04 time_varying_fields_and_maxwells_equations

📘 Topic 4: Time-Varying Fields & Maxwell’s Equations

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🗺️ Big-Picture Study Map

Up to this point, we have studied Electrostatics (fields produced by stationary charges) and Magnetostatics (fields produced by steady currents). These fields were completely uncoupled; an electric field had no relationship to a magnetic field.

In this chapter, we enter the domain of Electrodynamics where fields vary with time. A time-varying magnetic field acts as a source for an electric field (Faraday’s Law), and a time-varying electric field acts as a source for a magnetic field (Maxwell’s Displacement Current). This coupling gives rise to self-sustaining electromagnetic waves that propagate through space.

This chapter is highly mathematical but extremely structured. There are exactly five core pillars you must master to secure an A+:

1. The Continuity Equation (Conservation of Charge)
2. Displacement Current (Maxwell’s mathematical masterstroke)
3. Maxwell’s Equations (The four pillars of classical electromagnetics)
4. Dynamic Boundary Conditions (Why they are the same as static cases)
5. Poynting’s Theorem (Electromagnetic power and energy flow)

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1. Conservation of Charge & The Continuity Equation

1.1 Physical Principle

The Principle of Conservation of Charge states that electric charge can neither be created nor destroyed; it can only be transported from one region to another.

For any arbitrary closed volume $V$ bounded by a surface $S$, if charge is flowing out through the surface (creating a current $I_{out}$), the total charge $Q$ remaining inside the volume must decrease at exactly the same rate.

1.2 Mathematical Derivation

1. Let the total current leaving the closed volume $V$ through surface $S$ be $I_{out}$. Mathematically, this is the surface integral of the current density vector $\vec{J}$: $I_{out}={\oint }_S\vec{J}\cdot d\vec{s}$

2. According to the conservation of charge, this outgoing current must equal the negative rate of change of the enclosed charge $Q$ with respect to time: $I_{out}=-\frac{dQ}{dt}$ Combining these two expressions: ${\oint }_S\vec{J}\cdot d\vec{s}=-\frac{dQ}{dt}$

3. Express the total enclosed charge $Q$ in terms of its volume charge density ${\rho }_v$: $Q={\int }_V{\rho }_{v}dv$ Substitute this into the conservation equation: ${\oint }_S\vec{J}\cdot d\vec{s}=-\frac{d}{dt}{\int }_V{\rho }_{v}dv$

4. Assuming that the volume $V$ is stationary (rigid and not moving), we can bring the time derivative inside the integral. Since ${\rho }_v$ can be a function of both space coordinates and time, the total derivative becomes a partial derivative: ${\oint }_S\vec{J}\cdot d\vec{s}=-{\int }_V\frac{\partial {\rho }_v}{\partial t}dv$

5. Apply the Divergence Theorem to the closed surface integral on the left-hand side to transform it into a volume integral: ${\oint }_S\vec{J}\cdot d\vec{s}={\int }_V(\nabla \cdot \vec{J})dv$ Now equate both volume integrals: ${\int }_V(\nabla \cdot \vec{J})dv=-{\int }_V\frac{\partial {\rho }_v}{\partial t}dv$ ${\int }_V\left(\nabla \cdot \vec{J}+\frac{\partial {\rho }_v}{\partial t}\right)dv=0$

6. Since this relationship must hold true for any arbitrary, infinitesimally small volume $V$, the integrand itself must be identically zero. This yields the differential form of the Continuity Equation: $\nabla \cdot \vec{J}=-\frac{\partial {\rho }_v}{\partial t}$

1.3 Physical Significance & KCL Relation

• Physical Interpretation: The divergence of current density at any point is equal to the time rate of decrease of the volume charge density at that point. If current diverges ($\nabla \cdot \vec{J}>0$), charge density inside is depleting.
• The Steady Current Case: For steady (direct) currents, charges do not accumulate or deplete at any point over time, which means $\frac{\partial {\rho }_v}{\partial t}=0$. Under this condition, the continuity equation reduces to: $\nabla \cdot \vec{J}=0$
  • This shows that steady electric currents are divergence-less (solenoidal). They must flow in closed loops with no sources or sinks.
  • Taking the surface integral over a junction: ${\oint }_S\vec{J}\cdot d\vec{s}=0⟹{\sum }_j{I}_j=0$
  • This is the exact field-theory equivalent of Kirchhoff’s Current Law (KCL), which states that the algebraic sum of currents entering/leaving a junction is zero.

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📝 Verbatim PYQs & Solutions (Continuity Equation)

Question 1 [PYQ: 2015, 2018, 2022 - 10 Marks]

State the principle of conservation of charge. Based on this principle, derive the continuity equation, $\nabla \cdot \vec{J}=-\partial {\rho }_v/\partial t$, where the symbols have their usual meanings. Also write down the physical significance of this equation.

Model Answer:

• Statement: State the Principle of Conservation of Charge exactly as in Section 1.1.
• Derivation: Provide steps 1 through 6 from Section 1.2, clearly labeling each mathematical operator (Divergence Theorem, conversion of total to partial derivative).
• Physical Significance: Explain the divergence of current density and show its reduction to KCL under steady-state conditions ($\partial {\rho }_v/\partial t=0$), exactly as outlined in Section 1.3.

Question 2 [PYQ: 2019 - 10 Marks]

State the principle of conservation of charge. Through this principle, deduce the continuity equation, $\nabla \cdot \vec{J}=-\frac{\partial {\rho }_v}{\partial t}$, where the symbols have their usual meanings. Also, write down the physical significance of this equation.

Model Answer: (This is identical to the 2015/2018/2022 question. Write out the full derivation and physical significance as described above).

Question 3 [PYQ: 2021, 2025 - 10 Marks]

Explain the law of conservation of charge. Starting from this law, derive the continuity equation. $\nabla \cdot J=-\frac{\partial {\rho }_v}{\partial t}$. Where the symbols have their usual meanings. Briefly discuss its physical interpretation.

Model Answer: (Identical derivation. Ensure that you define every symbol: $\vec{J}$ = current density in ${\text{A/m}}^2$, ${\rho }_v$ = volume charge density in ${\text{C/m}}^3$, $t$ = time in seconds, and $\nabla \cdot$ = divergence operator).

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2. Displacement Current Density

2.1 The Inadequacy of Ampere’s Static Law

In magnetostatics, Ampere’s circuital law is written in differential form as: $\nabla \times \vec{H}=\vec{J}$

If we take the divergence of both sides of this equation, we get: $\nabla \cdot (\nabla \times \vec{H})=\nabla \cdot \vec{J}$

Using the fundamental vector identity where the divergence of the curl of any vector field is always zero ($\nabla \cdot \nabla \times \vec{A}\equiv 0$), the left-hand side becomes identically zero: $0=\nabla \cdot \vec{J}$

However, this contradicts the fundamental Continuity Equation for time-varying fields ($\nabla \cdot \vec{J}=-\frac{\partial {\rho }_v}{\partial t}$). It only holds true for steady-state static fields where $\frac{\partial {\rho }_v}{\partial t}=0$.

Thus, Ampere’s static law is mathematically inconsistent for time-varying fields.

2.2 Maxwell’s Insertion of Displacement Current

To resolve this inconsistency, James Clerk Maxwell hypothesized that a missing term ${\vec{J}}_d$ (displacement current density) must be added to the right-hand side of Ampere’s law: $\nabla \times \vec{H}=\vec{J}+{\vec{J}}_d$

Let us determine ${\vec{J}}_d$ mathematically using the continuity equation:

1. Take the divergence of both sides: $\nabla \cdot (\nabla \times \vec{H})=\nabla \cdot (\vec{J}+{\vec{J}}_d)$ $0=\nabla \cdot \vec{J}+\nabla \cdot {\vec{J}}_d⟹\nabla \cdot {\vec{J}}_d=-\nabla \cdot \vec{J}$

2. Substitute the Continuity Equation ($\nabla \cdot \vec{J}=-\frac{\partial {\rho }_v}{\partial t}$) into the relation: $\nabla \cdot {\vec{J}}_d=\frac{\partial {\rho }_v}{\partial t}$

3. According to Gauss’s Law, the volume charge density can be written as the divergence of the electric flux density (${\rho }_v=\nabla \cdot \vec{D}$). Substitute this in: $\nabla \cdot {\vec{J}}_d=\frac{\partial }{\partial t}(\nabla \cdot \vec{D})$

4. Since spatial and temporal derivatives are independent, we can swap their order on the right-hand side: $\nabla \cdot {\vec{J}}_d=\nabla \cdot \left(\frac{\partial \vec{D}}{\partial t}\right)$

5. Removing the divergence operator from both sides, we get the definition of Displacement Current Density (${\vec{J}}_d$): ${\vec{J}}_d=\frac{\partial \vec{D}}{\partial t}=ϵ\frac{\partial \vec{E}}{\partial t}$

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2.3 High-Yield Derivation: Displacement Current in a Capacitor under AC Excitation

This is a classic exam question designed to test your physical understanding of displacement current. You must prove that inside a capacitor, the displacement current $i_d$ is exactly equal to the conduction current $i_c$ in the external connecting wires.

[Insert Figure: Parallel plate capacitor connected to an AC source V(t). Show conduction current i_c in wires, electric field lines E between plates, and a closed loop passing through the wire vs between the plates. Ref: Cheng Figure 7-1 / 01 Static Electric Field PDF]

Step-by-Step Proof:

1. Consider a parallel-plate capacitor with plate area $A$ and plate separation $d$. Let it be energized by a time-harmonic AC voltage source: $v(t)=V_0\sin \omega t$

2. The time-varying electric field $\vec{E}$ generated between the plates (neglecting fringing effects at the edges) is: $\vec{E}(t)={\hat{a}}_n\frac{v(t)}{d}={\hat{a}}_n\frac{V_0}{d}\sin \omega t$

3. The electric flux density $\vec{D}$ in the dielectric medium ($ϵ$) between the plates is: $\vec{D}(t)=ϵ\vec{E}(t)={\hat{a}}_n\frac{ϵV_0}{d}\sin \omega t$

4. Calculate the displacement current density ${\vec{J}}_d$: ${\vec{J}}_d=\frac{\partial \vec{D}}{\partial t}={\hat{a}}_n\frac{ϵV_0\omega }{d}\cos \omega t$

5. The total displacement current $i_d$ passing through the area $A$ of the plates is: $i_d={\int }_A{\vec{J}}_d\cdot d\vec{s}=J_{d}A=\frac{ϵA}{d}{V}_0\omega \cos \omega t$

6. We recognize that the capacitance of a parallel plate capacitor is $C=\frac{ϵA}{d}$. Substitute this into our expression: $i_d=C{V}_0\omega \cos \omega t=C\frac{dv(t)}{dt}$

7. Now, let us find the conduction current $i_c$ flowing in the external wires. From basic circuit theory: $i_c(t)=C\frac{dv(t)}{dt}$

8. Comparing the two equations: $i_d(t)=i_c(t)$

• Conclusion: This mathematically proves that displacement current is not a physical current of moving charges, but rather a time-varying electric field that behaves exactly like a current. It bridges the gap between the plates of a capacitor, ensuring that current remains continuous throughout the closed loop.

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📝 Verbatim PYQs & Solutions (Displacement Current)

Question 1 [PYQ: 2016, 2017 - 9 Marks]

Define displacement current. Determine the displacement current in between two parallel plates of a capacitor energized by an alternating current source.

Model Answer:

• Definition: Write the exact definition of displacement current from the glossary: “A fictitious current proportional to the time rate of change of the electric displacement field ($\vec{D}$), introduced by Maxwell to make Ampere’s law consistent for time-varying scenarios.”
• Derivation: Reproduce the complete parallel plate capacitor proof from Section 2.3. Explain clearly that $i_d=i_c$ to get full marks.

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3. Maxwell’s Equations

Maxwell’s equations are the fundamental postulates of classical electromagnetics. They summarize all electric and magnetic phenomena into four elegant equations.

3.1 The Master Table (Time-Varying Fields)

Equation Name	Differential Form	Integral Form	Physical Significance / Experimental Law
Gauss’s Law for Electricity	$\nabla \cdot \vec{D}={\rho }_v$	${\oint }_S\vec{D}\cdot d\vec{s}=Q_{enc}$	Total outward electric flux through a closed surface equals the enclosed charge. Electric fields originate on positive charges and terminate on negative charges.
Gauss’s Law for Magnetism	$\nabla \cdot \vec{B}=0$	${\oint }_S\vec{B}\cdot d\vec{s}=0$	The net magnetic flux through any closed surface is zero. Magnetic monopoles do not exist; magnetic flux lines always form closed loops.
Faraday’s Law of Induction	$\nabla \times \vec{E}=-\frac{\partial \vec{B}}{\partial t}$	${\oint }_C\vec{E}\cdot d\vec{l}=-\frac{d}{dt}{\int }_S\vec{B}\cdot d\vec{s}$	A time-varying magnetic field induces a circulating, non-conservative electric field (EMF). The negative sign represents Lenz’s Law.
Ampere’s Circuital Law	$\nabla \times \vec{H}=\vec{J}+\frac{\partial \vec{D}}{\partial t}$	${\oint }_C\vec{H}\cdot d\vec{l}=I+{\int }_S\frac{\partial \vec{D}}{\partial t}\cdot d\vec{s}$	Circulating magnetic fields are generated by both conduction currents ($\vec{J}$) and time-varying electric flux (displacement currents).

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3.2 High-Yield Derivation: Deducing Maxwell’s Equations from Fundamental Static Equations

Examiners frequently ask you to derive/deduce Maxwell’s equations starting from the fundamental equations of electrostatics and magnetostatics by incorporating Faraday’s Law and the Continuity Equation.

Step-by-Step Mathematical Deduction:

1. Postulate I (Gauss’s Law for Electrostatics): Start with the static postulate $\nabla \cdot \vec{D}={\rho }_v$. Since time-varying electric fields still originate and terminate on electric charges, this relationship remains unmodified in the dynamic case: $\nabla \cdot \vec{D}={\rho }_v\text{(Maxwell’s 1st Equation)}$

2. Postulate II (Gauss’s Law for Magnetostatics): Start with the magnetostatic postulate $\nabla \cdot \vec{B}=0$. Since no physical experiment has ever isolated a single magnetic monopole, the divergence of magnetic flux density must remain zero under time-varying conditions: $\nabla \cdot \vec{B}=0\text{(Maxwell’s 2nd Equation)}$

3. Postulate III (Faraday’s Law Modification):

   • In electrostatics, the electric field is conservative: $\nabla \times \vec{E}=0$.
   • However, Faraday’s experimental discovery showed that a time-varying magnetic field induces an electromotive force (EMF) around a closed loop: $\text{EMF}={\oint }_C\vec{E}\cdot d\vec{l}=-\frac{d\Phi }{dt}$
   • Express the magnetic flux as $\Phi ={\int }_S\vec{B}\cdot d\vec{s}$: ${\oint }_C\vec{E}\cdot d\vec{l}=-\frac{d}{dt}{\int }_S\vec{B}\cdot d\vec{s}$
   • Assuming a stationary loop, pull the derivative inside the integral as a partial derivative: ${\oint }_C\vec{E}\cdot d\vec{l}=-{\int }_S\frac{\partial \vec{B}}{\partial t}\cdot d\vec{s}$
   • Apply Stokes’s Theorem to the left-hand side: ${\int }_S(\nabla \times \vec{E})\cdot d\vec{s}=-{\int }_S\frac{\partial \vec{B}}{\partial t}\cdot d\vec{s}$
   • Equating the integrands yields Faraday’s dynamic law: $\nabla \times \vec{E}=-\frac{\partial \vec{B}}{\partial t}\text{(Maxwell’s 3rd Equation)}$

4. Postulate IV (Ampere’s Circuital Law Modification):

   • In magnetostatics, Ampere’s law is $\nabla \times \vec{H}=\vec{J}$.
   • Apply the vector identity $\nabla \cdot (\nabla \times \vec{H})=0⟹\nabla \cdot \vec{J}=0$.
   • However, for time-varying fields, the Continuity Equation states that $\nabla \cdot \vec{J}=-\frac{\partial {\rho }_v}{\partial t}$.
   • To make Ampere’s law mathematically compatible, introduce displacement current density ${\vec{J}}_d$ such that $\nabla \times \vec{H}=\vec{J}+{\vec{J}}_d$.
   • Take the divergence of both sides: $\nabla \cdot (\nabla \times \vec{H})=\nabla \cdot \vec{J}+\nabla \cdot {\vec{J}}_d⟹0=-\frac{\partial {\rho }_v}{\partial t}+\nabla \cdot {\vec{J}}_d$ $\nabla \cdot {\vec{J}}_d=\frac{\partial {\rho }_v}{\partial t}$
   • Use Gauss’s Law (${\rho }_v=\nabla \cdot \vec{D}$): $\nabla \cdot {\vec{J}}_d=\frac{\partial }{\partial t}(\nabla \cdot \vec{D})=\nabla \cdot \left(\frac{\partial \vec{D}}{\partial t}\right)$
   • This yields ${\vec{J}}_d=\frac{\partial \vec{D}}{\partial t}$. Substitute this back to get: $\nabla \times \vec{H}=\vec{J}+\frac{\partial \vec{D}}{\partial t}\text{(Maxwell’s 4th Equation)}$

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📝 Verbatim PYQs & Solutions (Maxwell’s Equations)

Question 1 [PYQ: 2015, 2021, 2023, 2024, 2025 - 10 Marks]

Write down the differential and integral form of Maxwell’s equations with their physical significance.

Model Answer:

• Recreate the complete Master Table from Section 3.1 verbatim. Ensure all vector signs ($\vec{D},\vec{E},\vec{H},\vec{B},\vec{J}$) and differential surface/line elements ($d\vec{s},d\vec{l}$) are perfectly written.

Question 2 [PYQ: 2016, 2017 - 8 Marks]

Write down the differential form and integral form of the Maxwell’s equations and identify each equation with proper experimental law.

Model Answer:

• Recreate the differential and integral columns of the Master Table (Section 3.1) and clearly label the associated experimental law in a dedicated column:
  1. Gauss’s Law for Electricity.
  2. No isolated magnetic charge / Gauss’s Law for Magnetism.
  3. Faraday’s Law of Electromagnetic Induction.
  4. Ampere’s Circuital Law as modified by Maxwell.

Question 3 [PYQ: 2018, 2023 - 13 Marks]

Deduce Maxwell’s equations from the four fundamental governing equations of electrostatics and magnetostatics.

Model Answer:

• Follow the exact deduction procedure detailed in Section 3.2. Begin with the un-modified static equations, explain why Gauss’s Laws remain unchanged, and show step-by-step modifications of Faraday’s and Ampere’s Laws.

Question 4 [PYQ: 2019 - 10 Marks]

Derive Maxwell’s equation from fundamental electrostatic and magnetostatic expressions by incorporating Faraday’s law of electromagnetic induction and continuity equation.

Model Answer: (This is identical to Question 3 above. Focus heavily on Section 3.2, ensuring deep explanations for step 3 and step 4).

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4. Dynamic Boundary Conditions

4.1 Why dynamic boundary conditions are identical to static boundary conditions

A very common, highly conceptual A+ question asks: “Why are the boundary conditions for electromagnetic fields exactly the same as the boundary conditions for static electric and static magnetic fields?”

The Scaling Proof (As $\Delta h\to 0$):

Boundary conditions are derived directly from the integral form of Maxwell’s equations by constructing a hypothetical loop or pillbox across the interface of two media.

[Insert Figure: Interface boundary between Medium 1 and Medium 2. Show a thin rectangular loop abcd of width \Delta w and height \Delta h crossing the boundary. Also show a thin cylindrical pillbox of height \Delta h and surface area \Delta A crossing the boundary. Ref: 01 Static Electric Field PDF]

Let us analyze Faraday’s Law in integral form for a rectangular loop $abcd$: ${\oint }_C\vec{E}\cdot d\vec{l}=-{\int }_S\frac{\partial \vec{B}}{\partial t}\cdot d\vec{s}$

1. Let the loop have a tangential width $\Delta w$ along the interface and a normal height $\Delta h$ perpendicular to the interface.
2. The surface area of this loop is $S=\Delta w\cdot \Delta h$.
3. Now, we contract the loop to the boundary by taking the limit as the height approaches zero ($\Delta h\to 0$).
4. The surface area of the loop shrinks to zero ($S\to 0$).
5. Since the magnetic flux density $\vec{B}$ and its time rate of change $\frac{\partial \vec{B}}{\partial t}$ must remain physically finite in any real material, the magnetic flux term on the right-hand side of Faraday’s Law must go to zero: ${\lim }_{\Delta h\to 0}\left(-{\int }_S\frac{\partial \vec{B}}{\partial t}\cdot d\vec{s}\right)=0$
6. Therefore, the dynamic term vanishes, and the equation reduces to the static conservative form: ${\oint }_C\vec{E}\cdot d\vec{l}=0⟹E_{1t}=E_{2t}$

The same scaling proof applies to Ampere’s Law for the dynamic displacement current term: ${\lim }_{\Delta h\to 0}\left({\int }_S\frac{\partial \vec{D}}{\partial t}\cdot d\vec{s}\right)=0$ This forces Ampere’s Law to reduce to its magnetostatic form at the interface boundary: ${\oint }_C\vec{H}\cdot d\vec{l}=I_{enc}⟹H_{1t}-H_{2t}=J_{sn}$

Since the boundary conditions depend solely on instantaneous field values and local surface properties (surface charge/current densities), they remain completely unchanged in static and dynamic cases.

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4.2 Dynamic Boundary Conditions Equations & Derivations

1. Tangential Electric Field ($E_t$):

Applying $\oint \vec{E}\cdot d\vec{l}=0$ around loop $abcd$ as $\Delta h\to 0$: $E_{1t}\cdot \Delta w-E_{2t}\cdot \Delta w=0⟹E_{1t}=E_{2t}$ In vector form: ${\hat{a}}_{n2}\times ({\vec{E}}_1-{\vec{E}}_2)=0$

2. Normal Electric Flux Density ($D_n$):

Applying Gauss’s Law $\oint \vec{D}\cdot d\vec{s}=Q_{enc}$ on a thin pillbox as height $\Delta h\to 0$: $D_{1n}\cdot \Delta A-D_{2n}\cdot \Delta A={\rho }_s\cdot \Delta A⟹D_{1n}-D_{2n}={\rho }_s$ In vector form: ${\hat{a}}_{n2}\cdot ({\vec{D}}_1-{\vec{D}}_2)={\rho }_s$

3. Tangential Magnetic Field ($H_t$):

Applying Ampere’s Law $\oint \vec{H}\cdot d\vec{l}=I_{enc}$ around loop $abcd$ as $\Delta h\to 0$: $H_{1t}\cdot \Delta w-H_{2t}\cdot \Delta w=I_{enc}=J_s\cdot \Delta w⟹H_{1t}-H_{2t}=J_s$ In vector form: ${\hat{a}}_{n2}\times ({\vec{H}}_1-{\vec{H}}_2)={\vec{J}}_s$

4. Normal Magnetic Flux Density ($B_n$):

Applying Gauss’s Law for magnetism $\oint \vec{B}\cdot d\vec{s}=0$ on a thin pillbox as $\Delta h\to 0$: $B_{1n}\cdot \Delta A-B_{2n}\cdot \Delta A=0⟹B_{1n}=B_{2n}$ In vector form: ${\hat{a}}_{n2}\cdot ({\vec{B}}_1-{\vec{B}}_2)=0$

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4.3 Special Boundary Case Classifications

Case A: Interface between two Lossless Dielectric Media

A lossless dielectric has zero conductivity (${\sigma }_1={\sigma }_2=0$). Consequently, there are no free charges or currents residing on the surface (${\rho }_s=0$, ${\vec{J}}_s=0$).

• Tangential components: Continuous $E_{1t}=E_{2t}\text{and}{H}_{1t}=H_{2t}$
• Normal components: Continuous $D_{1n}=D_{2n}⟹{ϵ}_1{E}_{1n}={ϵ}_2{E}_{2n}$ $B_{1n}=B_{2n}⟹{\mu }_1{H}_{1n}={\mu }_2{H}_{2n}$

Case B: Interface between a Dielectric Medium (1) and a Perfect Conductor (2)

An ideal perfect conductor has infinite conductivity (${\sigma }_2\to \infty$). Inside a perfect conductor, all electromagnetic fields must vanish identically (${\vec{E}}_2=0,{\vec{H}}_2=0,{\vec{D}}_2=0,{\vec{B}}_2=0$). Substituting these zeros into the general equations yields:

• Tangential Electric Field: $E_{1t}=E_{2t}=0⟹E_{1t}=0$ (The tangential electric field on a conductor surface is zero).
• Normal Electric Flux Density: $D_{1n}-0={\rho }_s⟹D_{1n}={\rho }_s$ (All normal electric flux terminating on the conductor is equal to the surface charge density).
• Tangential Magnetic Field: ${\hat{a}}_{n2}\times {\vec{H}}_1={\vec{J}}_s⟹H_{1t}=J_s$ (A surface current sheet exists, equal to the tangential magnetic field).
• Normal Magnetic Flux Density: $B_{1n}=0$ (No magnetic flux can enter a perfect conductor).

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📝 Verbatim PYQs & Solutions (Boundary Conditions)

Question 1 [PYQ: 2020 - 10 Marks]

Why the boundary conditions for electromagnetic fields are same to the boundary conditions for static electric and static magnetic field.

Model Answer:

• Follow the exact scaling proof provided in Section 4.1.
• Provide the mathematical limits showing why the time-derivative terms ($\frac{\partial \vec{B}}{\partial t}$ and $\frac{\partial \vec{D}}{\partial t}$) vanish as the integration loop/pillbox height $\Delta h\to 0$.

Question 2 [PYQ: 2018, 2021, 2022, 2023, 2024 - 10 Marks]

Write down the boundary equations for both electric field vectors and magnetic field vectors. Hence derive (i) boundary conditions between two lossless media and (ii) boundary conditions between a dielectric media and perfect conductor.

Model Answer:

• Part 1: Write down the four general vector boundary equations from Section 4.2.
• Part 2 (i): Provide the derivations and simplified continuous relations for two lossless media ($\sigma =0⟹{\rho }_s=0,{\vec{J}}_s=0$) from Section 4.3 (Case A).
• Part 2 (ii): Provide the dielectric-perfect conductor boundary equations from Section 4.3 (Case B), explaining physically why ${\vec{E}}_2=0,{\vec{H}}_2=0$ inside the conductor.

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5. Poynting’s Theorem & Electromagnetic Power Flow

5.1 Physical Meaning of Poynting’s Theorem

Poynting’s theorem is a statement of the conservation of energy for electromagnetic fields. It states that the net electromagnetic power flowing out of a closed volume must equal the rate of decrease of stored electromagnetic energy inside that volume minus the ohmic power dissipated as heat due to material conductivity.

5.2 Step-by-Step Derivation of Poynting’s Theorem

We begin with Maxwell’s two time-varying curl equations:

1. $\nabla \times \vec{E}=-\frac{\partial \vec{B}}{\partial t}=-\mu \frac{\partial \vec{H}}{\partial t}$ (Equation 1)
2. $\nabla \times \vec{H}=\vec{J}+\frac{\partial \vec{D}}{\partial t}=\sigma \vec{E}+ϵ\frac{\partial \vec{E}}{\partial t}$ (Equation 2)

Now, we utilize the fundamental vector identity: $\nabla \cdot (\vec{E}\times \vec{H})=\vec{H}\cdot (\nabla \times \vec{E})-\vec{E}\cdot (\nabla \times \vec{H})$ (Equation 3)

1. Let us evaluate the first term on the right-hand side by taking the dot product of Equation 1 with $\vec{H}$: $\vec{H}\cdot (\nabla \times \vec{E})=\vec{H}\cdot \left(-\mu \frac{\partial \vec{H}}{\partial t}\right)=-\mu \vec{H}\cdot \frac{\partial \vec{H}}{\partial t}$ Since $\vec{H}\cdot \frac{\partial \vec{H}}{\partial t}=\frac{1}{2}\frac{\partial }{\partial t}(\vec{H}\cdot \vec{H})=\frac{1}{2}\frac{\partial H^2}{\partial t}$, we can write: $\vec{H}\cdot (\nabla \times \vec{E})=-\frac{1}{2}\mu \frac{\partial H^2}{\partial t}=-\frac{\partial }{\partial t}\left(\frac{1}{2}\mu H^2\right)$ (Equation 4)

2. Now evaluate the second term on the right-hand side by taking the dot product of Equation 2 with $\vec{E}$: $\vec{E}\cdot (\nabla \times \vec{H})=\vec{E}\cdot \left(\vec{J}+ϵ\frac{\partial \vec{E}}{\partial t}\right)=\vec{E}\cdot \vec{J}+ϵ\vec{E}\cdot \frac{\partial \vec{E}}{\partial t}$ Using the same calculus trick where $\vec{E}\cdot \frac{\partial \vec{E}}{\partial t}=\frac{1}{2}\frac{\partial E^2}{\partial t}$: $\vec{E}\cdot (\nabla \times \vec{H})=\vec{J}\cdot \vec{E}+\frac{\partial }{\partial t}\left(\frac{1}{2}ϵE^2\right)$ (Equation 5)

3. Substitute Equation 4 and Equation 5 back into the vector identity (Equation 3): $\nabla \cdot (\vec{E}\times \vec{H})=\left[-\frac{\partial }{\partial t}\left(\frac{1}{2}\mu H^2\right)\right]-\left[\vec{J}\cdot \vec{E}+\frac{\partial }{\partial t}\left(\frac{1}{2}ϵE^2\right)\right]$ $\nabla \cdot (\vec{E}\times \vec{H})=-\vec{J}\cdot \vec{E}-\frac{\partial }{\partial t}\left(\frac{1}{2}ϵE^2+\frac{1}{2}\mu H^2\right)$

4. Take the volume integral over a closed volume $V$: ${\int }_V\nabla \cdot (\vec{E}\times \vec{H})dv=-{\int }_V\vec{J}\cdot \vec{E}dv-\frac{\partial }{\partial t}{\int }_V\left(\frac{1}{2}ϵE^2+\frac{1}{2}\mu H^2\right)dv$

5. Apply the Divergence Theorem to the volume integral on the left-hand side to convert it into a closed surface integral: ${\oint }_S(\vec{E}\times \vec{H})\cdot d\vec{s}=-{\int }_V\vec{J}\cdot \vec{E}dv-\frac{\partial }{\partial t}{\int }_V\left(\frac{1}{2}ϵE^2+\frac{1}{2}\mu H^2\right)dv$

Rearranging the terms: ${\oint }_S(\vec{E}\times \vec{H})\cdot d\vec{s}=-\frac{\partial }{\partial t}{\int }_V\left(\frac{1}{2}ϵE^2+\frac{1}{2}\mu H^2\right)dv-{\int }_V\sigma E^{2}dv$

Physical Interpretation of each term:

• ${\oint }_S(\vec{E}\times \vec{H})\cdot d\vec{s}$: Total electromagnetic power flowing out of the closed volume through surface $S$ [W].
• $-\frac{\partial }{\partial t}{\int }_V\left(\frac{1}{2}ϵE^2+\frac{1}{2}\mu H^2\right)dv$: Rate of decrease of the electric and magnetic energy stored in the volume [W].
• ${\int }_V\sigma E^{2}dv$: Ohmic power dissipated as heat in the medium due to conduction current [W].

5.3 Defining Poynting Vectors

• Instantaneous Poynting Vector ($\vec{\mathcal{P}}$): Defines the power density vector at any instant: $\vec{\mathcal{P}}=\vec{E}\times \vec{H}\left[{\text{W/m}}^2\right]$
  • The direction of $\vec{\mathcal{P}}$ points in the direction of wave propagation.
• Time-Average Poynting Vector (${\vec{\mathcal{P}}}_{ave}$): Since fields in practice are sinusoidal (time-harmonic), we care about the average power flowing over a cycle: ${\vec{\mathcal{P}}}_{ave}=\frac{1}{2}\text{Re}\left\{{\vec{E}}_s\times {\vec{H}}_s^{\ast }\right\}\left[{\text{W/m}}^2\right]$ (where ${\vec{H}}_s^{\ast }$ is the complex conjugate of the magnetic field phasor).

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5.4 Verification of Poynting’s Theorem on a DC Conducting Wire

This is an outstanding, highly critical derivation often tested in A+ exams. You must verify Poynting’s Theorem for a solid conductor carrying a DC current.

[Insert Figure: A cylindrical wire of length L and radius b carrying DC current I. Show electric field E pointing axially along z-axis, magnetic field H wrapping circumferentially around the boundary surface, and Poynting Vector pointing radially inward. Ref: 01 Static Electric Field PDF]

Complete Derivation & Verification:

1. Consider a straight cylindrical wire of length $L$, radius $b$, and conductivity $\sigma$ carrying a steady DC current $I$ along the +z-axis.

2. The current density $\vec{J}$ inside the wire is uniform: $\vec{J}={\hat{a}}_z\frac{I}{\pi b^2}$

3. By Ohm’s law ($\vec{J}=\sigma \vec{E}$), the electric field inside and on the surface of the wire is: $\vec{E}={\hat{a}}_z\frac{I}{\sigma \pi b^2}$ We know the total resistance of the wire is $R=\frac{L}{\sigma \pi b^2}$. Thus, we can write: $\vec{E}={\hat{a}}_z\frac{IR}{L}$

4. According to Ampere’s Law, the magnetic field intensity $\vec{H}$ on the surface of the wire ($r=b$) is directed circumferentially: $\vec{H}={\hat{a}}_{\varphi }\frac{I}{2\pi b}$

5. Calculate the Poynting Vector $\vec{\mathcal{P}}$ on the surface of the wire: $\vec{\mathcal{P}}=\vec{E}\times \vec{H}=\left({\hat{a}}_z\frac{IR}{L}\right)\times \left({\hat{a}}_{\varphi }\frac{I}{2\pi b}\right)$ Using the cross product ${\hat{a}}_z\times {\hat{a}}_{\varphi }=-{\hat{a}}_r$: $\vec{\mathcal{P}}=-{\hat{a}}_r\frac{I^{2}R}{2\pi bL}$

6. The Physical Meaning of the Negative Sign: The negative sign $-{\hat{a}}_r$ indicates that the electromagnetic power is flowing radially inward from the surrounding space into the wire surface!

7. To find the total power $P$ entering the wire, integrate the Poynting vector over the closed surface area of the wire cylinder of length $L$: $P=-{\oint }_S\vec{\mathcal{P}}\cdot d\vec{s}=-{\int }_0^L{\int }_0^{2\pi }\left(-{\hat{a}}_r\frac{I^{2}R}{2\pi bL}\right)\cdot \left({\hat{a}}_{r}bd\varphi dz\right)$ $P=\frac{I^{2}R}{2\pi bL}{\int }_0^{L}dz{\int }_0^{2\pi }bd\varphi =\frac{I^{2}R}{2\pi bL}(L)(2\pi b)$ $P=I^{2}R$

8. Verification: The total power entering the wire through its surface is exactly equal to $I^{2}R$, which is the precise rate of ohmic dissipation (heat loss) inside the wire. Poynting’s Theorem is verified!

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📝 Verbatim PYQs & Solutions (Poynting’s Theorem)

Question 1 [PYQ: 2015, 2023 - 9/10 Marks]

State and explain Poynting’s theorem with necessary equations. Also denotes the pointing vector $\mathcal{P}$.

Model Answer:

• Statement: Write the physical statement of Poynting’s Theorem from Section 5.1.
• Equations: Provide the mathematical volume-integral Poynting equation from Section 5.2, explicitly defining each term.
• Vector notation: Define $\vec{\mathcal{P}}=\vec{E}\times \vec{H}$ in ${\text{W/m}}^2$.

Question 2 [PYQ: 2016, 2022, 2024 - 10/13 Marks]

Find the equation of the total power flowing in a closed surface due to electromagnetic waves at any instant.

Model Answer:

• This asks for the complete, step-by-step vector calculus derivation of Poynting’s theorem. Write out steps 1 through 5 from Section 5.2 in precise detail to secure full marks.

Question 3 [PYQ: 2017 - 8 Marks]

Find the Poynting vector on the surface of a long straight conducting wire (of radius b and conductivity $\sigma$) that carries a direct current I as shown in Fig. 8(b). Also verify Poynting’s theorem. [Fig 8(b) is a cylindrical wire sketch]

Model Answer:

• Perform the complete radial-inward integration derivation detailed in Section 5.4. Ensure you explicitly explain the physical meaning of the $-{\hat{a}}_r$ direction.

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⚡ A+ Exam Hacks & Common Pitfalls

1. The “Closed vs Open” Integral Trap:
   • Pitfall: Writing a closed circle on the surface integral of Poynting’s theorem when defining the power through a cross-section.
   • A+ Hack: Remember that the total power leaving a closed volume must use a closed surface integral (${\oint }_S$). However, if the question asks for the power crossing a flat, open surface (such as a aperture or a cross-section of a waveguide), you must use an open surface integral (${\int }_S$).
2. Vector cross products direction:
   • Pitfall: Writing $\vec{\mathcal{P}}=\vec{H}\times \vec{E}$.
   • A+ Hack: Order matters! Poynting vector is strictly $\vec{E}\times \vec{H}$. Swapping the order yields a negative sign, representing power propagation in the opposite direction.
3. The “Time-Harmonic Factor of 1/2”:
   • Pitfall: Forgetting the factor of $1/2$ when calculating time-average power using phasors.
   • A+ Hack: In real-time domain, the average of ${\cos }^2(\omega t)$ is $1/2$. Thus, the phasor formula is strictly ${\vec{\mathcal{P}}}_{ave}=\frac{1}{2}\text{Re}\{{\vec{E}}_s\times {\vec{H}}_s^{\ast }\}$. Missing this factor of 2 is a common error in numericals.