

ek perfect reflector par apply hota hai, jahan momentum change double ho jaata hai. Dekho Radiation pressure.

Spot the error: "Magnitude

∣ E \times B ∣ = E_{x} B_{x} + E_{y} B_{y} + E_{z} B_{z}

Yeh expression dot product hai. Cross-product magnitude

E B sin θ

hoti hai; perpendicular wave fields ke liye yeh simply

E B

hoti hai.

Spot the error: "Kyunki energy wire ke neeche flow karti hai,

S

resistor ke andar wire ke along point karta hai."

Resistor ke andar/aas-paas

E

wire ke along hota hai aur

B

uske around ghoomta hai, isliye

E \times B

radially inward point karta hai — energy surrounding fields se sides ke through andar aati hai, exactly

I^{2} R

deliver karke. Neeche figure dekho.

Spot the error: "Humne

S

ka form assume karke Maxwell's equations mein plug kiya."

Nahi —

S

derivation se nikla hai. Maxwell ke curl laws ko ek vector identity ke saath combine karne par flux term

\frac{1}{μ _{0}} E \times B

honi hi padti hai; humne ise kabhi posit nahi kiya.

Why questions

S

kyun

E

aur

B

dono ke perpendicular hona chahiye?

Kyunki yeh ek cross product ki tarah define kiya gaya hai, aur koi bhi cross product apne dono inputs ke perpendicular hota hai — jo is physical fact se match karta hai ki plane wave energy us ek direction mein carry karta hai jo dono fields ke perpendicular hai.

Intensity mein

\frac{1}{2}

kyun aata hai lekin instantaneous

S

mein nahi?

Instantaneously

S = ε_{0} c E^{2}

hota hai jahan

E = E_{0} cos (k x - ω t)

. Intensity magnitude ka time average hai, aur

⟨ cos^{2} ⟩ = \frac{1}{2}

hai, isliye average peak ka aadha hota hai.

Hum instantaneous

S

quote karne ki jagah time-average kyun lete hain?

Optical frequencies (

\sim 1 0^{15}

Hz) kisi bhi detector se kahin zyada fast hain; jo hum actually measure karte hain woh hai kai cycles mein deliver ki gayi energy, jo average hoti hai.

Vector identity

\nabla \cdot (E \times B) = B \cdot (\nabla \times E) - E \cdot (\nabla \times B)

key step kyun hai?

Yeh do alag curl terms (do Maxwell curl laws se) ko ek single vector ke divergence mein repackage karta hai — aur woh single vector exactly

S

hai. Parent derivation iska use

\frac{1}{μ _{0}} [E \cdot (\nabla \times B) - B \cdot (\nabla \times E)]

- \nabla \cdot S

mein fold karne ke liye karta hai; general form ke liye dekho Vector calculus identities.

Light ka magnetic field aam taur par ignore kyun kiya jaata hai jab ki woh "equal partner" hai?

Equal energy mein hai, equal force-per-charge mein nahi: magnetic force

q v B

, electric force

q E

se roughly

v / c

factor se chhoti hoti hai, aur

B_{0} = E_{0} / c

use ek tiny bana deta hai (jaise sunlight ke liye

\sim 3 \times 1 0^{- 6}

T).

S = u c

specifically plane wave ke liye kyun hold karta hai?

Wave mein energy simply

c

speed se saath chalti hai: energy density

u

jo area

A

c d t

length mein sweep hoti hai woh flux

u c

deti hai. Yeh static fields ke liye nahi chalega jahan energy

c

par translate nahi kar rahi. Dekho Energy density of electric and magnetic fields aur EM wave equation.

Edge cases

Edge case: Purely static, uniform electric field (koi magnetic field nahi) ke liye

S

kya hoga?

Zero.

B = 0

ke saath,

E \times B = 0

— stored electric energy wahan baithe rehti hai, koi flux nahi.

Edge case: Us instant par

S

kya hoga jab

E = 0

ho (standing wave ka node, ya zero-crossing)?

Us instant par zero, kyunki

S \propto E B

aur

E = 0

. Standing wave ke liye time-average

⟨ S ⟩

bhi har jagah zero hota hai — koi net energy transport nahi hoti, sirf aage-peechhe slosh hoti hai. Neeche figure mein yeh nodes mark hain.

Edge case: Standing wave mein, kya intensity

\frac{1}{2} ε_{0} c E_{0}^{2}

hoti hai?

Nahi. Woh formula travelling wave ke liye hai. Pure standing wave koi net energy flow carry nahi karta, isliye iska net intensity zero hota hai chahe local

u

bada hi kyun na ho.

Edge case: Do beams of equal intensity ek point par cross karti hain — kya total

S

sirf magnitudes ka sum hai?

Nahi; pehle field vectors add karo, phir totals se

S

banao. Kyunki

S

fields mein quadratic hai, interference (constructive ya destructive) result ko change kar deta hai — magnitudes simply add nahi hote.

Edge case: Agar wave ki propagation direction reverse kar do toh

S

ka kya hoga?

Direction reverse karne se

B

ka sign

E

ke relative flip ho jaata hai (ya

E

B

ke relative), isliye

E \times B

reverse ho jaata hai —

S

ab naye direction mein point karta hai, jaisa hona chahiye.

Edge case: Absorbing medium (complex permittivity) mein, time-averaged

S

wave ke penetrate karne par kya karta hai?

Iska magnitude depth ke saath exponentially decay karta hai — "lost" flux exactly wahi energy hai jo

- J \cdot E

term material mein heat ki tarah dump karti hai. Isliye

\nabla \cdot ⟨ S ⟩ \neq = 0

: divergence local absorption measure karta hai.

Edge case: Dispersive lekin lossless medium mein, kya energy phir bhi wave ki phase speed par travel karti hai?

Nahi — energy (aur

⟨ S ⟩

) group velocity par move karti hai, jo phase velocity se kaafi alag ho sakti hai.

S = u c

relation mein actual energy-transport speed use honi chahiye,

c

nahi.

Recall Traps ki ek-line summary

Flux not density ( $S$ W/m² hai), cross not dot, $E = c B$ not $E = B$ , aur intensity ka $\frac{1}{2}$ woh hero hai jise tum baar baar bhool jaate ho.