Vector Analysis and Coordinate Systems Questions and Answers

Answer: A

Correct answer: the divergence of a scalar

Divergence takes a vector and returns a scalar, so it has nothing to operate on when given a scalar. The valid operations are gradient on a scalar, and divergence and curl on a vector - and the Laplacian, which is the divergence of a gradient and so applies to a scalar by way of a vector.

Answer: B

A uniform field has the same value everywhere, so every spatial derivative of it is zero and both its divergence and curl vanish, which makes its closed line integral vanish too. The field itself is a non-zero constant.

Answer: C

The cylindrical divergence is \( (1/\rho)\partial(\rho A_\rho)/\partial\rho \), and the bracket here is the constant k, so the derivative is zero. This is the field of an infinite line charge, whose divergence vanishes everywhere except on the line itself.

Answer: D

The radial field of a coaxial cable is a function of the radial distance alone, so a three-dimensional problem becomes one-dimensional. In Cartesian coordinates the same field would depend on both x and y in an awkward way.

Answer: A

This identity is what turns Maxwell's equations into the wave equation: substituting it and using the vanishing divergence in a source-free region leaves the Laplacian of the field equal to its second time derivative multiplied by a constant.

Answer: B

The two arc lengths on the surface are \( a \, d\theta \) and \( a \sin\theta \, d\phi \), and their product is the element. Integrating it over the full angular ranges gives \( 4\pi a^2 \), the surface area of the sphere.

Answer: C

Correct answer: its curl vanish throughout the region

A field expressible as \( \nabla V \) has zero curl, since the curl of any gradient vanishes identically, and in a simply connected region the converse holds too. The divergence is unconstrained: it equals \( \nabla^2 V \), which may be anything the sources require.

Answer: D

In a curvilinear system the coordinate differentials are not all lengths and the unit vectors turn with position, and both effects appear as the metric factors of the operator. Only in Cartesian coordinates do those factors reduce to unity.

Answer: A

Correct answer: 3

\[ |\mathbf{r}| = \sqrt{1^2+2^2+2^2} = \sqrt{9} = 3. \] The triple 1, 2, 2 is one of the small integer combinations giving an exact result, along with 2, 3, 6 and 1, 4, 8, and it is worth recognising when it appears.

Answer: B

The projection on a unit vector is the dot product with it, which picks out the corresponding component. Dividing by the magnitude of the vector would instead give the direction cosine of 0.8, which is a different quantity.