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  • The figure shows an amperian path  Part <img alt="ABC" src="https://entrancecorner.oncodecogs.com/gif.latex?ABC"

The figure shows an amperian pathABCDA  Part ABC is in the vertical plane PSTUwhile part  CDA  is in the horizontal plane PQRS . The direction of circulation along the path is shown by an arrow near point Band D. 

   \oint \vec{B}.\vec{dl} for this path according to Ampere’s law will be :  

                                                              

Option: 1

\left ( i_{1} -i_{2}+i_{_{3}}\right )\mu _{0}

 

 


Option: 2

\left (- i_{1} +i_{2} )\mu _{0}


Option: 3

i_{3}\mu _{0}


Option: 4

\left ( i_{1}+i_{2} \right )\mu _{0}


Answers (1)

best_answer

\oint _{ABCDA}\vec{B}\vec{d}l = \mu _{0}\left ( i_{1} +i_{3}+i_{2}-i_{3}\right )=\mu _{0}\left ( i_{1}+i_{2} \right )

[Since for the given direction of circulation i_{3} entering at PSTU is positive while i_{3} at PQRS is negative]

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Rakesh

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