For A = x i + y j + z k, each component depends only on its corresponding coordinate. All cross-partial derivatives in ∇×A are zero, so the field is curl-free (irrotational).
Formula / Key Relation
∇×A = 0
A = ∇[(x²+y²+z²)/2]
By Stokes:
∮ A·dl = ∬(∇×A)·dS = 0
Detailed Explanation
Concept & ReasoningCompute curl directly:∇×A = (∂Az/∂y - ∂Ay/∂z)i + (∂Ax/∂z - ∂Az/∂x)j + (∂Ay/∂x - ∂Ax/∂y)k.For Ax=x, Ay=y, Az=z, every cross derivative is zero. Hence curl A = 0. In fact A = ∇[(x²+y²+z²)/2], so it is a conservative gradient field. Stokes' theorem is consistent with this: the circulation around any closed contour in a simply connected region is zero.How to Solve It in the ExamComponents x,y,z with no cross-coordinate dependence usually make curl checks very quick.Important Exam PointWrite the determinant/cross-derivative form once and inspect which derivatives vanish.Related Revision PathThis question sits in the revision path Electromagnetic Fields →…
Concept & Reasoning
Compute curl directly: ∇×A = (∂Az/∂y - ∂Ay/∂z)i + (∂Ax/∂z - ∂Az/∂x)j + (∂Ay/∂x - ∂Ax/∂y)k. For Ax=x, Ay=y, Az=z, every cross derivative is zero. Hence curl A = 0. In fact A = ∇[(x²+y²+z²)/2], so it is a conservative gradient field. Stokes' theorem is consistent with this: the circulation around any closed contour in a simply connected region is zero.
How to Solve It in the Exam
Components x,y,z with no cross-coordinate dependence usually make curl checks very quick.
Important Exam Point
Write the determinant/cross-derivative form once and inspect which derivatives vanish.
Related Revision Path
This question sits in the revision path Electromagnetic Fields → Vector Calculus → Curl .
Quick Trick
Components x,y,z with no cross-coordinate dependence usually make curl checks very quick.
Common Mistake
Divergence is 3 here, so the field is not solenoidal; curl-free does not mean divergence-free.
Exam Tip
Write the determinant/cross-derivative form once and inspect which derivatives vanish.
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