Changing E & B Fields - Physics 2

1. Which of the following correctly describes what Maxwell's equations predict about this configuration?
   Answer: This field configuration cannot represent a free electromagnetic wave because a spatially uniform, time-varying electric field requires a spatially varying magnetic field (via Faraday's law), yet a spatially uniform $\vec{E}$ with no spatial variation cannot sustain the necessary curl of $\vec{B}$ via Ampere's law in a self-consistent traveling-wave solution. When you see a question about electromagnetic fields and Maxwell's equations, your first instinct should be to check self-consistency: does the proposed field simultaneously satisfy all four Maxwell equations, not just one or two? For a spatially uniform electric field $\vec{E}(t) = E_0\sin(\omega t)\hat{x}$, start with Faraday's law: $\nabla \times \vec{B} = -\mu_0\epsilon_0\frac{\partial \vec{E}}{\partial t}$... wait — actually Faraday's law states $\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}$. Since $\vec{E}$ has no spatial variation, $\nabla \times \vec{E} = 0$, which means $\frac{\partial \vec{B}}{\partial t} = 0$, so $\vec{B}$ must be static or zero. But then Ampere's law requires $\nabla \times \vec{B} = \mu_0\epsilon_0\frac{\partial \vec{E}}{\partial t} \neq 0$. A static or zero $\vec{B}$ cannot have a nonzero curl — contradiction. The configuration cannot sustain itself, confirming B is correct. A is wrong because it assumes a time-varying $\vec{E}$ automatically produces a self-consistent wave — it ignores that spatial variation is required for a traveling wave solution; without it, the equations are contradictory, not wave-like. C is wrong because Maxwell's equations absolutely permit time-varying electric fields without free charges (displacement current is the whole point of Ampere's modification) — the issue here is spatial uniformity, not the presence of charges. D is wrong because there is no conduction current in free space to cancel the displacement current — this "cancellation" scenario is fabricated and physically meaningless here. Study tip: When evaluating field configurations, always check all relevant Maxwell equations together. Self-consistency requires satisfying Faraday's and Ampere's law simultaneously — passing one doesn't guarantee passing both.
2. In a region of free space, a magnetic field is observed to be changing at a rate $\partial \vec{B}/\partial t \neq 0$. A student argues: 'This changing magnetic field causes an electric field, and that induced electric field then causes the magnetic field to change further, so the two fields take turns causing each other — this is why electromagnetic waves propagate.' Which of the following best identifies the flaw (if any) in this causal description?
   Answer: The description is misleading: Maxwell's equations are simultaneous coupled partial differential equations, so $\vec{E}$ and $\vec{B}$ are not causally sequential but rather mutually consistent solutions to those equations; propagation arises from the spatial structure (curl) required by both laws together, not from a temporal cause-and-effect relay. When you see a question about electromagnetic wave propagation, resist the intuitive but misleading picture of fields "taking turns" causing each other like a relay race. The real question is whether Maxwell's equations describe a temporal sequence or a simultaneous constraint. Faraday's law ($\nabla \times \vec{E} = -\partial \vec{B}/\partial t$) and the Ampere-Maxwell law ($\nabla \times \vec{B} = \mu_0\epsilon_0\,\partial \vec{E}/\partial t$) are coupled partial differential equations that must be satisfied at every point, at every instant, simultaneously. There is no moment where $\vec{B}$ changes first, "waits," and then $\vec{E}$ responds. Both fields are part of a single self-consistent solution. What actually drives propagation is the spatial curl structure — the way each field's circulation in space links to the other field's time variation — not a sequential handoff. This makes C the correct answer. A is wrong because it treats Maxwell's equations as describing a strict temporal sequence, which they do not. The equations impose simultaneous consistency, not a step-by-step causal chain. B introduces a fictitious mechanism. In free space there are no currents at all — that's the whole point of the displacement current term. Changing $\vec{B}$ induces a curling $\vec{E}$, not a conduction current. D is wrong for two reasons: it reverses no meaningful sequence (there is none), and it incorrectly treats $\vec{B}$ as strictly secondary with no feedback — exactly the flawed sequential thinking the question warns against. Study tip: On physics-2 exams, any answer claiming Maxwell's equations imply a before/after causal order between $\vec{E}$ and $\vec{B}$ is almost certainly wrong — the equations are simultaneous, not sequential.
3. How does the peak displacement current through Capacitor 2 compare to the peak displacement current through Capacitor 1?
   Answer: The peak displacement current through Capacitor 2 is twice that through Capacitor 1, because displacement current is proportional to $\partial E/\partial t$, which scales with frequency, so doubling the frequency doubles the peak rate of change of $\vec{E}$. When you see a question about displacement current in capacitors driven at different frequencies, your anchor should be Maxwell's displacement current formula: $I_d = \epsilon_0 \frac{\partial \Phi_E}{\partial t} = \epsilon_0 A \frac{\partial E}{\partial t}$. The key insight is that displacement current depends on the rate of change of the electric field, not its peak value. If the voltage across the capacitor varies as $V(t) = V_0 \sin(2\pi f t)$, then the electric field between the plates is $E(t) = \frac{V_0}{d}\sin(2\pi f t)$. Taking the time derivative gives $\frac{\partial E}{\partial t} = \frac{2\pi f V_0}{d}\cos(2\pi f t)$. The peak value of this derivative is $\frac{2\pi f V_0}{d}$, which scales linearly with frequency. Doubling the frequency doubles the peak $\partial E/\partial t$, and therefore doubles the peak displacement current. That confirms B is correct. Choice A is wrong because it conflates the peak electric field (which is the same for both capacitors) with the peak rate of change of the field — those are very different quantities. C is wrong because it invents an incorrect energy-based argument; displacement current has no $f^2$ or square-root dependence. D gets the reactance concept partially right — yes, $X_C$ decreases with frequency — but then misapplies it. Lower reactance means more current flows, not less, so the reasoning contradicts itself and leads to the wrong conclusion. Your study tip: whenever displacement current appears, immediately write $I_d \propto \partial E/\partial t$ — that derivative is where frequency enters, and it's the most commonly missed step on this type of question.
4. In this modified universe, which of the following would be the most significant consequence for the propagation of electromagnetic disturbances?
   Answer: No self-sustaining electromagnetic wave could propagate in vacuum: without the displacement current, a changing $\vec{B}$ could induce $\vec{E}$ (via Faraday's law), but that induced $\vec{E}$ could not in turn regenerate $\vec{B}$ (since changing $\vec{E}$ no longer drives $\vec{B}$), breaking the mutual sustaining mechanism; any disturbance would decay rather than propagate. Whenever you see a question about electromagnetic wave propagation, think about the mutual feedback loop that sustains a wave: a changing $\vec{B}$ induces $\vec{E}$ (Faraday's law), and a changing $\vec{E}$ regenerates $\vec{B}$ (Maxwell's displacement current term). Remove either link, and the chain breaks. In this modified universe, Maxwell's correction — the displacement current $\mu_0\epsilon_0 \frac{\partial \vec{E}}{\partial t}$ — is gone. A changing $\vec{B}$ can still induce $\vec{E}$ via Faraday's law, but that induced $\vec{E}$, even if it oscillates, produces **no corresponding $\vec{B}$. The feedback loop is severed after one step. Without that regeneration, the disturbance cannot sustain itself across space — it simply decays. This is exactly what C** describes, making it the correct answer. A is tempting because conduction currents in wires do drive $\vec{B}$ through ordinary Ampere's law — but the question concerns propagation of the disturbance itself. Even in conductors, a traveling wave requires the same mutual sustaining mechanism; wire transmission doesn't bypass the fundamental coupling problem. B is physically flawed. Faraday's law doesn't allow $\vec{B}$ to "self-induce" — it explicitly requires a changing $\vec{B}$ to produce $\vec{E}$, not to sustain $\vec{B}$ itself. A pure magnetic wave with no coupled $\vec{E}$ has no mechanism to persist. D contains a real mathematical observation — removing the second time derivative does yield a Laplace-type equation — but "infinite speed" misrepresents the physics. Solutions would be instantaneous field configurations, not propagating waves at all. Your study tip: always trace the two-step feedback loop ($\vec{B} \rightarrow \vec{E} \rightarrow \vec{B}$) when analyzing EM wave questions. If either arrow is broken, no vacuum propagation is possible.
5. Which physicist's position is more nearly correct, and what is the key physical reasoning?
   Answer: Physicist B is more nearly correct: without the displacement current, Ampere's law reads $\nabla \times \vec{B} = \mu_0 \vec{J}$; in vacuum ($\vec{J} = 0$), this forces $\nabla \times \vec{B} = 0$ at all times, preventing the spatial variation in $\vec{B}$ needed to drive a changing $\vec{E}$ via the wave equation. Combined with Faraday's law alone, one obtains $\nabla^2 \vec{E} = 0$ rather than a wave equation, so no wave propagation is possible. When tackling questions about Maxwell's equations, your anchor should always be: what does each equation actually constrain? Write out the modified (incomplete) system and ask what it forces on the fields. Without the displacement current, Ampere's law in vacuum becomes $\nabla \times \vec{B} = \mu_0 \vec{J} = 0$. This is the critical constraint — it says the curl of $\vec{B}$ is identically zero everywhere in vacuum at all times. Now apply Faraday's law: $\nabla \times \vec{E} = -\partial \vec{B}/\partial t$. Take the curl of both sides and use the vector identity $\nabla \times (\nabla \times \vec{E}) = \nabla(\nabla \cdot \vec{E}) - \nabla^2 \vec{E}$. In vacuum, $\nabla \cdot \vec{E} = 0$, so you get $-\nabla^2 \vec{E} = -\partial(\nabla \times \vec{B})/\partial t = 0$. The result is $\nabla^2 \vec{E} = 0$ — Laplace's equation, not a wave equation. No oscillatory propagating solutions exist. Physicist B is correct, and D captures exactly this reasoning. Choice A fails because restricting to charge-free regions doesn't rescue the situation — the constraint $\nabla \times \vec{B} = 0$ already applies in vacuum, and that's precisely what kills the waves. Choice B contains a subtle but fatal error: you cannot derive a wave equation for $\vec{B}$ from Faraday's law alone plus the divergence equations. You'd need to take the curl of Faraday's law, which reintroduces $\nabla \times \vec{B}$ — and without displacement current, that term vanishes. Choice C is a distractor built on a misconception. The issue isn't that $\epsilon_0$ disappears from the math; it's that the field-coupling structure collapses entirely. Study tip: When evaluating modified Maxwell's equations, always substitute the constraint back into the curl-of-curl derivation step by step — that's where wave equations live or die.
6. Maxwell added the displacement current term $\mu_0 \epsilon_0 \frac{\partial \vec{E}}{\partial t}$ to Ampere's law. Consider a long solenoid whose current is increasing at a constant rate $dI/dt = k$ (constant). Inside the solenoid, $\vec{B}$ is increasing uniformly. Which of the following statements about the fields inside the solenoid is correct, and demonstrates the relationship between changing fields in this context?
   Answer: The increasing $\vec{B}$ induces a circular $\vec{E}$ field inside the solenoid via Faraday's law. Since $\vec{E}$ is time-independent ($dI/dt$ is constant, so $d^2B/dt^2 = 0$), the displacement current $\epsilon_0 \partial \vec{E}/\partial t = 0$ inside the solenoid, and Ampere's law is satisfied by the conduction current alone without any additional magnetic field contribution from the displacement current. When you see a question combining Faraday's law and Maxwell's displacement current, your job is to carefully track what is changing and at what rate — don't just assume "changing fields always cascade." Here's the chain of reasoning for a solenoid with constant $dI/dt = k$. Since current increases linearly, $\vec{B}$ inside increases linearly: $\vec{B} = \mu_0 n k t$. By Faraday's law, this changing $\vec{B}$ induces a circular electric field $\vec{E}$ inside the solenoid. You can show $E \propto r \cdot (d\vec{B}/dt)$, which is constant in time because $d\vec{B}/dt = \mu_0 n k = \text{constant}$. Now apply Maxwell's displacement current: $\epsilon_0 \frac{\partial \vec{E}}{\partial t} = 0$, since $\vec{E}$ itself is not changing. Therefore, no displacement current exists inside the solenoid, and Ampere's law is fully satisfied by the conduction current alone. Answer A captures this precisely. Answer B is wrong because it claims this situation creates a "self-reinforcing field enhancement." The displacement current is zero inside, so no additional $\vec{B}$ is generated — there's no runaway feedback loop here. Answer C contains the critical misconception: it assumes $\vec{E}$ is changing in time because $\vec{B}$ is changing. But a uniformly changing $\vec{B}$ produces a constant $\vec{E}$, so $\partial \vec{E}/\partial t = 0$. You need the second time derivative of $\vec{B}$ to be nonzero for displacement current to appear. Answer D is simply false — Faraday's law guarantees an induced $\vec{E}$ inside any region of changing flux, regardless of symmetry. Study tip: Always ask "what is the time derivative of $\vec{E}$, not just $\vec{E}$ itself" when evaluating displacement current — constant rate of change means zero displacement current.
7. Near the antenna (in the near-field zone), there exist electric and magnetic fields that fall off faster than $1/r$. A student claims: 'These near-field components, even though they don't carry energy to infinity, still satisfy Maxwell's equations including the requirement that changing $\vec{E}$ fields produce $\vec{B}$ fields and vice versa.' Which of the following most precisely evaluates this claim?
   Answer: The claim is correct: near-field components are full solutions to Maxwell's equations, and the mutual relationship between $\partial \vec{E}/\partial t$ and $\vec{B}$ (and $\partial \vec{B}/\partial t$ and $\vec{E}$) holds in the near zone just as in the far zone, even though the phase relationship and radial dependence differ from radiation fields. Whenever you see a question about near-field versus far-field electromagnetic behavior, the key concept to anchor on is this: Maxwell's equations are universal. They don't have a "near-field version" and a "far-field version" — the same four equations govern all electromagnetic phenomena at every point in space. The complete solution for a radiating dipole antenna contains multiple terms with different radial dependencies ($1/r$, $1/r^2$, $1/r^3$). These terms coexist and together form one self-consistent solution. The near-field terms — which dominate close to the source and decay faster than $1/r$ — are not approximations bolted onto the solution; they are genuine components of the full solution. Because they oscillate in time (the source is oscillating at $\omega$), they absolutely have nonzero $\partial\vec{E}/\partial t$ and $\partial\vec{B}/\partial t$. Ampere's law and Faraday's law therefore link them, just as in the far field. The distinction is that near-field energy sloshes back and forth reactively rather than propagating outward — but Maxwell's equations still govern this behavior completely. Answer A captures this precisely. Answer B is wrong because it creates a false division: Maxwell's equations are not restricted to radiation fields. Near-field terms are not quasi-static unless you're making a deliberate approximation. Answer C is wrong because Faraday's law in differential form — $\nabla \times \vec{E} = -\partial\vec{B}/\partial t$ — requires no physical circuit. It applies everywhere in free space. Answer D is wrong because the near-field components of an oscillating source do vary in time; they are not static Coulomb or Biot-Savart fields. Study tip: When a question asks whether a physical law "applies" in some region, your default should be yes — fundamental laws like Maxwell's equations hold everywhere unless an explicit approximation is stated.
8. An electromagnetic wave propagates in the $+\hat{z}$ direction in vacuum. At a particular point in space and time, the electric field is $\vec{E} = E_0 \hat{x}$. Using Faraday's law applied to the wave, which of the following correctly gives the instantaneous magnetic field $\vec{B}$ at that same point and time, and correctly identifies the reasoning step that uniquely determines its direction?
   Answer: $\vec{B} = (E_0/c)\hat{y}$, determined by applying Faraday's law: $\nabla \times \vec{E} = -\partial \vec{B}/\partial t$; for a plane wave $\vec{E} = E_0\cos(kz - \omega t)\hat{x}$, evaluating the curl yields $\partial \vec{B}/\partial t$ in the $+\hat{y}$ direction, which integrates to give $\vec{B} = (E_0/c)\hat{y}$ — consistent with the wave propagating in $+\hat{z}$. When you see a question asking you to determine both the magnetic field direction and the reasoning behind it in an EM wave, your instinct should be to reach for Faraday's law — it's the direct mathematical link between $\vec{E}$ and $\vec{B}$ in a propagating wave. For a plane wave $\vec{E} = E_0\cos(kz - \omega t)\hat{x}$, Faraday's law states $\nabla \times \vec{E} = -\partial \vec{B}/\partial t$. Taking the curl of $\vec{E}$, the only nonzero term is $\partial E_x/\partial z = -kE_0\sin(kz-\omega t)$, which contributes to the $\hat{y}$ component. This gives $\partial \vec{B}/\partial t$ pointing in $+\hat{y}$, and integrating confirms $\vec{B} = (E_0/c)\hat{y}$. That's exactly what B describes — a complete, rigorous derivation. A reaches the correct numerical answer but uses flawed logic. Invoking the Poynting vector to determine $\vec{B}$ is circular: the Poynting vector tells you energy flow after you already know both fields. It's a verification tool, not a derivation tool. C misapplies Lenz's law, which governs induced EMFs in circuits opposing changes in magnetic flux — it has no role here. The sign in Faraday's law is handled by the curl calculation, not by an opposition principle involving the electric field direction directly. D is simply false. In a transverse EM wave, $\vec{E}$ and $\vec{B}$ are perpendicular to each other and to the propagation direction — never parallel. Study tip: Always distinguish between tools that derive a result (Faraday's law, Maxwell's equations) versus tools that verify it (Poynting vector). Questions on this exam often trap students who confuse consistency checks with causal reasoning.
9. A plane electromagnetic wave in vacuum has angular frequency $\omega$ and wave vector $k = \omega/c$. The wave's electric field amplitude is $E_0$. Consider two statements about the relationship between the fields: Statement 1: The magnetic field amplitude $B_0 = E_0/c$ follows from requiring that the Poynting vector have the correct dimensions of intensity. Statement 2: The magnetic field amplitude $B_0 = E_0/c$ follows from applying Faraday's law to the plane wave, which gives $k E_0 = \omega B_0$, and since $k/\omega = 1/c$, one obtains $B_0 = E_0/c$. Which of the following correctly evaluates both statements?
   Answer: Statement 1 is incorrect and Statement 2 is correct: dimensional analysis of the Poynting vector cannot uniquely determine $B_0$ in terms of $E_0$ because $\vec{S} = (1/\mu_0)(\vec{E}\times\vec{B})$ has correct dimensions for any ratio $E_0/B_0$ with units of m/s, so Statement 1 merely shows consistency but does not derive the amplitude relationship. Statement 2 correctly derives $B_0 = E_0/c$ from Maxwell's equations. When evaluating claims about where a physics result comes from, you must carefully distinguish between deriving something and merely confirming consistency with it. That distinction is exactly what this question tests. Statement 2 is the genuine derivation. Faraday's law, $\nabla \times \vec{E} = -\partial \vec{B}/\partial t$, applied to a plane wave $\vec{E} = E_0\hat{y}\cos(kx - \omega t)$ gives a spatial derivative on the left (pulling down a factor of $k$) and a time derivative on the right (pulling down a factor of $\omega$), yielding $kE_0 = \omega B_0$. Since $k = \omega/c$, this directly gives $B_0 = E_0/c$. This is a true first-principles derivation from Maxwell's equations — D is correct. Statement 1 fails because dimensional analysis cannot derive a specific physical relationship — it can only check consistency. The Poynting vector $\vec{S} = \frac{1}{\mu_0}(\vec{E} \times \vec{B})$ has units of W/m² for any ratio $E_0/B_0$ that carries units of m/s. You could hypothetically have $B_0 = 2E_0/c$ or $B_0 = 0.5E_0/c$ and the dimensions would still work. Dimensional analysis tells you the ratio has units of speed, not that the speed must equal $c$. Answer A is wrong because it incorrectly validates Statement 1 as a genuine derivation. Answer B is wrong because Faraday's law alone is sufficient to derive the amplitude ratio — you don't need Ampere's law simultaneously. Answer C is wrong because $\partial \vec{B}/\partial t$ for a cosine wave yields a sine, but the amplitude relationship $kE_0 = \omega B_0$ is extracted correctly regardless of phase. Study tip: On exam questions involving "derivations," always ask whether the reasoning actually forces a unique answer or merely permits it — dimensional analysis almost never derives; it only constrains.

Start this quiz