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Figure shows an Amperian path ABCDA. Par...

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 point B and at D. `ointvecB.dvecl` 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_3mu_0`

D

`(I_1+I_2)mu_0`

Text Solution

Verified by Experts

The correct Answer is:
D
(d) `oint vecB.dvecl=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]
Alternative solution: `oint_(ABCDA) vecB.vec(dl)=oint_(ABCA)vecB.vec(dl)+oint _(CDAC) vecB.vec(dl)`
`=mu_0(I_1+I_3)+mu_0(I_2-I_3)=mu_0(I_1+I_2)`
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