GATE 2020 EC Q.2: solution and answer

GATE 2020 · ECQ.2 · 1 mark · MCQ

For a vector field $A$ , which one of the following is FALSE?

(A)
$A$ is solenoidal if $\nabla \cdot A = 0$ .
(B)
$\nabla \times A$ is another vector field.
(C)
$A$ is irrotational if $\nabla^{2} A = 0$ .
(D)
$\nabla \times (\nabla \times A) = \nabla (\nabla \cdot A) - \nabla^{2} A$

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Answer and solution

ANSWER

Option (C) is FALSE. A field is irrotational when $\nabla \times A = 0$ , not when $\nabla^{2} A = 0$ .

Official answer: CSource: Official answer key, GATE 2020 EC (IIT Delhi)

Check each statement against its definition. We want the one that is FALSE.
(A) By definition, a field is solenoidal (divergence-free) when $\nabla \cdot A = 0$ . TRUE.
(B) The curl has three components, $\nabla \times A = (\frac{\partial A _{z}}{\partial y} - \frac{\partial A _{y}}{\partial z}) \hat{i} + (\frac{\partial A _{x}}{\partial z} - \frac{\partial A _{z}}{\partial x}) \hat{j} + (\frac{\partial A _{y}}{\partial x} - \frac{\partial A _{x}}{\partial y}) \hat{k},$ so it is itself a vector field. TRUE.
(C) By definition, a field is irrotational when its curl vanishes: $\nabla \times A = 0$ . The condition $\nabla^{2} A = 0$ is Laplace's equation, which makes each component harmonic. It is not the irrotationality condition. FALSE.
Counter-example for (C): take $A = y \hat{i}$ . Every second derivative of $y$ is zero, so $\nabla^{2} A = 0$ . But the $\hat{k}$ component of the curl is $\frac{\partial A _{y}}{\partial x} - \frac{\partial A _{x}}{\partial y} = 0 - 1 = - 1$ , so $\nabla \times A = - \hat{k} \neq = 0$ and the field is not irrotational.
It fails the other way too. $A = 3 x^{2} \hat{i}$ has $\nabla \times A = 0$ , since $A_{x}$ depends only on $x$ and $A_{y} = A_{z} = 0$ . So it is irrotational, but $\nabla^{2} A = 6 \hat{i} \neq = 0$ . The Laplacian condition is neither sufficient nor necessary for irrotationality.
(D) This is the standard curl-of-curl identity, $\nabla \times (\nabla \times A) = \nabla (\nabla \cdot A) - \nabla^{2} A$ , where $\nabla^{2} A$ is the vector Laplacian. TRUE.
The only false statement is (C).

KEY IDEA

Each field condition belongs to one operator. Solenoidal means $\nabla \cdot A = 0$ (divergence). Irrotational means $\nabla \times A = 0$ (curl). Harmonic means $\nabla^{2} A = 0$ (Laplacian). The curl-of-curl identity $\nabla \times (\nabla \times A) = \nabla (\nabla \cdot A) - \nabla^{2} A$ is standard. Neither of the last two conditions implies the other: $y \hat{i}$ is harmonic but not irrotational, and $3 x^{2} \hat{i}$ is irrotational but not harmonic.

WATCH OUT FOR

Taking $\nabla^{2} A = 0$ for the irrotational condition because it looks like it involves the curl through the identity in (D). It doesn't: $\nabla^{2} A = 0$ is Laplace's equation, and $\nabla \times A = 0$ is what makes a field irrotational. Another trap is picking (D) because of its sign. The minus sign in front of $\nabla^{2} A$ is correct.

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GATE 2020 EC Q.2: solution and answer

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