Figure shows an Amperian path ABCDA. Part ABC is in vertical plane PSTU while part CDA is in horizontal plane POQRS. Direction of circulation along the path is shown by an arrow near point B and at D.
$\oint \vec{B} \cdot d \vec{ℓ}$ for this path according to Ampere’s law will be

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$(I_{1} - I_{2} + I_{3}) μ_{0}$

$I_{3} μ_{0}$

$(- I_{1} + I_{2}) μ_{0}$

$(I_{1} + I_{2}) μ_{0}$

answer is D.

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Detailed Solution

$\oint \vec{B} \cdot \vec{d} ℓ = μ_{0} (I_{1} + I_{3} + I_{2} - I_{3}) = μ_{0} (I_{1} + I_{2}) [Since for the given direction of circulation I_{3} enteringat P S T U is positive while I_{3} at P Q R S is negative] A l t e r n a t i v e s o l u t i o n : \oint_{A B C D A} \vec{B} \cdot \vec{d l} = \oint_{A B C A} \vec{B} \cdot \vec{d l} + \oint_{C D A C} \vec{B} \cdot \vec{d l} = μ_{0} (I_{1} + I_{3}) + μ_{0} (I_{2} - I_{3}) = μ_{0} (I_{1} + I_{2})$