Questions

# 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.$\oint \stackrel{\to }{B}·d\stackrel{\to }{\ell }$ for this path according to Ampere’s law will be

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a
I1-I2+I3μ0
b
-I1+I2μ0
c
I3μ0
d
I1+I2μ0

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detailed solution

Correct option is D

∮B→·d→ℓ=μ0I1+I3+I2-I3=μ0I1+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→ =μ0I1+I3+μ0I2-I3=μ0I1+I2

A cylindrical wire of radius R is carrying uniformly distributed current i over its cross-section. If a circular loop of radius r is taken as amperian loop, then the variation value of  $\oint \stackrel{\to }{B}·\stackrel{\to }{d\ell }$ over this loop with radius ‘r” of loop will be best represented by