Physics 2 Quiz: Gausss Law For Magnetism

What this quiz covers

This quiz focuses on Gausss Law For Magnetism, giving you a quick way to practice the rules, question types, and explanations that matter most for Physics 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $\Phi_{B,1} = 0$ and $\Phi_{B,2} = 0$, because Gauss's law for magnetism requires the net flux through any closed surface to be zero, regardless of how many dipoles are enclosed or where they are located relative to the surface. (correct answer)
2. $\Phi_{B,1} = 0$ and $\Phi_{B,2} \neq 0$, because the second dipole lies between the surfaces and thus contributes to the flux through $S_2$ but not through $S_1$, analogous to how charge outside a Gaussian surface contributes to the net electric flux.
3. $\Phi_{B,1} \neq 0$ and $\Phi_{B,2} \neq 0$, because each enclosed dipole produces a net outward flux proportional to its magnetic moment, making the total flux additive in the same way enclosed charges add in Gauss's law for electricity.
4. $\Phi_{B,1} \neq 0$ and $\Phi_{B,2} = 0$, because the two dipoles oriented inside $S_2$ cancel each other's contributions only when both are enclosed, while the single dipole inside $S_1$ produces a nonzero net flux through the smaller surface.

Question 2

1. The claim is consistent with Gauss's law only if the time-varying electric field inside the surface produces a displacement current that accounts for the nonzero flux, since Maxwell's extension of Ampère's law permits such contributions.
2. The claim directly violates Gauss's law for magnetism, because that law requires the net magnetic flux through any closed surface to equal zero regardless of what sources are enclosed, implying the measurement contains an error. (correct answer)
3. The claim is plausible because Gauss's law for magnetism only constrains the flux when the closed surface encloses no net current; the presence of current-carrying loops inside the surface can produce a nonzero net flux.
4. The claim is consistent with Gauss's law for magnetism if the bar magnets are oriented such that more field lines exit the surface than enter it, since the law applies only to monopole sources and bar magnets are dipoles that can create asymmetric flux.

Question 3

1. The region enclosed by the surface contains no magnetic monopoles of any sign, since a zero surface integral unambiguously requires the absence of all magnetic charge within the enclosed volume.
2. The net magnetic charge enclosed is zero, but this does not rule out the presence of equal and opposite magnetic charges inside the surface, nor does it require the magnetic field to be zero everywhere on the surface. (correct answer)
3. The magnetic field $\vec{B}$ is zero at every point on the surface, because if any field lines penetrated the surface, they would have to originate from a monopole source inside, producing a nonzero integral.
4. The surface must enclose an even number of monopoles, since monopoles are created in particle–antiparticle pairs and the integral being zero implies a symmetric pairing of all enclosed sources.

Question 4

1. Student X is fully correct. The Biot–Savart law, when applied to all possible current configurations, mathematically requires that field lines form closed loops, from which $\oint \vec{B} \cdot d\vec{A} = 0$ follows as a theorem rather than a postulate.
2. Student Y is fully correct. The Biot–Savart law describes the field of a current element but says nothing about the topology of field lines or the existence of monopoles, so the condition $\nabla \cdot \vec{B} = 0$ must be postulated separately as an empirical law.
3. Both students are partially correct: $\nabla \cdot \vec{B} = 0$ is derivable from the Biot–Savart law for magnetostatic fields produced by steady currents in simply connected geometries, but it must be separately postulated to hold in time-varying situations or in regions that may contain monopoles. (correct answer)
4. Student X is correct only for infinite, straight current-carrying wires; for more complex current distributions such as loops or solenoids, the Biot–Savart law does not guarantee closed field lines, and Gauss's law for magnetism must be imposed independently.

Question 5

1. The net flux is nonzero because the face inside the solenoid experiences a strong perpendicular field while the face outside experiences essentially zero field, creating an imbalance that reveals the solenoid acts as a localized source of magnetic flux.
2. The net flux is zero, consistent with Gauss's law for magnetism. Although the two perpendicular faces have unequal contributions (strong inward or outward flux on the inside face, near-zero on the outside face), the four faces parallel to the axis must collectively carry the compensating flux to maintain a zero total. (correct answer)
3. The net flux is zero only if the solenoid is infinitely long; for a finite solenoid, fringe fields at the ends break the symmetry required by Gauss's law, producing a small but nonzero net flux through the rectangular surface.
4. The net flux is zero, and this result implies that the magnetic field must be zero everywhere on all six faces of the Gaussian surface, since the only way a closed-surface integral can vanish is if the integrand is zero at each point.

Question 6

1. Gauss's law for magnetism would be modified to $\oint \vec{B} \cdot d\vec{A} = \mu_0 q_{m,enc}$, directly analogous to Gauss's law for electricity, and Faraday's law would simultaneously require a magnetic current density term $\vec{J}_m$ to maintain self-consistency of the full set of Maxwell's equations. (correct answer)
2. Faraday's law would need to gain a magnetic current density term $\vec{J}_m$, and Gauss's law for magnetism would remain $\oint \vec{B} \cdot d\vec{A} = 0$ because monopoles affect only the induction of electric fields, not the divergence of $\vec{B}$.
3. Only a continuity equation for magnetic charge ($\nabla \cdot \vec{J}_m + \partial \rho_m / \partial t = 0$) would need to be added as a new equation; Gauss's law for magnetism and the other three Maxwell equations would remain unchanged since monopoles obey separate conservation laws independent of the field equations.
4. Gauss's law for magnetism would be modified to $\oint \vec{B} \cdot d\vec{A} = \mu_0 q_{m,enc}$, but Faraday's law would be unaffected because the interaction between moving magnetic monopoles and electric fields is mediated entirely by the Lorentz force, which is a separate equation and does not alter the field equations themselves.

Question 7

1. The field is not physically possible. The $x$-component increases with $x$ and the $y$-component becomes more negative with $y$, so field lines are diverging in the $xy$-plane, indicating a net source of magnetic flux at every point — a violation of $\nabla \cdot \vec{B} = 0$.
2. The field is not physically possible. Computing $\nabla \cdot \vec{B} = \frac{\partial}{\partial x}\!\left(\frac{B_0 x}{a}\right) + \frac{\partial}{\partial y}\!\left(-\frac{B_0 y}{a}\right) = \frac{B_0}{a} + \frac{B_0}{a} = \frac{2B_0}{a}$, which is nonzero everywhere, violating the requirement that $\nabla \cdot \vec{B} = 0$.
3. The field is physically possible. Evaluating $\nabla \cdot \vec{B} = \frac{\partial}{\partial x}\!\left(\frac{B_0 x}{a}\right) + \frac{\partial}{\partial y}\!\left(-\frac{B_0 y}{a}\right) = \frac{B_0}{a} - \frac{B_0}{a} = 0$, satisfying $\nabla \cdot \vec{B} = 0$ everywhere, as required by Gauss's law for magnetism. (correct answer)
4. The field is physically possible only along the line $y = x$. At all other points, the magnitudes of the $x$- and $y$-components differ, so their contributions to the divergence do not cancel, producing a nonzero $\nabla \cdot \vec{B}$ away from that line.

Question 8

1. The second student is correct. The vector potential $\vec{A}$ is a purely mathematical convenience introduced to simplify calculations in specific geometries. The condition $\nabla \cdot \vec{B} = 0$ is a separate empirical statement about the absence of monopoles, and the two are logically independent: one can accept either statement without the other, as they arise from different physical and mathematical considerations.
2. The first student is correct in all cases. By the Helmholtz decomposition theorem, any vector field with zero divergence in any domain — whether simply or multiply connected — can always be expressed globally as the curl of another vector field. Thus $\nabla \cdot \vec{B} = 0$ directly and rigorously implies the existence of a globally defined $\vec{A}$ such that $\vec{B} = \nabla \times \vec{A}$ throughout all of space.
3. The second student is correct. While $\nabla \cdot \vec{B} = 0$ is consistent with writing $\vec{B} = \nabla \times \vec{A}$, this representation also requires that $\vec{B}$ be irrotational ($\nabla \times \vec{B} = 0$) throughout the region of interest — a condition not guaranteed by Gauss's law alone and, in fact, violated wherever currents flow.
4. Both students are partially correct. The condition $\nabla \cdot \vec{B} = 0$ guarantees that $\vec{A}$ exists locally (on any contractible region) by the Poincaré lemma, so the first student is right in that domain. However, on multiply connected domains — such as the region surrounding an infinite solenoid, as in the Aharonov–Bohm effect — topological obstructions can prevent a globally single-valued $\vec{A}$ even when $\nabla \cdot \vec{B} = 0$ holds everywhere, vindicating the second student's caution about treating $\vec{A}$ as universally implied. (correct answer)

Question 9

1. The student's analogy fails because magnetic fields, unlike electric fields, obey the superposition principle differently near current-carrying conductors, so the surface integral cannot be evaluated without knowing the exact geometry of all currents.
2. The student's analogy fails because $\mu_0 I_{enc}$ already appears in Ampère's law, which governs the line integral of $\vec{B}$, not the surface integral; conflating these two integrals constitutes a fundamental dimensional and conceptual error.
3. The student confuses the source of electric fields (charges, which can be isolated) with the source of magnetic fields; because no isolated magnetic charges (monopoles) have been observed to exist, the correct magnetic Gauss's law has zero on the right-hand side. (correct answer)
4. The student's proposed law is dimensionally inconsistent: $\mu_0$ has units of T·m/A, so $\mu_0 I_{enc}$ has units of T·m, whereas the surface integral $\oint \vec{B} \cdot d\vec{A}$ has units of T·m². Because the two sides cannot be equal in any unit system, the analogy is invalid on purely dimensional grounds, independent of any physical argument about monopoles.

Question 10

1. The experiment is fundamentally flawed because Gauss's law for magnetism applies only to mathematical surfaces, not to physically realizable measurement surfaces, so any nonzero result would be an artifact of the discrete sampling method.
2. The magnetometer measures the total $\vec{B}$ field, which includes contributions from both the hypothetical monopole and any external dipole sources. Because $\oint \vec{B} \cdot d\vec{A} = 0$ for all dipole sources, the external contributions automatically cancel, making the only remaining concern the calibration accuracy of the magnetometer itself.
3. The experiment incorrectly uses a spherical surface; only a cubical Gaussian surface would sample the field uniformly enough to detect a monopole, since field lines from a point source are not perpendicular to a sphere at all but the topmost and bottommost points.
4. A nonzero numerical result could arise from ordinary magnetic dipoles (e.g., the Earth's field or nearby magnets) whose field lines are not perfectly canceled at the 1000 discrete sample points, making it impossible to distinguish a genuine monopole signal from numerical integration error without extraordinary precision. (correct answer)