1.

As per Ampere - Maxwell.s circuital law

Answer»

`ointvecB.vecdl=mu_(0)I_(C )`
`ointvecB.vecdl=mu_(0)(I_(C)+I_(D))`
`ointvecBxxvecdl=mu_(0)(I_(C )+I_(D))`
`oint vecBxxvecdl=mu_(0)I_(C )+in_(0)(dphi_(E))/(DT)`

Solution :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)]`


Discussion

No Comment Found

Related InterviewSolutions