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: Utilize vector calculus identities.Two fundamental vector identities can resolve this problem without direct computation:1. The curl of the gradient of any scalar field is identically zero: \(\vec{abla} \times (\vec{abla}\phi) = \vec{0}\).2. The divergence of the gradient equals the Laplacian: \(\vec{abla} \cdot (\vec{abla}\phi) = abla^2\phi\).
Step 2: Apply the identities to the problem at hand.- Curl of \(\vec{A}\): Given \(\vec{A} = \vec{abla}\phi\). Employing the first identity yields \(\vec{abla} \times \vec{A} = \vec{abla} \times (\vec{abla}\phi) = \vec{0}\) directly. Consequently, statement C is valid and D is invalid. A vector field that is the gradient of a scalar potential is defined as a conservative or irrotational field.- Divergence of \(\vec{A}\): Using the second identity requires computing the Laplacian of \(\phi\). \[ \vec{abla} \cdot \vec{A} = abla^2\phi = \frac{\partial^2\phi}{\partial x^2} + \frac{\partial^2\phi}{\partial y^2} + \frac{\partial^2\phi}{\partial z^2} \] The first partial derivatives are: \(\frac{\partial\phi}{\partial x} = y+z\), \(\frac{\partial\phi}{\partial y} = x+z\), \(\frac{\partial\phi}{\partial z} = y+x\). The second partial derivatives are: \[ \frac{\partial^2\phi}{\partial x^2} = 0, \quad \frac{\partial^2\phi}{\partial y^2} = 0, \quad \frac{\partial^2\phi}{\partial z^2} = 0 \] Thus, \(\vec{abla} \cdot \vec{A} = 0+0+0=0\). Statement A is valid and B is invalid.
Step 3: Determine the correct statements.Both statements A and C are confirmed to be true.