Displacement current — Maxwell's addition to Ampere's law: Worked examples

1.8.32 · D3 · Physics › Electromagnetism › Displacement current — Maxwell's addition to Ampere's law

Yeh practice arena hai parent topic ke liye. Kuch bhi solve karne se pehle, ek aisi habit notice karo jo tumhe har baar bachayegi: displacement current koi nayi cheez nahi hai jo har problem mein alag se yaad karni ho — yeh hamesha kuch giney-chuney case classes mein se ek hoti hai. Neeche hum har woh case list karte hain jo topic tumhare saamne rakh sakta hai, phir unhe ek-ek karke gira dete hain.

Do symbols jo tumhe aage padhne se pehle already apne karne chahiye (hum inhe earn karte hain, contract ke mutabiq):

$Φ_{E}$ = electric flux = "kitna electric field ek chosen flat area ke through ghusta hai". Socho $E$ ke arrows ek hoop mein ghus rahe hain; $Φ_{E}$ unhe count karta hai. Agar $E$ uniform ho, strength $E$ ho, aur area $A$ pe seedha lage, toh $Φ_{E} = E A$ (V·m).
$ε_{0} = 8.85 \times 1 0^{- 12}$ (units C²/(N·m²)) = permittivity of free space, ek fixed number jo nature "field growing" ko "current" mein convert karne ke liye use karti hai. Ise ek exchange rate samjho.

The scenario matrix

Har displacement-current problem exactly inhi cells mein se kisi ek mein aata hai. Worked examples pe cell ka tag laga hai jo woh hit karta hai.

Cell	Case class	Tumhe kya diya gaya hai	$I_{d}$ tak ka raasta
A	Tumhe flux rate seedha pata hai	$d Φ_{E} / d t$	$I_{d} = ε_{0} d Φ_{E} / d t$
B	Tumhe capacitor pe voltage ramp pata hai	$C, d V / d t$	$I_{d} = C d V / d t$
C	Tumhe charge rate / wire current pata hai	$d Q / d t = I$	$I_{d} = I$ (dono equal hain)
D	Geometry: plates ke beech $r$ radius pe $B$ nikalna hai	$I, R, r$	$B = \frac{μ _{0} I r}{2 π R ^{2}}$
E	Edge / limiting: $r = R$ , ya $r > R$ (bahar)	boundary radius	$r = R$ pe formula switch karo
F	Zero / degenerate: constant field, DC steady state	$d V / d t = 0$	$I_{d} = 0$
G	Sign / direction: discharging capacitor	$d Q / d t < 0$	$I_{d} < 0$ → $B$ reverse ho jaata hai
H	Real-world word problem	mixed clues	matching row chunno
I	Exam twist: area aur $d E / d t$ se field	$A, d E / d t$	$I_{d} = ε_{0} A d E / d t$

Ab hum A se I tak aath examples ke saath cover karte hain.

The worked examples

Worked example 1. (Cell A) Seedha flux rate se

Ek surface ke through electric flux $\frac{d Φ _{E}}{d t} = 4 \times 1 0^{6} V\cdotm/s$ ki rate se badh raha hai. Displacement current nikalo.

Forecast: padhne se pehle order of magnitude guess karo. (Hint: $ε_{0}$ bahut chhota hai, toh expect karo ek bahut chhota current.)

Definition likho: $I_{d} = ε_{0} \frac{d Φ _{E}}{d t}$ . Yeh step kyun? Cell A humein seedha flux rate deta hai — definition sabse seedha raasta hai; koi charges ya plates ki zaroorat nahi.
Plug in karo: $I_{d} = (8.85 \times 1 0^{- 12}) (4 \times 1 0^{6})$ . Yeh step kyun? Hum bas exchange rate $ε_{0}$ ko flux kitni tezi se badhta hai usse multiply kar rahe hain.
$I_{d} = 3.54 \times 1 0^{- 5} A$ .

Verify: units $= \frac{C ^{2}}{N\cdotm ^{2}} \cdot \frac{V\cdotm}{s}$ . Kyunki $1 V = 1 N\cdotm/C$ , yeh collapse hokar $C/s = A$ ban jaata hai. ✓ Aur yeh chhota hai, humara forecast sahi nikla.

Worked example 4. (Cell D) Geometry classic — plates ke

beech $B$ $R = 5 cm$ radius ki circular plates pe charging current $I = 2 A$ hai. Gap mein axis se $r = 3 cm$ radius pe magnetic field $B$ nikalo. Figure dekho.

Forecast: kya $B$ bahar poore wire ke around hone wale $B$ se bada hoga ya chhota? (Hint: sirf flux ka ek hissa enclosed hai.)

Symmetry se $\oint B \cdot d l = B (2 π r)$ . Yeh step kyun? Field axis ke around circles mein ghoomta hai jaise wire ke around (figure mein blue loop dekho) — $r$ radius ke circle pe constant magnitude.
Enclosed displacement current area ke saath scale karta hai: $I_{d, e n c} = I \frac{π r ^{2}}{π R ^{2}} = I \frac{r ^{2}}{R ^{2}}$ . Yeh step kyun? Plates ke beech field uniform hai, isliye flux area ke proportional hai; $r$ radius ka chhota loop total ka $r^{2} / R^{2}$ fraction pakadta hai.
Ampère–Maxwell: $B (2 π r) = μ_{0} I \frac{r ^{2}}{R ^{2}} \Rightarrow B = \frac{μ _{0} I r}{2 π R ^{2}}$ . Yeh step kyun? Hum circulation ko $μ_{0}$ times enclosed (displacement) current ke barabar rakhte hain aur $B$ solve karte hain.
Numbers: $B = \frac{( 4 π \times 1 0 ^{- 7} ) ( 2 ) ( 0.03 )}{2 π ( 0.05 ) ^{2}} = 4.8 \times 1 0^{- 6} T$ .

Verify: formula bilkul wahi hai jo uniform current-carrying wire ke andar $B$ ka hai, $B = μ_{0} I r / (2 π R^{2})$ — waada ki gayi symmetry ✓. Numerically $4.8 μ T$ .

Worked example 5. (Cell E) Edge aur bahar — limiting behaviour

Wahi plates Example 4 wali ( $R = 5 cm$ , $I = 2 A$ ). $B$ nikalo (a) bilkul rim pe $r = R$ pe, aur (b) gap ke bahar $r = 8 cm$ pe. Dono regimes ke liye figure dekho.

Forecast: kya $B$ forever badhta rehega jaise $r$ badhta hai, ya $r = R$ pe kuch switch hoga?

Andar ( $r \leq R$ ) hum Example 4 ka result use karte hain. $r = R$ pe: $B = \frac{μ _{0} I R}{2 π R ^{2}} = \frac{μ _{0} I}{2 π R}$ . Yeh step kyun? $r = R$ rakh do; dono $R$ aapas mein kuch cancel ho jaate hain. Yeh peak hai — rim ke baad enclose karne ke liye aur flux nahi bachta.
$B (R) = \frac{( 4 π \times 1 0 ^{- 7} ) ( 2 )}{2 π ( 0.05 )} = 8.0 \times 1 0^{- 6} T$ .
Bahar ( $r > R$ ) loop poora displacement current $I_{d} = I = 2 A$ enclose karta hai (figure mein green shading), isliye $B (2 π r) = μ_{0} I$ aur $B = \frac{μ _{0} I}{2 π r}$ . Yeh step kyun? Jab $r > R$ ho jaata hai tab tumne saara flux pakad liya; radius badhane se aur enclosed current nahi badhti, isliye $B$ ab $1/ r$ ki tarah girta hai.
$r = 8 cm$ pe: $B = \frac{( 4 π \times 1 0 ^{- 7} ) ( 2 )}{2 π ( 0.08 )} = 5.0 \times 1 0^{- 6} T$ .

Verify: rim pe dono formulas agree karni chahiye — andar wali branch $μ_{0} I r / (2 π R^{2})$ ko $r = R$ pe lagao, milega $μ_{0} I / (2 π R)$ , jo exactly bahar wali branch $μ_{0} I / (2 π r)$ ko $r = R$ pe lagate hain. Continuous, koi jump nahi ✓. Aur $B$ rim pe peak karta hai, phir decay (forecast confirmed) ✓.

Worked example 6. (Cell F) Degenerate case — poori tarah charged, kuch nahi badal raha

Wahi capacitor ab fully charged hai aur constant voltage pe baitha hai. Wire current zero hai. $I_{d}$ aur gap mein magnetic field nikalo.

Forecast: wire mein koi current nahi, toh kya gap mein koi field hone ki ummeed hai?

$\frac{d V}{d t} = 0 \Rightarrow \frac{d Φ _{E}}{d t} = 0$ . Yeh step kyun? Plates ke beech field fixed hai (charge ab move nahi kar raha), isliye uska flux frozen hai.
$I_{d} = ε_{0} \frac{d Φ _{E}}{d t} = 0$ . Yeh step kyun? Flux mein koi change nahi matlab koi displacement current nahi — yeh effect sirf tab exists karta hai jab field move kar raha ho.
Isliye $\oint B \cdot d l = μ_{0} (0 + 0) = 0$ , toh gap mein har jagah $B = 0$ .

Verify: wire ke saath consistent — ek fully charged, static capacitor ke paas compass nahi hilta (forecast confirmed: koi field nahi) ✓. Yeh "off" state hai jo har charging problem ko bookend karta hai.

Worked example 7. (Cell G) Sign / direction —

discharging capacitor Ek capacitor ek resistor ke through discharge karta hai. Uska stored charge $Q (t) = Q_{0} e^{- t / τ}$ ki tarah girta hai jahan $Q_{0} = 6 \times 1 0^{- 6} C$ aur $τ = 0.5 s$ hai. $t = 0$ pe gap mein displacement current nikalo, aur charging case ke relative uski direction batao.

Forecast: kya $I_{d}$ positive aayega ya negative? Sign ka physically kya matlab hai?

$I_{d} = \frac{d Q}{d t} = \frac{d}{d t} (Q_{0} e^{- t / τ}) = - \frac{Q _{0}}{τ} e^{- t / τ}$ . Yeh step kyun? Abhi bhi $I_{d} = d Q / d t$ ; discharging ka matlab bas yeh hai ki $Q$ ghatt raha hai, isliye derivative negative hai.
$t = 0$ pe: $I_{d} = - \frac{6 \times 1 0 ^{- 6}}{0.5} = - 1.2 \times 1 0^{- 5} A$ . Yeh step kyun? Exponential ( $e^{0} = 1$ ) ko start pe evaluate karo, jahan discharge sabse tez hoti hai.
Negative sign ka matlab hai ki plates ke beech electric field ab shrink kar raha hai badhne ki jagah. Isliye induced $B$ ulti direction mein circulate karta hai charging ke comparison mein. Yeh step kyun? $B$ ki direction $d Φ_{E} / d t$ ke sign ko follow karti hai; sign flip karo, curl flip ho jaata hai.

Verify: magnitude $1.2 \times 1 0^{- 5} A$ ; sign negative. $Q_{0} / τ$ ke units: $C/s = A$ ✓. Reversed field shrinking flux ki physics se match karta hai ✓.

Worked example 8. (Cells H + I) Word problem with a field-rate twist

Story: Ek engineer vacuum mein ek capacitor test kar rahi hai. Woh seedha current measure nahi kar sakti, lekin uska sensor report karta hai ki $L = 10 cm$ side wali do square plates ke beech uniform electric field $\frac{d E}{d t} = 5 \times 1 0^{9} V/(m\cdots)$ ki rate se badh raha hai. Gap ke across kaunsa displacement current flow kar raha hai?

Forecast: field rate bahut badi hai — lekin $ε_{0}$ tiny hai aur plates chhoti hain. Milliamps? Micro? Pehle guess karo.

Plate ka area: $A = L^{2} = (0.1)^{2} = 0.01 m^{2}$ . Yeh step kyun? Flux ke liye area chahiye; plates squares hain, isliye $A = L^{2}$ , $π R^{2}$ nahi.
Flux $Φ_{E} = E A$ (field uniform aur perpendicular hai), isliye $\frac{d Φ _{E}}{d t} = A \frac{d E}{d t}$ . Yeh step kyun? $A$ yahan constant hai — sirf $E$ change karta hai — isliye yeh derivative ke bahar factor ho jaata hai. Yeh Cell I hai.
$I_{d} = ε_{0} A \frac{d E}{d t} = (8.85 \times 1 0^{- 12}) (0.01) (5 \times 1 0^{9})$ . Yeh step kyun? Definition ko abhi banaye flux ke saath combine karo.
$I_{d} = 4.425 \times 1 0^{- 4} A \approx 0.44 mA$ .

Verify: units $\frac{C ^{2}}{N\cdotm ^{2}} \cdot m^{2} \cdot \frac{V}{m\cdots}$ . $V = N\cdotm/C$ use karne se milta hai $C/s = A$ ✓. Sub-milliamp — forecast se match karta hai ki tiny $ε_{0}$ giant field rate ko bhi kabu mein rakhta hai ✓. Note: koi charges vacuum ke across cross nahi kiye — pure Cell A/I physics, exactly wahi jo light ko possible banata hai.

Recall Self-check: clue ko cell se match karo

Sirf $d V / d t$ aur $C$ diya gaya hai ::: Cell B → $I_{d} = C d V / d t$ Sirf $d E / d t$ aur plate area diya gaya hai ::: Cell I → $I_{d} = ε_{0} A d E / d t$ $r > R$ pe $B$ poochha gaya ::: Cell E → $B = μ_{0} I / (2 π r)$ Capacitor fully charged, DC ::: Cell F → $I_{d} = 0$ $Q (t)$ ghatt raha hai ::: Cell G → $I_{d} < 0$ , field reversed

Connections

Parent topic
Ampère's Law — circulation rule jo hum har geometry example mein apply karte hain
Parallel Plate Capacitor — Cells B–G ka canonical stage
Gauss's Law — gap mein uniform field deta hai jo flux mein use hota hai
Faraday's Law of Induction — sign discussion ke liye mirror twin
Maxwell's Equations · Electromagnetic Waves · Speed of Light