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

Question

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

Infinity Learn · Accepted Answer

∮B→·d→ℓ$$=$$μ$$0I1+I3+I2-I3=$$μ$$0I1+I2$$  [Since for the given direction of circulation $$I3$$ enteringatPSTU  is positive while $$I3$$ at PQRS is negative] Alternative solution: ∮ABCDAB→·dl→$$=$$∮ABCAB→·dl→$$+$$∮CDACB→·dl→ $$=$$μ$$0I1+I3+$$μ$$0I2-I3=$$μ$$0I1+I2$$

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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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.∮B→·dℓ→ for this path according to Ampere’s law will be

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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