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

`ointvecB.vecdvecI=mu_(0)(i_(1)+i_(3)+i_(2)-i_(3))=mu_(0)(i_(1)+i_(2))`
[since for the given direction of circumlation `i_(3)` entering at `PSTU ABCDA` is positive while `i_(3)` at `PQRS` is negative]
`oint vec(B).vec(d)l=oint vec(B).vec(d)l+oint vec(B).vec(d)l=mu_(0)i_(1)+mu_(0)i_(2)=mu_(0)(i_(1)+i_(2))`
`ABCDA ABCA CDAC`