Axiom Physics | Class 12 - Displacement Current & Maxwell's Equations

⚡ 01. The Spark (Analogy)

Socho ek railway track ko jisme beech mein ek chaudi nadi (river) aati hai jahan koi patri nahi bichi. Track par chal rahi train ke passengers ek taraf utarte hain aur boat mein baithkar nadi cross karte hain, fir doosri taraf se nayi train aage nikal jaati hai. Agar koi observer sirf lohe ki patri (physical conduction current $I_c$) ginega, toh use lagega ki nadi ke beech mein connection toot gaya! James Clerk Maxwell ne yahi pehchana: capacitor ke plates ke beech khali space (vacuum) mein koi charge nahi behta, lekin charging ke dauran electric field tezi se badalta hai. Yeh changing electric field wahi kaam karta hai jo moving charge karta hai—yani magnetic field paida karna! Is virtual bridge ko unhone Displacement Current ($I_d$) kaha, jisne electromagnetism ke adhoore laws ko jodkar universe ki sabse khoobsurat equations banayi!

📐 02. The Core Law & Formulae

1. Displacement Current ($I_d$)

Current arising from time rate of change of electric flux:

$$I_d = \varepsilon_0 \frac{d\Phi_E}{dt} = \varepsilon_0 A \frac{dE}{dt}$$

SI Unit: $\text{Ampere (A)}$. Outside plates: $I_c = I, I_d = 0$. Inside capacitor gap: $I_c = 0, I_d = I$. Total current is continuous everywhere: $I = I_c + I_d$.

2. Ampere-Maxwell Law

Generalized line integral of magnetic field:

$$\oint \vec{B} \cdot d\vec{l} = \mu_0 (I_c + I_d) = \mu_0 I_c + \mu_0 \varepsilon_0 \frac{d\Phi_E}{dt}$$

Physical Signification: A magnetic field can be produced not only by moving physical charges (conduction current) but also by a time-varying electric field!

03. Interactive Visualizer: Capacitor Charging, Displacement Current & Magnetic Field Bench

Charging Current ($I_c$) 2.50 A

Capacitor Plate Radius ($R$) 10.0 cm

Observation Radius ($r$) 6.0 cm (Inside Gap)

Total Displacement Current ($I_d$):

2.50 A (Exact Match)

Rate of Flux Change ($d\Phi_E/dt$):

2.82 × 10¹¹ V·m/s

Induced Field at $r$ ($B$):

3.00 μT

Continuity Law Status:

I_c + I_d = Invariant

📝 04. Comprehensive Stepwise Derivations Blueprint

Primary Board Derivation

Inconsistency of Ampere's Circuital Law & Deduction of Displacement Current

Fig 8.1A: Ampere's law contradiction; Loop $C$ bounded by flat disk $S_1$ pierces current $I_c$, but pot-shaped surface $S_2$ passing through gap pierces zero conduction current.

1. Consider a parallel plate capacitor of plate area $A$ being charged by steady conduction current $I_c$.

2. Let us calculate the line integral $\oint \vec{B}\cdot d\vec{l}$ around a circular loop $C$ outside the capacitor plate:

• For a flat circular disk surface $S_1$ spanning loop $C$, the wire pierces it: $\oint \vec{B}\cdot d\vec{l} = \mu_0 I_c$.

• For a pot-shaped balloon surface $S_2$ having the exact same boundary loop $C$ but passing between the capacitor plates, no wire pierces it: $\oint \vec{B}\cdot d\vec{l} = 0$.

3. This gives two completely contradictory answers for the exact same loop $C$! To remove this logical inconsistency, Maxwell postulated that a changing electric field in the vacuum gap must produce a current equivalent to conduction current.

4. At any instant during charging, charge on the plates is $q(t)$. Uniform electric field between plates:

$$E = \frac{\sigma}{\varepsilon_0} = \frac{q}{\varepsilon_0 A}$$

5. Electric flux through surface $S_2$ inside the gap:

$$\Phi_E = E \cdot A = \left(\frac{q}{\varepsilon_0 A}\right) A = \frac{q}{\varepsilon_0} \implies q = \varepsilon_0 \Phi_E$$

6. Differentiating with respect to time $t$:

$$\frac{dq}{dt} = \varepsilon_0 \frac{d\Phi_E}{dt}$$

7. Since $\frac{dq}{dt} = I_c$ (rate of charging), this term has the exact dimensions of electric current, defined as the Displacement Current ($I_d$):

$$I_d = \varepsilon_0 \frac{d\Phi_E}{dt} \quad \text{and} \quad I_d \equiv I_c$$

Resolution of Inconsistency: While surface $S_1$ carries conduction current $I_c$, surface $S_2$ carries displacement current $I_d = \varepsilon_0 \frac{d\Phi_E}{dt}$. Both surfaces give identical magnetic circulation $\oint \vec{B}\cdot d\vec{l} = \mu_0 I$.

Derivation B

Magnetic Field Induced Inside Circular Capacitor Plates ($r \le R$ and $r \ge R$)

1. Consider circular capacitor plates of radius $R$. Conduction current charging the plates is $I_c$. Total displacement current through the entire gap is $I_d = I_c$.

2. Construct a circular Amperian loop of radius $r$ coaxial with the plates inside the gap. By cylindrical symmetry, $\vec{B}$ is tangential to the circle at all points:

$$\oint \vec{B}\cdot d\vec{l} = B (2\pi r) = \mu_0 I_{d,\text{enc}}$$

3. Case (i) Inside the plates ($r \le R$): Displacement current is uniformly distributed across plate area $\pi R^2$. The current enclosed by radius $r$ is:

$$I_{d,\text{enc}} = I_d \left( \frac{\pi r^2}{\pi R^2} \right) = I_d \frac{r^2}{R^2}$$

Substituting into the Ampere-Maxwell law:

$$B(2\pi r) = \mu_0 I_d \frac{r^2}{R^2} \implies B = \frac{\mu_0 I_d r}{2\pi R^2} \quad (B \propto r \text{ for } r \le R)$$

4. Case (ii) Outside the plates ($r \ge R$): The entire displacement current is enclosed ($I_{d,\text{enc}} = I_d = I_c$):

$$B(2\pi r) = \mu_0 I_d \implies B = \frac{\mu_0 I_d}{2\pi r} = \frac{\mu_0 I_c}{2\pi r} \quad \left(B \propto \frac{1}{r} \text{ for } r \ge R\right)$$

Core Insight: Exactly at the circumference ($r = R$), magnetic field reaches maximum value $B_{\max} = \frac{\mu_0 I_c}{2\pi R}$, completely identical to the field outside a straight wire carrying current $I_c$!

Foundational Synthesis

Maxwell's Four Master Equations of Electromagnetism

1. Gauss's Law for Electricity

$$\oint \vec{E}\cdot d\vec{A} = \frac{q_{\text{enc}}}{\varepsilon_0}$$

Electric charges act as sources and sinks of electric field; isolated positive and negative charges exist.

2. Gauss's Law for Magnetism

$$\oint \vec{B}\cdot d\vec{A} = 0$$

Isolated magnetic monopoles do not exist; magnetic lines are continuous closed unbroken loops.

3. Faraday-Lenz Law of Induction

$$\oint \vec{E}\cdot d\vec{l} = -\frac{d\Phi_B}{dt}$$

A time-varying magnetic field creates an induced non-electrostatic curling electric field.

4. Ampere-Maxwell Law

$$\oint \vec{B}\cdot d\vec{l} = \mu_0 I_c + \mu_0 \varepsilon_0 \frac{d\Phi_E}{dt}$$

A magnetic field is created by electric currents as well as by a time-varying electric field.

⚠️ 05. Red Alert Exam Trap Zone

Trap 1: Displacement Current in a Fully Charged Capacitor

*Board Question:* A capacitor is fully charged by a DC battery. What is the displacement current between the plates?
*Correct Answer:* ZERO ($I_d = 0$)! Once fully charged, charge $Q$ is constant $\implies$ electric field $E = Q/\varepsilon_0 A$ is constant $\implies d\Phi_E/dt = 0$. Displacement current exists ONLY while the capacitor is charging or discharging!

Trap 2: Does Displacement Current Dissipate Heat ($I^2R$)?

*Conceptual Trap:* Does displacement current produce Joule heat in the vacuum gap between plates?
*Correct Answer:* NO! Displacement current is not carried by physical charge carriers (electrons or ions); it has no collision mechanism with a lattice ($R = 0$). Hence, no Joule heating occurs ($I_d^2 R = 0$). However, it creates identical magnetic fields as conduction current!

Trap 3: Continuity of Current ($I_c$ vs $I_d$)

At any cross section along a circuit:
• Inside connecting lead wires: $I_c = I, I_d = 0$.
• Inside capacitor dielectric gap: $I_c = 0, I_d = I$.
• Total current $I_{\text{total}} = I_c + I_d$ is strictly continuous across all cross-sections of the entire circuit.

Trap 4: Symmetry of Nature in Maxwell's Equations

• Faraday's law states: Changing magnetic field $\to$ Electric field ($\oint \vec{E}\cdot d\vec{l} = -\frac{d\Phi_B}{dt}$).
• Maxwell's fourth law states: Changing electric field $\to$ Magnetic field ($\oint \vec{B}\cdot d\vec{l} = \mu_0\varepsilon_0\frac{d\Phi_E}{dt}$).
This mutual time-dependent regeneration of fields is what enables electromagnetic waves to propagate across empty space without any material medium!

💡 06. Handcrafted Exemplars

Board Classic Level

Problem: A parallel plate capacitor with circular plates of radius $R = 12\text{ cm}$ separated by distance $d = 5.0\text{ mm}$ is being charged by an external current of $0.15\text{ A}$. (a) Calculate the capacitance and the rate of change of potential difference between the plates. (b) Find the displacement current across the plates. (c) Determine the magnetic field induced between the plates at a distance of $r = 6.0\text{ cm}$ from the axis.

Solution:

1. Area $A = \pi R^2 = \pi (0.12\text{ m})^2 \approx 4.52 \times 10^{-2}\text{ m}^2$. Capacitance $C$:

$$C = \frac{\varepsilon_0 A}{d} = \frac{(8.854 \times 10^{-12}) \times (4.52 \times 10^{-2})}{5.0 \times 10^{-3}} \approx 8.01 \times 10^{-11}\text{ F} \approx 80.1\text{ pF}$$

Rate of change of voltage: Since $q = C V \implies I = C \frac{dV}{dt}$:

$$\frac{dV}{dt} = \frac{I}{C} = \frac{0.15\text{ A}}{8.01 \times 10^{-11}\text{ F}} \approx 1.87 \times 10^9\text{ V/s}$$

2. Displacement current $I_d$ equals conduction current:

$$I_d = I_c = 0.15\text{ Amperes}$$

3. Induced magnetic field at $r = 6.0\text{ cm} = 0.06\text{ m}$ (inside $R = 12\text{ cm}$):

$$B = \frac{\mu_0 I_d r}{2\pi R^2} = \frac{(4\pi \times 10^{-7}) \times 0.15 \times 0.06}{2\pi \times (0.12)^2} = \frac{2 \times 10^{-7} \times 0.009}{0.0144} = \frac{1.8 \times 10^{-9}}{0.0144}$$

$$B = 1.25 \times 10^{-7}\text{ Tesla} = 0.125\text{ }\mu\text{T}$$

JEE/NEET Competitive Trap

Problem: The electric field between the plates of a circular parallel plate capacitor of radius $R$ is changing at a uniform rate $\frac{dE}{dt} = K$. Show that the ratio of the magnetic field induced at distance $r_1 = R/2$ to that at distance $r_2 = 2R$ from the central axis is $1:1$.

Solution:

1. At an interior point ($r_1 = R/2 \le R$):

$$B_1(2\pi r_1) = \mu_0 \varepsilon_0 \left(\frac{dE}{dt}\right)(\pi r_1^2) \implies B_1 = \frac{\mu_0 \varepsilon_0 r_1}{2} K = \frac{\mu_0 \varepsilon_0 (R/2)}{2} K = \frac{\mu_0 \varepsilon_0 R}{4} K$$

2. At an exterior point ($r_2 = 2R \ge R$): The entire flux through plate area $\pi R^2$ contributes:

$$B_2(2\pi r_2) = \mu_0 \varepsilon_0 \left(\frac{dE}{dt}\right)(\pi R^2) \implies B_2 = \frac{\mu_0 \varepsilon_0 R^2}{2 r_2} K = \frac{\mu_0 \varepsilon_0 R^2}{2 (2R)} K = \frac{\mu_0 \varepsilon_0 R}{4} K$$

3. Comparing both fields:

$$\frac{B_1}{B_2} = \frac{\frac{\mu_0 \varepsilon_0 R}{4} K}{\frac{\mu_0 \varepsilon_0 R}{4} K} = \frac{1}{1}$$

Symmetry Insight: Inside the plates $B \propto r$, and outside $B \propto 1/r$. Thus, the induced field at half radius ($R/2$) exactly equals the field at double radius ($2R$)!

🗂️ 07. Quick Revision Flashcard Core Formulae

Displacement Current

$I_d = \varepsilon_0\frac{d\Phi_E}{dt}$

Ampere-Maxwell Law

$\oint B\cdot dl = \mu_0(I_c + I_d)$

Field inside plates ($r \le R$)

$B = \frac{\mu_0 I_d r}{2\pi R^2} \propto r$

Field outside ($r \ge R$)

$B = \frac{\mu_0 I_d}{2\pi r} \propto \frac{1}{r}$