Vector Calculus Questions and Answers

Answer: C

Expanding gives six mixed third derivatives that cancel in pairs, so

\[ \nabla \cdot (\nabla \times \mathbf{A}) \;=\; 0. \]

Every curl field is therefore solenoidal, which is why a solenoidal field can be represented by a vector potential and why magnetic flux density, having zero divergence, is written as the curl of one

Answer: B

The operator differentiates both factors, and the term acting on the scalar has to be contracted with the vector to stay a scalar:

\[ \nabla \cdot (\phi\mathbf{A}) \;=\; \phi\,(\nabla \cdot \mathbf{A}) + \mathbf{A} \cdot \nabla\phi. \]

The sign is positive, as in an ordinary product rule; the minus sign belongs to no standard identity of this kind

Answer: A

Solenoidal means divergence-free, so the flux out of every closed surface is zero and the field lines close on themselves rather than beginning or ending. A curl-free field is irrotational instead, and that is the one expressible as a gradient. A field may be either, both, or neither

Answer: D

Irrotational means \( \nabla \times \mathbf{F} = \mathbf{0} \), and since the curl of any gradient vanishes identically, a scalar potential is the natural representation. A vector potential belongs to the solenoidal case, where the divergence rather than the curl is zero

Answer: C

Only the component of the force along the motion does work, so each element contributes \( \mathbf{F} \cdot d\mathbf{r} \) and

\[ W \;=\; \int_{C} \mathbf{F} \cdot d\mathbf{r}. \]

Integrating the magnitude alone would count a sideways force as though it were pushing along the path, and a cross product would give a vector rather than an amount of work

Answer: B

On the path \( y = x \), so with \( x = t \) and \( y = t \) for \( t \) from zero to one,

\[ \int_{C}\left(y\,dx + x\,dy\right) \;=\; \int_{0}^{1}\left(t + t\right)dt \;=\; \left[t^{2}\right]_{0}^{1} \;=\; 1. \]

The field is the gradient of \( xy \), so the value is just \( xy \) evaluated between the ends, \( 1 - 0 \), whatever path is taken

Answer: A

Path independence, a vanishing circulation round every closed loop, the existence of a scalar potential, and an identically zero curl on a simply connected region are four equivalent statements. A vanishing divergence is a different condition altogether, describing a solenoidal rather than a conservative field

Answer: D

With \( \mathbf{F} = \nabla\phi \) the integrand becomes an exact differential, \( \nabla\phi \cdot d\mathbf{r} = d\phi \), so

\[ \int_{A}^{B} \mathbf{F} \cdot d\mathbf{r} \;=\; \phi(B) - \phi(A). \]

It vanishes only when the two endpoints share a potential value, which for a closed path is automatic but for an open one is not

Answer: C

Only the component crossing the surface counts, so each element contributes \( \mathbf{F} \cdot \mathbf{n}\,dS \) and

\[ \Phi \;=\; \iint_{S} \mathbf{F} \cdot \mathbf{n}\,dS. \]

A field lying wholly in the surface carries no flux through it, which the dot product records and an integral of the magnitude would not

Answer: B

On the sphere the field is radial with magnitude \( a \), and the outward normal is radial too, so \( \mathbf{F} \cdot \mathbf{n} = a \) everywhere:

\[ \Phi \;=\; a \times 4\pi a^{2} \;=\; 4\pi a^{3}. \]

The divergence theorem confirms it, since \( \nabla \cdot \mathbf{F} = 3 \) and \( 3 \times \tfrac{4}{3}\pi a^{3} = 4\pi a^{3} \)