Vector Analysis and Coordinate Systems Questions and Answers

Q: Which of the following is NOT a valid vector operation?

Correct answer: the divergence of a scalarDivergence 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.

Q: For the vector field \( \mathbf{A} \) to be expressible as the gradient of a scalar in a region, it is necessary that

Correct answer: its curl vanish throughout the regionA 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.

Q: The position vector of the point \( (1, 2, 2) \) has a magnitude of

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.

61.

Consider the following statements about coordinate systems.

1. The cylindrical system suits a long straight conductor 2. The spherical system suits a point charge 3. The Cartesian unit vectors vary with position

A.1, 2 and 3 are all correct statements about the three systems
B.2 and 3 only are correct
C.1 and 3 only are correct
D.1 and 2 only are correct

Answer: D

Choosing the system whose symmetry matches the problem is what makes a field calculation tractable. The Cartesian unit vectors are the only ones that do not vary with position, which is precisely why the other two systems have more complicated operator forms.

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62.

Which of the following is NOT a valid vector operation?

A.the divergence of a scalar
B.the gradient of a scalar
C.the curl of a vector
D.the divergence of a vector

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.

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63.

Which quantity would NOT be expected to vanish for a uniform vector field?

A.the line integral of the field round a closed path
B.the field itself
C.the divergence of the field
D.the curl of the field

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.

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64.

A field is given as \( \mathbf{A} = k\mathbf{a}_\rho/\rho \) in cylindrical coordinates. Its divergence away from the axis is

A.\( k/\rho \)
B.\( -k/\rho^2 \)
C.zero
D.\( k/\rho^2 \)

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.

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65.

Choosing cylindrical rather than Cartesian coordinates for a coaxial cable problem is worthwhile because

A.the unit vectors of the cylindrical system are constant vectors
B.the divergence takes a simpler form in cylindrical coordinates generally
C.Cartesian coordinates cannot represent a field with circular symmetry
D.the field depends on one variable only in that system

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.

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66.

The expression \( \nabla \times (\nabla \times \mathbf{A}) \) may be expanded as

A.\( \nabla(\nabla \cdot \mathbf{A}) - \nabla^2 \mathbf{A} \)
B.\( \nabla^2 \mathbf{A} - \nabla(\nabla \cdot \mathbf{A}) \)
C.zero
D.\( \nabla \cdot (\nabla \times \mathbf{A}) \)

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.

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67.

The differential surface element on a sphere of radius \( a \) is

A.\( a^2 \sin\theta \, da \, d\theta \)
B.\( a^2 \sin\theta \, d\theta \, d\phi \)
C.\( a \sin\theta \, d\theta \, d\phi \)
D.\( a^2 \, d\theta \, d\phi \)

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.

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68.

For the vector field \( \mathbf{A} \) to be expressible as the gradient of a scalar in a region, it is necessary that

A.both its curl and its divergence vanish throughout the region
B.its magnitude be constant throughout the region
C.its curl vanish throughout the region
D.its divergence vanish throughout the region

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.

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69.

A student computes the divergence of a spherical radial field using the Cartesian formula and obtains the wrong answer. The reason is that

A.the divergence is not defined in spherical coordinates at all
B.the radial unit vector has a magnitude that varies with radius
C.the Cartesian formula applies only to fields with a single component
D.the metric factors of the curvilinear system have been ignored

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.

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70.

The position vector of the point \( (1, 2, 2) \) has a magnitude of

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.

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