Displacement current — Maxwell's addition to Ampere's law: Question bank

1.8.32 · D5 · Physics › Electromagnetism › Displacement current — Maxwell's addition to Ampere's law

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Conduction current $I_{c}$ = actual charge ka move karna (electrons ka wire mein drift karna).
Displacement current $I_{d} = ε_{0} \frac{d Φ _{E}}{d t}$ = ek changing electric flux ka magnetic-field-banana wala effect — koi charge move nahi karta.
Poora law: $\oint B \cdot d l = μ_{0} (I_{c} + I_{d})$ Ampère's Law jo Maxwell ne correct kiya.

Error dhundo

"Gap ke across koi charge nahi jaata, isliye gap ke paas koi magnetic field exist nahi karta."

Error: conclusion

I_{d}

ko ignore karta hai. Gap ke paas ek compass actually deflect karta hai kyunki growing

E

μ_{0} I_{d}

supply karta hai jo wire se

μ_{0} I_{c}

ke exactly equal hota hai.

I_{d} = ε_{0} d Φ_{E} / d t

, aur flux

Φ_{E} = E \cdot A

hamesha hota hai, toh

I_{d} = ε_{0} A d E / d t

har situation mein."

Error:

Φ_{E} = E A

sirf tab hold karta hai jab

E

uniform ho aur

A

ke perpendicular ho. Generally

Φ_{E} = \int E \cdot d A

; tum

E

ko fringing ya tilted field ke liye integral se bahar nahi nikal sakte.

"Ampère's law fail karta hai kyunki humne ek bekar bulging surface choose ki; flat disc use karo aur problem khatam."

Error: ek valid law ko har surface ke liye ek hi answer dena chahiye loop par. Yeh ki flat disc "kaam karti hai" ek aise law ko nahi bachata jo equally-valid bulging surface par fail hota hai.

"Kyunki

I_{d} = I_{c}

ek capacitor ke liye, displacement current bas conduction current ka fancy rename hai."

Error: yeh is special case mein charge conservation ki wajah se equal hain, identical concepts nahi hain. Ek vacuum wave mein

I_{c} = 0

lekin

I_{d} \neq = 0

— rename karna wahan koi sense nahi banata.

"Displacement current density

J_{d} = ε_{0} E

hai."

Error: yeh field ka time derivative hota hai,

J_{d} = ε_{0} \partial E / \partial t

. Ek bada lekin constant

E

zero

J_{d}

deta hai.

"Plates ke beech

E = σ / ε_{0}

, toh

Φ_{E}

, plate separation

d

par depend karta hai."

Error: ideal parallel-plate field ke liye,

Φ_{E} = E A = Q / ε_{0}

sirf charge par depend karta hai,

d

par nahi — separation cancel ho jaata hai (dekho Parallel Plate Capacitor).

Left side sirf loop (boundary) par depend karti hai, toh right side bhi karni chahiye. Agar yeh surface ke saath change hoti, toh equation koi single number define nahi karti aur meaningless ho jaati.

I_{d}

, Maxwell's Equations ko sirf ek ko patch karne ki bajaye kyun complete karta hai?

Yeh electric aur magnetic induction ke beech symmetry restore karta hai, wo feedback loop close karta hai jo charon equations ko self-consistent aur wave-supporting banata hai.

Wire ke andar

I_{d}

negligible kyun hai lekin gap ke andar huge kyun hai?

Conductor mein

E \approx 0

(charges arrange ho jaate hain ise cancel karne ke liye), toh iska rate of change tiny hai; gap ke across poora voltage baitha hai, toh

E

aur iska growth large hain.

Edge cases

Jab capacitor fully charged ho jaata hai (current ruk jaata hai) toh us instant

I_{d}

kya hota hai?

Zero — charging ruk jaata hai, toh

d Q / d t = 0

, isliye

d Φ_{E} / d t = 0

aur

I_{d} = 0

. Plates ke beech magnetic field bhi us instant vanish ho jaata hai.

Charging plates ke beech exactly central axis (

r = 0

) par magnetic field kya hota hai?

Zero, symmetry ki wajah se:

B = μ_{0} I r / (2 π R^{2})

B (0) = 0

milta hai, bilkul waise jaise ek uniform current-carrying wire ke axis par.

Agar capacitor charge hone ki bajaye discharge ho toh

I_{d}

ka kya hoga?

Yeh

d Q / d t

ke saath sign reverse karta hai;

E

ab shrink karta hai, toh

d Φ_{E} / d t < 0

, aur induced

B

opposite way mein circulate karta hai — reversed wire current ke saath consistent.

Plates ke bilkul bahar (radius

r > R

par), kya formula

B = μ_{0} I r / (2 π R^{2})

abhi bhi hold karta hai?

Nahi — plate edge ke baad poora displacement current

I_{d} = I

enclosed ho jaata hai, toh

B = μ_{0} I / (2 π r)

, feeding wire ke around field se identical. Linear-in-

r

form sirf

r < R

ke liye hai.

Agar electric field grow karta hai lekin perfectly uniform aur infinite extent mein rehta hai, toh kya phir bhi

B

hoga?

Ek changing uniform

E

nonzero

J_{d}

define karta hai, lekin symmetry se ek infinite uniform source koi net circulating

B

produce nahi karta — geometry, sirf

I_{d}

nahi, field fix karta hai.

Pure vacuum mein jab light pass ho rahi ho, toh Ampère–Maxwell mein "current" kaun carry karta hai?

Purely

I_{d}

I_{c} = 0

kyunki koi charges nahi hain, toh

\oint B \cdot d l = μ_{0} ε_{0} d Φ_{E} / d t

akela wave sustain karta hai.

Jis moment EM wave ka

E

zero se pass hota hai, kya

I_{d}

bhi wahan zero hota hai?

Nahi —

E = 0

hai lekin iska time mein slope usi waqt sabse steepest hota hai, toh

d Φ_{E} / d t

(isliye

I_{d}

aur resulting

B

) apne maximum par hota hai. Field value aur iska rate of change 90° out of phase hain.

Connections

Ampère's Law — woh law jo ye traps stress-test karte hain
Faraday's Law of Induction — mirror term
Maxwell's Equations — completed set
Electromagnetic Waves — jahan $I_{c} = 0$ lekin $I_{d} \neq = 0$
Parallel Plate Capacitor — canonical geometry
Gauss's Law — $Φ_{E} = Q / ε_{0}$ supply karta hai
Speed of Light — emergent constant $1/ μ_{0} ε_{0}$

Question bank — Displacement current — Maxwell's addition to Ampere's law

True or false — justify karo

Error dhundo

Why questions

Edge cases

Connections