Practice 5: Maxwell's Equations

8 MCQ + 3 FRQ · Covers Maxwell's Four Equations, Displacement Current, EM Waves, and Unification of E&M

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Multiple Choice Questions

Q1. Which of Maxwell's equations states that there are no magnetic monopoles?

Correct! Gauss's Law for magnetism: \(\oint \vec{B} \cdot d\vec{A} = 0\). The net magnetic flux through any closed surface is zero — magnetic field lines form closed loops with no beginning or end.
Incorrect. The correct answer is (C). Gauss's Law for magnetism: \(\oint \vec{B} \cdot d\vec{A} = 0\) — no magnetic monopoles exist.

Q2. The displacement current term \(\mu_0\varepsilon_0\,d\Phi_E/dt\) in Ampere-Maxwell Law is needed because:

Correct! In a charging capacitor, Ampere's Law without displacement current gives \(B\) on one side of the plates but zero on the other (ambiguous). The displacement current \(\varepsilon_0\,d\Phi_E/dt\) resolves this inconsistency and ensures continuity of current.
Incorrect. The correct answer is (D). The displacement current resolves the inconsistency in Ampere's Law for circuits with capacitors — maintaining charge conservation.

Q3. An electromagnetic wave travels in vacuum. The E-field is given by \(E_y = E_0\sin(kx - \omega t)\). The B-field is:

Correct! In EM waves: \(\vec{E} \perp \vec{B} \perp \vec{v}\) (direction of propagation). \(\vec{E}\) is along \(y\), propagation is along \(x\), so \(\vec{B}\) must be along \(z\). Both are in phase: \(\sin(kx-\omega t)\) with \(E_0 = cB_0\).
Incorrect. The correct answer is (B). \(\vec{E} \perp \vec{B} \perp \hat{x}\) (propagation direction). \(\vec{E}\) along \(y\) → \(\vec{B}\) along \(z\). \(E_0/B_0 = c\).

Q4. The speed of electromagnetic waves in vacuum is \(c = 1/\sqrt{\mu_0\varepsilon_0}\). This result comes directly from:

Correct! Taking the curl of Faraday's Law and substituting from Ampere-Maxwell (or vice versa) yields the wave equation: \(\nabla^2\vec{E} = \mu_0\varepsilon_0\,\partial^2\vec{E}/\partial t^2\), giving wave speed \(v = 1/\sqrt{\mu_0\varepsilon_0} = c\).
Incorrect. The correct answer is (E). Combining Faraday's Law and Ampere-Maxwell Law gives the EM wave equation with \(c = 1/\sqrt{\mu_0\varepsilon_0}\).

Q5. A parallel-plate capacitor with circular plates of radius \(R = 5.0\ \text{cm}\) is being charged with current \(I = 2.0\ \text{A}\). The displacement current density between the plates is:

Correct! Displacement current \(I_d = I = 2.0\ \text{A}\). Area: \(A = \pi R^2 = \pi(0.05)^2 = 7.85\times10^{-3}\ \text{m}^2\). \(J_d = I_d/A = 2.0/(7.85\times10^{-3}) = 255\ \text{A/m}^2\).
Incorrect. The correct answer is (A) \(255\ \text{A/m}^2\). \(I_d = I = 2.0\ \text{A}\), \(J_d = I/A = 2.0/[\pi(0.05)^2] = 255\ \text{A/m}^2\).

Q6. In an EM wave, the Poynting vector \(\vec{S} = \frac{1}{\mu_0}\vec{E} \times \vec{B}\) represents:

Correct! Poynting vector \(\vec{S} = \vec{E} \times \vec{B}/\mu_0\) gives the instantaneous energy flux (W/m²) and points in the direction of energy propagation. For a sinusoidal wave, the time-averaged intensity is \(I = S_{\text{avg}} = \frac{1}{2}c\varepsilon_0 E_0^2\).
Incorrect. The correct answer is (B). \(\vec{S}\) is the energy flux vector — power per unit area propagating in direction of \(\vec{E} \times \vec{B}\).

Q7. Which of Maxwell's equations, in integral form, describes how a changing magnetic field produces an electric field?

Correct! Faraday's Law: \(\oint \vec{E} \cdot d\vec{\ell} = -\frac{d\Phi_B}{dt}\). A changing magnetic flux induces a circulating (non-conservative) electric field.
Incorrect. The correct answer is (D). Faraday's Law: \(\oint \vec{E} \cdot d\vec{\ell} = -d\Phi_B/dt\). Changing B → induced E.

Q8. An EM wave has peak electric field \(E_0 = 100\ \text{V/m}\). The average intensity (power per unit area) is:

Correct! \(I = S_{\text{avg}} = \frac{1}{2}c\varepsilon_0 E_0^2 = \frac{1}{2}(3\times10^8)(8.85\times10^{-12})(100)^2 = 13.3\ \text{W/m}^2\).
Incorrect. The correct answer is (C) \(13.3\ \text{W/m}^2\). \(I = \frac{1}{2}c\varepsilon_0 E_0^2 = 0.5(3\times10^8)(8.85\times10^{-12})(10^4) = 13.3\ \text{W/m}^2\).

Free Response Questions

FRQ

F1: Displacement Current in a Capacitor

A parallel-plate capacitor with circular plates of radius \(R = 0.10\ \text{m}\) and separation \(d = 2.0\ \text{mm}\) is connected to an AC source: \(V(t) = V_0\sin(\omega t)\) with \(V_0 = 100\ \text{V}\) and \(\omega = 500\ \text{rad/s}\).

  1. Find the conduction current \(I(t)\) in the wires.
  2. Calculate the displacement current between the plates and show it equals the conduction current.
  3. Use the Ampere-Maxwell Law to find \(B\) at \(r = 0.05\ \text{m}\) (between plates) as a function of time.
  4. Find the maximum value of this magnetic field.
Show Solution
(a): \(C = \varepsilon_0 A/d = \varepsilon_0\pi R^2/d = (8.85\times10^{-12})\pi(0.10)^2/(0.002) = 1.39\times10^{-10}\ \text{F}\). \(Q = CV = CV_0\sin(\omega t)\). \(I = dQ/dt = CV_0\omega\cos(\omega t) = (1.39\times10^{-10})(100)(500)\cos(500t) = 6.95\times10^{-6}\cos(500t)\ \text{A}\).
(b): \(E = V/d = (V_0/d)\sin(\omega t)\). \(\Phi_E = EA = (V_0/d)\pi R^2\sin(\omega t)\). \(I_d = \varepsilon_0\,d\Phi_E/dt = \varepsilon_0(V_0/d)\pi R^2\omega\cos(\omega t) = (\varepsilon_0\pi R^2/d)V_0\omega\cos(\omega t) = CV_0\omega\cos(\omega t) = I(t)\). ✓
(c): Ampere-Maxwell: \(\oint \vec{B} \cdot d\vec{\ell} = \mu_0 I_{d,\text{enc}}\). For \(r < R\): \(I_{d,\text{enc}} = I_d(\pi r^2)/(\pi R^2) = I_d(r^2/R^2)\). \(B(2\pi r) = \mu_0 I_d(r^2/R^2)\) → \(B = \frac{\mu_0 I_d r}{2\pi R^2} = \frac{\mu_0 CV_0\omega r}{2\pi R^2}\cos(\omega t)\).
(d): At \(r = 0.05\ \text{m}\): \(B_{\text{max}} = \frac{(4\pi\times10^{-7})(6.95\times10^{-6})(0.05)}{2\pi(0.10)^2} = 6.95\times10^{-12}\ \text{T}\).
FRQ

F2: Deriving the EM Wave Equation

Starting from Maxwell's equations in vacuum (no charges, no currents):

  1. Write all four Maxwell equations in vacuum.
  2. Take the curl of Faraday's Law and substitute from the Ampere-Maxwell Law to derive the wave equation for \(\vec{E}\).
  3. Identify the wave speed and show that \(c = 1/\sqrt{\mu_0\varepsilon_0}\).
  4. Explain why this result was historically significant — what did Maxwell predict?
Show Solution
(a) Vacuum Maxwell: (1) \(\nabla \cdot \vec{E} = 0\) (2) \(\nabla \cdot \vec{B} = 0\) (3) \(\nabla \times \vec{E} = -\partial\vec{B}/\partial t\) (4) \(\nabla \times \vec{B} = \mu_0\varepsilon_0\,\partial\vec{E}/\partial t\).
(b): \(\nabla \times (\nabla \times \vec{E}) = \nabla \times (-\partial\vec{B}/\partial t) = -\frac{\partial}{\partial t}(\nabla \times \vec{B}) = -\frac{\partial}{\partial t}(\mu_0\varepsilon_0\,\partial\vec{E}/\partial t) = -\mu_0\varepsilon_0\,\partial^2\vec{E}/\partial t^2\). Using vector identity and \(\nabla \cdot \vec{E} = 0\): \(\nabla \times (\nabla \times \vec{E}) = \nabla(\nabla \cdot \vec{E}) - \nabla^2\vec{E} = -\nabla^2\vec{E}\). So \(\nabla^2\vec{E} = \mu_0\varepsilon_0\,\partial^2\vec{E}/\partial t^2\).
(c): Wave equation: \(\partial^2\vec{E}/\partial x^2 = \mu_0\varepsilon_0\,\partial^2\vec{E}/\partial t^2\). Speed: \(v = 1/\sqrt{\mu_0\varepsilon_0} = 1/\sqrt{(4\pi\times10^{-7})(8.85\times10^{-12})} = 2.998\times10^8\ \text{m/s} = c\).
(d): Maxwell predicted that light is an electromagnetic wave — unifying optics with electricity and magnetism. The speed calculated from purely electromagnetic constants (\(\mu_0, \varepsilon_0\)) matched the measured speed of light, which was a monumental discovery.
FRQ

F3: EM Wave Properties

A plane EM wave in vacuum has \(\vec{E}(x,t) = 50\,\hat{y}\sin(2.0\times10^7 x - 6.0\times10^{15} t)\ \text{V/m}\).

  1. Determine the direction of propagation, wavelength, and frequency.
  2. Write the corresponding magnetic field \(\vec{B}(x,t)\).
  3. Calculate the Poynting vector \(\vec{S}(x,t)\) and the average intensity.
  4. What is the radiation pressure this wave would exert on a perfectly absorbing surface?
Show Solution
(a): Propagation: \(+x\) direction (sign of \(kx-\omega t\)). \(k = 2.0\times10^7\ \text{rad/m}\). \(\lambda = 2\pi/k = 2\pi/(2.0\times10^7) = 3.14\times10^{-7}\ \text{m} = 314\ \text{nm}\) (UV). \(f = \omega/(2\pi) = 6.0\times10^{15}/(2\pi) = 9.55\times10^{14}\ \text{Hz}\). Verify: \(c = \lambda f = (3.14\times10^{-7})(9.55\times10^{14}) = 3.0\times10^8\ \text{m/s}\).
(b): \(\vec{E} \times \vec{B}\) gives direction of propagation (\(+x\)). \(\vec{E}\) along \(y\) → \(\vec{B}\) along \(z\). \(B_0 = E_0/c = 50/(3\times10^8) = 1.67\times10^{-7}\ \text{T}\). \(\vec{B} = 1.67\times10^{-7}\,\hat{z}\sin(2.0\times10^7 x - 6.0\times10^{15} t)\ \text{T}\).
(c): \(\vec{S} = \frac{1}{\mu_0}\vec{E} \times \vec{B} = \frac{E_0 B_0}{\mu_0}\sin^2(kx - \omega t)\,\hat{x} = \frac{(50)(1.67\times10^{-7})}{4\pi\times10^{-7}}\sin^2(\ldots)\,\hat{x} = 6.63\sin^2(\ldots)\,\hat{x}\ \text{W/m}^2\). Average: \(I = S_{\text{avg}} = \frac{1}{2} \times 6.63 = 3.32\ \text{W/m}^2\).
(d): Perfect absorber: \(P_{\text{rad}} = I/c = 3.32/(3\times10^8) = 1.11\times10^{-8}\ \text{Pa}\). (Perfect reflector would be twice this.)

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