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According to Ampere's circuital law...

According to Ampere's circuital law

A

`oint bar(B). bar(dI) = mu_(0) (i_(C ) + epsilon_(0) (d phi_E)/(dt))`

B

`oint bar(B). bar(dI) = mu_(0) epsilon_(0) (d phi_E)/(dt)`

C

`oint bar(B). bar(dI) = mu_(0) epsilon_(0) i`

D

`oint bar(B). bar(dI) = mu_(0)(i_C (d phi_E)/(dt) +i_(O) ) `

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The correct Answer is:
A
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