Question:
The vector field $\vec{V} = e^{x}\sin y~i + e^{x}\cos y~j$ is}
The vector field $\vec{V} = e^{x}\sin y~i + e^{x}\cos y~j$ is}
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Fields that satisfy both conditions are often solutions to Laplace's equation.
Updated On: Jul 3, 2026
- Solenoidal but not Irrotational
- Irrotational but not Solenoidal
- Both Solenoidal and Irrotational
- neither Solenoidal nor Irrotational
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The Correct Option is C
Solution and Explanation
Step 1: Concept
Test for divergence ($\nabla \cdot \vec{V}$) and curl ($\nabla \times \vec{V}$).
Step 2: Meaning
$\nabla \cdot \vec{V} = \frac{\partial(e^x \sin y)}{\partial x} + \frac{\partial(e^x \cos y)}{\partial y} = e^x \sin y - e^x \sin y = 0$ (Solenoidal).
Step 3: Analysis
$\nabla \times \vec{V} = (\frac{\partial(e^x \cos y)}{\partial x} - \frac{\partial(e^x \sin y)}{\partial y})k = (e^x \cos y - e^x \cos y)k = 0$ (Irrotational).
Step 4: Conclusion
Since both divergence and curl are zero, the field is both solenoidal and irrotational. Final Answer: (C)
Test for divergence ($\nabla \cdot \vec{V}$) and curl ($\nabla \times \vec{V}$).
Step 2: Meaning
$\nabla \cdot \vec{V} = \frac{\partial(e^x \sin y)}{\partial x} + \frac{\partial(e^x \cos y)}{\partial y} = e^x \sin y - e^x \sin y = 0$ (Solenoidal).
Step 3: Analysis
$\nabla \times \vec{V} = (\frac{\partial(e^x \cos y)}{\partial x} - \frac{\partial(e^x \sin y)}{\partial y})k = (e^x \cos y - e^x \cos y)k = 0$ (Irrotational).
Step 4: Conclusion
Since both divergence and curl are zero, the field is both solenoidal and irrotational. Final Answer: (C)
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