Maxwell's Equations

The four fundamental equations of electromagnetism

Maxwell's Equations

The Four Maxwell's Equations

1. Gauss's Law (Electric)

Integral form: ∮E⃗⋅dA⃗=Qencϵ0\oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\epsilon_0}

Differential form: ∇⋅E⃗=ρϵ0\nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0}

Electric field diverges from charges.

2. Gauss's Law for Magnetism

Integral form: ∮B⃗⋅dA⃗=0\oint \vec{B} \cdot d\vec{A} = 0

Differential form: ∇⋅B⃗=0\nabla \cdot \vec{B} = 0

No magnetic monopoles; magnetic field lines are closed loops.

3. Faraday's Law

Integral form: ∮E⃗⋅dl⃗=−dΦBdt\oint \vec{E} \cdot d\vec{l} = -\frac{d\Phi_B}{dt}

Differential form: ∇×E⃗=−∂B⃗∂t\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}

Changing magnetic field creates electric field (circulation).

4. Ampere-Maxwell Law

Integral form: ∮B⃗⋅dl⃗=μ0Ienc+μ0ϵ0dΦEdt\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc} + \mu_0\epsilon_0\frac{d\Phi_E}{dt}

Differential form: ∇×B⃗=μ0J⃗+μ0ϵ0∂E⃗∂t\nabla \times \vec{B} = \mu_0\vec{J} + \mu_0\epsilon_0\frac{\partial \vec{E}}{\partial t}

Current and changing electric field create magnetic field (circulation).

Displacement Current

Maxwell's addition to Ampere's law:

Id=ϵ0dΦEdtI_d = \epsilon_0\frac{d\Phi_E}{dt}

Displacement current density: J⃗d=ϵ0∂E⃗∂t\vec{J}_d = \epsilon_0\frac{\partial \vec{E}}{\partial t}

Why needed:

In charging capacitor, no conduction current between plates, but changing E⃗\vec{E} creates magnetic field as if current flowed.

Ensures ∇⋅J⃗+∂ρ∂t=0\nabla \cdot \vec{J} + \frac{\partial \rho}{\partial t} = 0 (charge conservation).

Symmetry in Maxwell's Equations

In vacuum (ρ=0\rho = 0, J⃗=0\vec{J} = 0):

∇⋅E⃗=0,∇⋅B⃗=0\nabla \cdot \vec{E} = 0, \quad \nabla \cdot \vec{B} = 0

∇×E⃗=−∂B⃗∂t,∇×B⃗=μ0ϵ0∂E⃗∂t\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}, \quad \nabla \times \vec{B} = \mu_0\epsilon_0\frac{\partial \vec{E}}{\partial t}

Nearly symmetric! Differences:

  • No magnetic monopoles (no ρm\rho_m)
  • Factor μ0ϵ0\mu_0\epsilon_0 in Ampere-Maxwell

Poynting Vector

Energy flux in electromagnetic field:

S⃗=1μ0E⃗×B⃗\vec{S} = \frac{1}{\mu_0}\vec{E} \times \vec{B}

Units: W/m² (power per area)

Direction: Direction of energy propagation

Intensity: I=∣S⃗∣avgI = |\vec{S}|_{avg}

Energy Density

Electric field: uE=12ϵ0E2u_E = \frac{1}{2}\epsilon_0 E^2

Magnetic field: uB=12μ0B2u_B = \frac{1}{2\mu_0}B^2

Total: u=uE+uB=12ϵ0E2+12μ0B2u = u_E + u_B = \frac{1}{2}\epsilon_0 E^2 + \frac{1}{2\mu_0}B^2

Continuity Equation

Energy conservation:

∂u∂t+∇⋅S⃗=−J⃗⋅E⃗\frac{\partial u}{\partial t} + \nabla \cdot \vec{S} = -\vec{J} \cdot \vec{E}

Rate of energy change + energy flux out = work done by field on charges.

Historical Significance

Maxwell's equations unified electricity and magnetism, predicted electromagnetic waves, led to:

  • Radio
  • Radar
  • Modern telecommunications
  • Understanding of light as electromagnetic wave

📚 Practice Problems

1Problem 1hard

❓ Question:

A parallel-plate capacitor with circular plates (radius R = 0.05 m, separation d = 2.0 mm) is being charged. At a certain instant, the electric field between the plates is increasing at rate dE/dt = 1.0 × 10¹² V/(m·s). Find: (a) the displacement current between the plates, (b) the magnetic field at radius r = 0.03 m between the plates, and (c) verify this satisfies Maxwell-Ampère law.

💡 Show Solution

Given:

  • R = 0.05 m
  • d = 2.0 mm = 0.002 m
  • dE/dt = 1.0 × 10¹² V/(m·s)
  • r = 0.03 m
  • ε₀ = 8.85 × 10⁻¹² F/m
  • μ₀ = 4π × 10⁻⁷ T·m/A

(a) Displacement current:

Id=ε0dΦEdt=ε0AdEdtI_d = \varepsilon_0 \frac{d\Phi_E}{dt} = \varepsilon_0 A \frac{dE}{dt}

where A=πR2=π(0.05)2=7.85×10−3A = \pi R^2 = \pi(0.05)^2 = 7.85 \times 10^{-3} m²

Id=(8.85×10−12)(7.85×10−3)(1.0×1012)I_d = (8.85 \times 10^{-12})(7.85 \times 10^{-3})(1.0 \times 10^{12})

Id=69.5 AI_d = \boxed{69.5 \text{ A}}

(b) Magnetic field at r = 0.03 m:

Using Maxwell-Ampère law with cylindrical symmetry: ∮B⃗⋅dl⃗=μ0Id,enc\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{d,enc}

Displacement current through circle of radius r: Id,enc=Idπr2πR2=Idr2R2I_{d,enc} = I_d \frac{\pi r^2}{\pi R^2} = I_d \frac{r^2}{R^2}

B(2πr)=μ0Idr2R2B(2\pi r) = \mu_0 I_d \frac{r^2}{R^2}

B=μ0Idr2πR2B = \frac{\mu_0 I_d r}{2\pi R^2}

B=(4π×10−7)(69.5)(0.03)2π(0.05)2B = \frac{(4\pi \times 10^{-7})(69.5)(0.03)}{2\pi(0.05)^2}

B=(2×10−7)(69.5)(0.03)(0.05)2B = \frac{(2 \times 10^{-7})(69.5)(0.03)}{(0.05)^2}

B=1.67×10−4 T=0.167 mTB = \boxed{1.67 \times 10^{-4} \text{ T} = 0.167 \text{ mT}}

(c) Verify Maxwell-Ampère:

∮B⃗⋅dl⃗=B(2πr)=(1.67×10−4)(2π)(0.03)\oint \vec{B} \cdot d\vec{l} = B(2\pi r) = (1.67 \times 10^{-4})(2\pi)(0.03)

=3.14×10−5 T\cdotpm= 3.14 \times 10^{-5} \text{ T·m}

μ0Id,enc=(4π×10−7)(69.5)(0.03)2(0.05)2\mu_0 I_{d,enc} = (4\pi \times 10^{-7})(69.5)\frac{(0.03)^2}{(0.05)^2}

=(4π×10−7)(69.5)(0.36)=3.14×10−5 T\cdotpm= (4\pi \times 10^{-7})(69.5)(0.36) = 3.14 \times 10^{-5} \text{ T·m} ✓

The circulating magnetic field is produced by the changing electric flux!

2Problem 2hard

❓ Question:

A parallel-plate capacitor with circular plates (radius R = 0.05 m, separation d = 2.0 mm) is being charged. At a certain instant, the electric field between the plates is increasing at rate dE/dt = 1.0 × 10¹² V/(m·s). Find: (a) the displacement current between the plates, (b) the magnetic field at radius r = 0.03 m between the plates, and (c) verify this satisfies Maxwell-Ampère law.

💡 Show Solution

Given:

  • R = 0.05 m
  • d = 2.0 mm = 0.002 m
  • dE/dt = 1.0 × 10¹² V/(m·s)
  • r = 0.03 m
  • ε₀ = 8.85 × 10⁻¹² F/m
  • μ₀ = 4π × 10⁻⁷ T·m/A

(a) Displacement current:

Id=ε0dΦEdt=ε0AdEdtI_d = \varepsilon_0 \frac{d\Phi_E}{dt} = \varepsilon_0 A \frac{dE}{dt}

where A=πR2=π(0.05)2=7.85×10−3A = \pi R^2 = \pi(0.05)^2 = 7.85 \times 10^{-3} m²

Id=(8.85×10−12)(7.85×10−3)(1.0×1012)I_d = (8.85 \times 10^{-12})(7.85 \times 10^{-3})(1.0 \times 10^{12})

Id=69.5 AI_d = \boxed{69.5 \text{ A}}

(b) Magnetic field at r = 0.03 m:

Using Maxwell-Ampère law with cylindrical symmetry: ∮B⃗⋅dl⃗=μ0Id,enc\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{d,enc}

Displacement current through circle of radius r: Id,enc=Idπr2πR2=Idr2R2I_{d,enc} = I_d \frac{\pi r^2}{\pi R^2} = I_d \frac{r^2}{R^2}

B(2πr)=μ0Idr2R2B(2\pi r) = \mu_0 I_d \frac{r^2}{R^2}

B=μ0Idr2πR2B = \frac{\mu_0 I_d r}{2\pi R^2}

B=(4π×10−7)(69.5)(0.03)2π(0.05)2B = \frac{(4\pi \times 10^{-7})(69.5)(0.03)}{2\pi(0.05)^2}

B=(2×10−7)(69.5)(0.03)(0.05)2B = \frac{(2 \times 10^{-7})(69.5)(0.03)}{(0.05)^2}

B=1.67×10−4 T=0.167 mTB = \boxed{1.67 \times 10^{-4} \text{ T} = 0.167 \text{ mT}}

(c) Verify Maxwell-Ampère:

∮B⃗⋅dl⃗=B(2πr)=(1.67×10−4)(2π)(0.03)\oint \vec{B} \cdot d\vec{l} = B(2\pi r) = (1.67 \times 10^{-4})(2\pi)(0.03)

=3.14×10−5 T\cdotpm= 3.14 \times 10^{-5} \text{ T·m}

μ0Id,enc=(4π×10−7)(69.5)(0.03)2(0.05)2\mu_0 I_{d,enc} = (4\pi \times 10^{-7})(69.5)\frac{(0.03)^2}{(0.05)^2}

=(4π×10−7)(69.5)(0.36)=3.14×10−5 T\cdotpm= (4\pi \times 10^{-7})(69.5)(0.36) = 3.14 \times 10^{-5} \text{ T·m} ✓

The circulating magnetic field is produced by the changing electric flux!

3Problem 3hard

❓ Question:

State all four Maxwell's equations in both integral and differential forms. Then explain: (a) which equation shows that magnetic monopoles don't exist, (b) which equation was modified by Maxwell's displacement current, and (c) how these equations predict electromagnetic waves.

💡 Show Solution

Maxwell's Equations:

1. Gauss's Law (Electric):

Integral: ∮E⃗⋅dA⃗=Qencε0\oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\varepsilon_0}

Differential: ∇⋅E⃗=ρε0\nabla \cdot \vec{E} = \frac{\rho}{\varepsilon_0}

2. Gauss's Law (Magnetic):

Integral: ∮B⃗⋅dA⃗=0\oint \vec{B} \cdot d\vec{A} = 0

Differential: ∇⋅B⃗=0\nabla \cdot \vec{B} = 0

3. Faraday's Law:

Integral: ∮E⃗⋅dl⃗=−dΦBdt\oint \vec{E} \cdot d\vec{l} = -\frac{d\Phi_B}{dt}

Differential: ∇×E⃗=−∂B⃗∂t\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}

4. Maxwell-Ampère Law:

Integral: ∮B⃗⋅dl⃗=μ0I+μ0ε0dΦEdt\oint \vec{B} \cdot d\vec{l} = \mu_0 I + \mu_0 \varepsilon_0 \frac{d\Phi_E}{dt}

Differential: ∇×B⃗=μ0J⃗+μ0ε0∂E⃗∂t\nabla \times \vec{B} = \mu_0 \vec{J} + \mu_0 \varepsilon_0 \frac{\partial \vec{E}}{\partial t}

(a) No magnetic monopoles:

Gauss's Law for Magnetism (∇⋅B⃗=0\nabla \cdot \vec{B} = 0)

This says magnetic field lines form closed loops - they never start or end. Electric field lines start on + charges and end on - charges, but there are no magnetic "charges" (monopoles).

(b) Maxwell's modification:

Ampère's Law → Maxwell-Ampère Law

Maxwell added the displacement current term: Id=ε0dΦEdtI_d = \varepsilon_0 \frac{d\Phi_E}{dt}

This was needed for consistency - changing electric fields produce magnetic fields, just as changing magnetic fields (Faraday) produce electric fields. Symmetry!

(c) Electromagnetic waves:

Take curl of Faraday's law: ∇×(∇×E⃗)=−∂∂t(∇×B⃗)\nabla \times (\nabla \times \vec{E}) = -\frac{\partial}{\partial t}(\nabla \times \vec{B})

Using Maxwell-Ampère (in vacuum, J = 0): ∇×(∇×E⃗)=−μ0ε0∂2E⃗∂t2\nabla \times (\nabla \times \vec{E}) = -\mu_0 \varepsilon_0 \frac{\partial^2 \vec{E}}{\partial t^2}

Using vector identity and ∇⋅E⃗=0\nabla \cdot \vec{E} = 0: ∇2E⃗=μ0ε0∂2E⃗∂t2\nabla^2 \vec{E} = \mu_0 \varepsilon_0 \frac{\partial^2 \vec{E}}{\partial t^2}

This is the wave equation! Wave speed: c=1μ0ε0=3.0×108 m/sc = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} = 3.0 \times 10^8 \text{ m/s}

Similarly for B. Maxwell predicted light is EM waves!

4Problem 4hard

❓ Question:

State all four Maxwell's equations in both integral and differential forms. Then explain: (a) which equation shows that magnetic monopoles don't exist, (b) which equation was modified by Maxwell's displacement current, and (c) how these equations predict electromagnetic waves.

💡 Show Solution

Maxwell's Equations:

1. Gauss's Law (Electric):

Integral: ∮E⃗⋅dA⃗=Qencε0\oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\varepsilon_0}

Differential: ∇⋅E⃗=ρε0\nabla \cdot \vec{E} = \frac{\rho}{\varepsilon_0}

2. Gauss's Law (Magnetic):

Integral: ∮B⃗⋅dA⃗=0\oint \vec{B} \cdot d\vec{A} = 0

Differential: ∇⋅B⃗=0\nabla \cdot \vec{B} = 0

3. Faraday's Law:

Integral: ∮E⃗⋅dl⃗=−dΦBdt\oint \vec{E} \cdot d\vec{l} = -\frac{d\Phi_B}{dt}

Differential: ∇×E⃗=−∂B⃗∂t\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}

4. Maxwell-Ampère Law:

Integral: ∮B⃗⋅dl⃗=μ0I+μ0ε0dΦEdt\oint \vec{B} \cdot d\vec{l} = \mu_0 I + \mu_0 \varepsilon_0 \frac{d\Phi_E}{dt}

Differential: ∇×B⃗=μ0J⃗+μ0ε0∂E⃗∂t\nabla \times \vec{B} = \mu_0 \vec{J} + \mu_0 \varepsilon_0 \frac{\partial \vec{E}}{\partial t}

(a) No magnetic monopoles:

Gauss's Law for Magnetism (∇⋅B⃗=0\nabla \cdot \vec{B} = 0)

This says magnetic field lines form closed loops - they never start or end. Electric field lines start on + charges and end on - charges, but there are no magnetic "charges" (monopoles).

(b) Maxwell's modification:

Ampère's Law → Maxwell-Ampère Law

Maxwell added the displacement current term: Id=ε0dΦEdtI_d = \varepsilon_0 \frac{d\Phi_E}{dt}

This was needed for consistency - changing electric fields produce magnetic fields, just as changing magnetic fields (Faraday) produce electric fields. Symmetry!

(c) Electromagnetic waves:

Take curl of Faraday's law: ∇×(∇×E⃗)=−∂∂t(∇×B⃗)\nabla \times (\nabla \times \vec{E}) = -\frac{\partial}{\partial t}(\nabla \times \vec{B})

Using Maxwell-Ampère (in vacuum, J = 0): ∇×(∇×E⃗)=−μ0ε0∂2E⃗∂t2\nabla \times (\nabla \times \vec{E}) = -\mu_0 \varepsilon_0 \frac{\partial^2 \vec{E}}{\partial t^2}

Using vector identity and ∇⋅E⃗=0\nabla \cdot \vec{E} = 0: ∇2E⃗=μ0ε0∂2E⃗∂t2\nabla^2 \vec{E} = \mu_0 \varepsilon_0 \frac{\partial^2 \vec{E}}{\partial t^2}

This is the wave equation! Wave speed: c=1μ0ε0=3.0×108 m/sc = \frac{1}{\sqrt{\mu_0 \varepsilon_0}} = 3.0 \times 10^8 \text{ m/s}

Similarly for B. Maxwell predicted light is EM waves!

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