• # question_answer 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 points B and D. $\oint{\mathbf{B}dI\,\,l}$for this path according to Ampere's law will be                                       A)  $(-{{i}_{1}}-{{i}_{2}}+{{i}_{3}}){{\mu }_{0}}$      B)  $(-{{i}_{1}}+{{i}_{2}}){{\mu }_{0}}$ C)  ${{i}_{3}}{{\mu }_{0}}$                                               D)  $({{i}_{1}}+{{i}_{2}}){{\mu }_{0}}$

$\oint{\mathbf{B}.\,}dl={{\mu }_{0}}({{i}_{1}}+{{i}_{3}}+{{i}_{2}}-{{i}_{3}})={{\mu }_{0}}({{i}_{1}}+{{i}_{2}})$ [Since, for the given direction of circulation${{i}_{3}}$entering at PSTU is positive while${{i}_{3}}$at PQRS is negative]