Figure shows an Amperian path ABCDA. Part ABC is in vertical plane PSTU while part CDA is in horizontal plane PQRS. Direction of circulation along the path is shown by an arrow near point B and at D. `ointvecB.dvecl` for this path according to Ampere's law will be

Options

A. `(i_(1)+i_(2)+i_(3))mu_(0)`
B. `(-i_(1)+i_(2))mu_(0)`
C. `(i_(3)mu_(0)`
D. `(i_(1)+i_(2))mu_(0)`

Answer

D

Explanation

If the loop current is coming out it should be taken as positive and if it is going in it is taken as negative.
`i_(3)` is coming out and again going in so its net contribution will be zero.
According to Ampere.s law
`ointvecB.vec(dl)=mu_(0)(Sigmai_("inside"))`
`=mu_(0)(i_(1)+i_(2)+i_(3)-i_(3))`
`=(mu_(0)(i_(1)+i_(2))`