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As per Ampere - Maxwell.s circuital law...

As per Ampere - Maxwell.s circuital law

A

`ointvecB.vecdl=mu_(0)I_(C )`

B

`ointvecB.vecdl=mu_(0)(I_(C)+I_(D))`

C

`ointvecBxxvecdl=mu_(0)(I_(C )+I_(D))`

D

`oint vecBxxvecdl=mu_(0)I_(C )+in_(0)(dphi_(E))/(dt)`

Text Solution

Verified by Experts

The correct Answer is:
B
As per Ampere - Maxwell.s modified circuital law
`ointvecB.vecdl=mu_(0)(I_(C )+I_(D))=mu_(0)[I_(C )+in_(0)(dphi_(E))/(dt)]`
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