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If \vec{A} = ay \hat{x} + bx \hat{y}, then curl will be:

Option: 1

(a - b) \hat{z}


Option: 2

(b - a) \hat{z}


Option: 3

(a + b) \hat{z}


Option: 4

(a - b) \hat{x}


Answers (1)

best_answer

If 

\vec{A} = ay \hat{x} + bx \hat{y}

\vec{\bigtriangledown } \times \vec{A} = curl = \begin{bmatrix} &\hat{i} & \hat{j} &\hat{k} \\ &\frac{\delta}{\delta x} &\frac{\delta}{\delta y} & \frac{\delta}{\delta z}\\ &ay & bx &0 \\ \end{bmatrix}

\vec{\bigtriangledown } \times \vec{A} = \hat{i} [\frac{\delta}{\delta y}(0) - \frac{\delta}{\delta z}bx] - \hat{j} [\frac{\delta}{\delta x}(0) - \frac{\delta}{\delta z}ay] + \hat{k} [\frac{\delta}{\delta x}(0) - \frac{\delta}{\delta y}ay]

\vec{\bigtriangledown } \times \vec{A} = \hat{i} [0] - \hat{j} [0] + \hat{k} [b - a]

\vec{\bigtriangledown } \times \vec{A} = [b - a]\hat{z}

Posted by

rishi.raj

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