CBSE Class 12-science Answered
v=e^xyz(i+j+k) find curl v
Asked by abhimeena1011 | 18 Jan, 2022, 10:43: AM
Given: v = exyz(i+j+k)
![curl left parenthesis straight v right parenthesis space equals space open vertical bar table row straight i straight j straight k row cell fraction numerator partial differential over denominator partial differential straight x end fraction end cell cell fraction numerator partial differential over denominator partial differential straight y end fraction end cell cell fraction numerator partial differential over denominator partial differential straight z end fraction end cell row cell straight e to the power of xyz end cell cell straight e to the power of xyz end cell cell straight e to the power of xyz end cell end table close vertical bar equals straight i open parentheses fraction numerator partial differential over denominator partial differential straight y end fraction open parentheses straight e to the power of xyz close parentheses minus fraction numerator partial differential over denominator partial differential straight z end fraction open parentheses straight e to the power of xyz close parentheses close parentheses minus straight j open parentheses fraction numerator partial differential over denominator partial differential straight x end fraction open parentheses straight e to the power of xyz close parentheses minus fraction numerator partial differential over denominator partial differential straight z end fraction open parentheses straight e to the power of xyz close parentheses close parentheses plus straight k open parentheses fraction numerator partial differential over denominator partial differential straight x end fraction open parentheses straight e to the power of xyz close parentheses minus fraction numerator partial differential over denominator partial differential straight y end fraction open parentheses straight e to the power of xyz close parentheses close parentheses](https://images.topperlearning.com/topper/tinymce/cache/eeab5492a98b23976cce64978b2a0687.png)
Answered by Renu Varma | 25 Jan, 2022, 09:56: AM
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