A vector field F has nonzero circulation ∮_C F · dr around some closed loop C. What can you conclude?
AF must be undefined at some point inside the loop
BF cannot be expressed as the gradient of any scalar potential function
CF has constant magnitude throughout the enclosed region
DThe work done by F along any open path in the region is zero
For a conservative field F = ∇f, the work integral depends only on endpoints: ∫_C F · dr = f(end) − f(start). Around any closed loop, start = end, so circulation = f(start) − f(start) = 0. Nonzero circulation therefore proves the field is not conservative — no scalar potential can exist. Option A is tempting (Stokes' theorem relates singularities to circulation in specific cases) but nonzero circulation alone doesn't require a singularity inside; it simply rules out conservativeness.
Question 2 Multiple Choice
A particle moves from point P to point Q along two different paths in a vector field F. The work done is 10 J on the first path and 7 J on the second. What does this imply?
AF is conservative but the particle moved faster on one path, changing the work done
BF is conservative, and f(Q) − f(P) = 10 J along the first path and 7 J along the second
CF is not conservative — work is path-dependent, so no scalar potential exists for F
DThe particle must have moved in a closed loop on one of the paths
Path independence of work is the defining property of conservative fields. If F = ∇f, the work from P to Q must equal f(Q) − f(P) regardless of which path is taken — any two paths give the same answer. Different work values (10 J vs. 7 J) for the same endpoints proves the work is path-dependent, which proves F is not conservative. Options A and B misunderstand conservation: for conservative fields, speed and path shape are irrelevant — only endpoints matter.
Question 3 True / False
For a conservative vector field, the circulation around any closed curve is zero.
TTrue
FFalse
Answer: True
This follows directly from the Fundamental Theorem of Line Integrals: if F = ∇f, then ∫_C F · dr = f(end) − f(start). For a closed curve, the endpoint is the same as the starting point, so the integral equals f(start) − f(start) = 0. Zero circulation is therefore a necessary consequence of conservativeness. (It is also sufficient under appropriate conditions, which Green's theorem makes precise.)
Question 4 True / False
The work done by a conservative force field along a path depends on the length and shape of that path.
TTrue
FFalse
Answer: False
This is precisely what 'conservative' means: work depends only on the starting and ending points, never on the path connecting them. ∫_C ∇f · dr = f(end) − f(start) — a formula with no reference to the path. A longer, curving path gives the same work as a straight-line shortcut between the same two points. Gravitational and electrostatic force fields are conservative for this reason: the energy gained or lost depends only on the height difference or potential difference, not on the route taken.
Question 5 Short Answer
Explain why nonzero circulation around a closed loop proves that a vector field is not conservative.
Think about your answer, then reveal below.
Model answer: If F were conservative (F = ∇f), the work integral around any closed loop would equal f(end) − f(start) = f(start) − f(start) = 0. A nonzero circulation value means the integral around the closed loop is not zero, which contradicts the requirement for a conservative field. Therefore the field cannot have a scalar potential, and is not conservative.
The argument is a proof by contradiction: assume conservativeness → circulation = 0. Contrapositive: circulation ≠ 0 → not conservative. This is exactly why circulation is the diagnostic tool for non-conservativeness, and why Green's theorem (which equates circulation to an area integral of the curl) is so powerful — curl measures how much the field 'rotates' locally, and nonzero curl anywhere inside a loop produces nonzero circulation around it.