Question

If \(\vec{A}=\vec{\nabla}\phi\) and \(\phi = xy+yz+zx\), then the true statements are:
A. \(\vec{\nabla}\cdot\vec{A}=0\)
B. \(\vec{\nabla}\cdot\vec{A}\neq0\)
C. \(\vec{\nabla}\times\vec{A}=\vec{0}\)
D. \(\vec{\nabla}\times\vec{A}\neq\vec{0}\)
Choose the correct answer from the option given below:

• A. A and C only
• B. A and D only
• C. B and D only
• D. B and C only

Hint:

Knowing the identities \(\nabla \times (\nabla \phi) = 0\) and \(\nabla \cdot (\nabla \times \vec{F}) = 0\) can save you a lot of time. If a vector field is given as a gradient of a scalar (\(\vec{A} = \vec{\nabla}\phi\)), its curl must be zero. If it's given as the curl of another vector (\(\vec{B} = \vec{\nabla} \times \vec{F}\)), its divergence must be zero.

Answer 1

The Correct Option is A

Step 1: Use vector calculus identities. There are two fundamental vector identities that can solve this problem without direct calculation: 1. The curl of the gradient of any scalar field is always zero: \(\vec{\nabla} \times (\vec{\nabla}\phi) = \vec{0}\). 2. The divergence of the gradient is the Laplacian: \(\vec{\nabla} \cdot (\vec{\nabla}\phi) = \nabla^2\phi\).
Step 2: Apply the identities to the given problem. - Curl of \(\vec{A}\): We are given \(\vec{A} = \vec{\nabla}\phi\). Using the first identity, we immediately know that \(\vec{\nabla} \times \vec{A} = \vec{\nabla} \times (\vec{\nabla}\phi) = \vec{0}\). Therefore, statement C is true and D is false. A vector field that is the gradient of a scalar potential is called a conservative or irrotational field. - Divergence of \(\vec{A}\): Using the second identity, we need to calculate the Laplacian of \(\phi\). \[ \vec{\nabla} \cdot \vec{A} = \nabla^2\phi = \frac{\partial^2\phi}{\partial x^2} + \frac{\partial^2\phi}{\partial y^2} + \frac{\partial^2\phi}{\partial z^2} \] First derivatives are: \(\frac{\partial\phi}{\partial x} = y+z\), \(\frac{\partial\phi}{\partial y} = x+z\), \(\frac{\partial\phi}{\partial z} = y+x\). Second derivatives are: \[ \frac{\partial^2\phi}{\partial x^2} = 0, \quad \frac{\partial^2\phi}{\partial y^2} = 0, \quad \frac{\partial^2\phi}{\partial z^2} = 0 \] Therefore, \(\vec{\nabla} \cdot \vec{A} = 0+0+0=0\). Statement A is true and B is false.
Step 3: Conclude the correct statements. Both statements A and C are true.