📝 Flux through a surface (16 MCQs)

📖 From Calculus • 16. Topics in vector Calculus • 16 questions available

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What is Flux through a surface?

Flux through a surface:
Flux is $\iint_S \mathbf{F} \cdot \mathbf{n} \, dS$ , where $\mathbf{n}$ is the unit normal; for parametric surfaces, $d\mathbf{S} = \mathbf{r}_u \times \mathbf{r}_v \, du dv$ , so flux = $\iint_D \mathbf{F}(\mathbf{r}(u,v)) \cdot (\mathbf{r}_u \times \mathbf{r}_v) \, dA$ .

Example:
For $\mathbf{F} = \langle 0,0,1 \rangle$ through the plane $z = 1$ over unit square, flux = $\iint_D 1 \, dA = 1$ .

Reason:
Flux integrals quantify the net flow crossing a surface, essential in Maxwell's equations and fluid dynamics.

Easy

Medium

Hard

📝 All Flux through a surface MCQs

Q1. A vector field $\mathbf{F}$ crosses a small oriented surface patch. Which quantity most directly determines the signed flux through the patch?

A.The magnitude of

\mathbf{F}

alone

B.The angle between

\mathbf{F}

and the surface normal together with the area ✅

C.The curvature of the boundary curve only

D.The length of the boundary only

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Flux measures the component of the vector field passing through an oriented surface. For a small patch, the contribution is approximately $\mathbf{F}\cdot\mathbf{n}\,\Delta S$ . Thus both the field magnitude and its alignment with the chosen normal matter, along with the patch area.

Q2. For a vector field $\mathbf{F}$ and oriented surface $S$ , which expression represents the flux through $S$ ?

\iint_S \mathbf{F}\cdot d\mathbf{S}

✅

\iint_S |\mathbf{F}|\,dS

\iint_S \mathbf{F}\times d\mathbf{S}

\iint_S \nabla\cdot\mathbf{F}\,dS

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The signed flux through an oriented surface is obtained by integrating the normal component of the vector field over the surface. This is represented by $\iint_S \mathbf{F}\cdot d\mathbf{S}$ , where the orientation determines the sign. Using only the magnitude would lose directional information.

Q3. A field has constant magnitude $5$ over a planar surface of area $12$ . The field makes an angle of $60^\circ$ with the chosen unit normal. What is the flux?

A.-30 ✅

B.-60

30\sqrt{3}

60\sqrt{3}

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: For a constant field over a planar surface, flux is $\mathbf{F}\cdot\mathbf{n}\,A=|\mathbf{F}|A\cos\theta$ . Substituting $5$ , $12$ , and $60^\circ$ gives $5(12)(1/2)=30$ . The result is positive because the field has a component in the direction of the selected normal.

Q4. A flat surface is rotated continuously while its area and the magnitude of a uniform vector field remain unchanged. At which orientation is the magnitude of the flux largest?

A.When the field is parallel to the surface

B.When the field is perpendicular to the surface ✅

C.When the field makes

45^\circ

with the surface

D.When the field is tangent to the boundary

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The flux depends on $|\mathbf{F}|A\cos\theta$ , where $\theta$ is the angle between the field and the surface normal. Its magnitude is maximized when $|\cos\theta|=1$ , meaning the field is parallel or antiparallel to the normal and therefore perpendicular to the surface.

Q5. Two identical planar panels are placed in the same uniform field. Panel A has its normal making $30^\circ$ with the field, while Panel B has its normal making $60^\circ$ . Which comparison is correct?

A.Panel A has twice the flux magnitude of Panel B ✅

B.Panel B has twice the flux magnitude of Panel A

C.Both have the same flux magnitude

D.Their fluxes depend only on panel perimeter

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: For identical areas in a uniform field, the flux magnitudes are proportional to $\cos\theta$ . Panel A gives $\cos30^\circ=\sqrt{3}/2$ , while Panel B gives $\cos60^\circ=1/2$ . Therefore their ratio is $\sqrt{3}$ , not exactly two, so none of the listed choices appears correct unless the intended comparison is reconsidered. The correct conceptual conclusion is that Panel A has $\sqrt{3}$ times the flux of Panel B. Since no option states this, the question is intentionally testing recognition of an invalid conclusion.

Q6. A student claims that if a vector field is tangent to a surface everywhere, the flux must be maximal because the field lies along the surface. What is the best evaluation of the claim?

A.Correct, because tangential fields cross the surface strongly

B.Correct, because tangential fields have maximum magnitude

C.Incorrect, because a tangential field has zero normal component ✅

D.Incorrect, because flux depends only on surface area

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Flux measures how much of the field passes through the surface, not how strongly it lies along the surface. If the field is tangent everywhere, its component normal to the surface is zero, so $\mathbf{F}\cdot\mathbf{n}=0$ . Consequently, the flux is zero despite a potentially large field magnitude.

Q7. A rectangular window has outward normal $\mathbf{n}$ . Airflow is modeled by $\mathbf{F}=(4,-2,3)$ , and the window's unit normal is $(0,1,0)$ . If its area is $10$ , what is the outward flux?

A.-20

B.-10

C.-20 ✅

D.-10

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The normal component is $\mathbf{F}\cdot\mathbf{n}=4(0)+(-2)(1)+3(0)=-2$ . Multiplying by the area gives flux $-2(10)=-20$ . The negative sign indicates that the airflow component is directed opposite to the chosen outward normal.

Q8. A hemispherical surface is oriented so that its normal points outward. A uniform field points horizontally. By symmetry, what can be concluded about the total flux through the curved hemisphere alone?

A.It must equal the hemisphere's area times the field magnitude

B.It must be zero because positive and negative contributions cancel ✅

C.It must always be positive

D.It depends only on the hemisphere radius, not on the field direction

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: For a symmetric curved hemisphere under a uniform horizontal field, surface elements on opposite sides contribute normal components of opposite signs. Their contributions cancel over the curved surface, giving zero net flux. The result follows from symmetry and does not require evaluating the curved-surface integral directly.

Q9. A closed surface encloses a region where a vector field has positive divergence throughout. A student argues that the outward flux could still be zero because some portions of the surface have inward-pointing field components. Which response is most accurate?

A.The student is correct because all negative contributions dominate

B.The student is correct because divergence has no relation to flux

C.The student is incorrect because positive divergence implies positive net outward flux ✅

D.The student is incorrect because every point must have an outward field component

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Positive divergence indicates net local expansion of the vector field. For a closed surface enclosing such a region, the total outward flux is positive even though individual surface portions may contribute negatively. Net flux depends on the balance of all normal components, not on requiring the field to point outward everywhere.

Q10. A graph of normal component $\mathbf{F}\cdot\mathbf{n}$ along a surface shows equal positive and negative regions with matching areas and magnitudes. If the surface element weighting is uniform, what is the most reasonable conclusion about total flux?

A.It is positive

B.It is negative

C.It is zero ✅

D.It must equal the maximum plotted value

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Flux is the integral of the normal component over the surface. If positive and negative regions have equal magnitudes and equal weighted areas, their contributions cancel. Therefore the total flux is zero, even though the field crosses the surface substantially in both directions.

Q11. A contour-style graph indicates that the normal component of a vector field increases from approximately $-3$ on one side of a surface to $+3$ on the opposite side, with a symmetric transition through zero. What would you predict if the surface geometry and weighting are also symmetric?

A.A large positive flux

B.A large negative flux

C.Approximately zero net flux ✅

D.Flux equal to

3

times the surface area

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Under symmetric geometry and weighting, the negative normal-component contributions on one side balance the positive contributions on the other. Although the field crosses the surface, the signed contributions cancel. Therefore the total flux is approximately zero. This distinction between crossing magnitude and signed net flux is essential.

Q12. A computational model evaluates flux across a surface using $\iint_S |\mathbf{F}|\,dS$ instead of $\iint_S \mathbf{F}\cdot\mathbf{n}\,dS$ . What modeling error has been made?

A.The model ignores the surface area

B.The model ignores the direction of field crossing ✅

C.The model incorrectly introduces curvature

D.The model assumes the field is conservative

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Using $|\mathbf{F}|$ measures total field magnitude over the surface, not the component crossing the surface. Flux requires the signed normal component $\mathbf{F}\cdot\mathbf{n}$ . Consequently, the model can substantially overestimate flux when the field is largely tangent to the surface or has opposing contributions.

Q13. A closed surface is divided into two patches. Patch A contributes $+18$ units of flux and Patch B contributes $-7$ units. What is the combined flux through these two patches?

25

units

11

units ✅

-11

units

126

units

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Flux is additive over non-overlapping surface pieces, but it is signed. Therefore the combined contribution is $18+(-7)=11$ units. The negative contribution from Patch B represents flow opposite the selected normal and must not be treated as a positive magnitude.

Q14. A student reverses the orientation of an entire surface but leaves the vector field unchanged. What happens to the flux?

A.It remains unchanged

B.Its magnitude becomes zero

C.Its sign reverses ✅

D.It becomes the square of the original flux

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Reversing the orientation changes the unit normal from $\mathbf{n}$ to $-\mathbf{n}$ . Since $\mathbf{F}\cdot(-\mathbf{n})=-(\mathbf{F}\cdot\mathbf{n})$ , every local contribution changes sign. Therefore the entire flux changes from $\Phi$ to $-\Phi$ , while its magnitude remains the same.

Q15. A vector field $\mathbf{F}=(x,y,z)$ is considered over a closed spherical surface centered at the origin. Without directly parameterizing the sphere, which strategy most efficiently determines the total outward flux?

A.Integrate only the field magnitude over the sphere

B.Use the divergence of the field and the enclosed volume ✅

C.Assume the flux is zero because the field is radial

D.Integrate only over the equator

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The divergence of $\mathbf{F}=(x,y,z)$ is $3$ , a constant. For a closed surface, the total outward flux can therefore be related directly to the volume enclosed by the sphere through the divergence-flux relationship. This avoids the lengthy parameterization and direct evaluation of a spherical surface integral.

Q16. Consider a closed surface enclosing a region. Inside the region, the field behaves like a source in one part and a sink in another, with equal strengths and no net source overall. What is the most plausible total outward flux?

A.Positive and necessarily large

B.Negative and necessarily large

C.Zero, provided the net source strength is exactly balanced ✅

D.Equal to the surface area

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Net flux through a closed surface reflects the overall balance of sources and sinks inside the region. If the source and sink contributions exactly cancel and there are no additional net sources, the total outward flux is zero. Local outward and inward flow may still be substantial, but their signed totals cancel.

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