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Question

# Figure shows an Amperian path ABCDA. Part ABC is in verical 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 D.∮→B.d→l for this path according to Ampere's law will be :

A
(I1I2+I3)μ0
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B
(I1+I2)μ0
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C
I3μ0
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D
(I1+I2)μ0
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Solution

## The correct option is C (I1+I2)μ0μ0(Ienc)=∮¯B.¯dlArea vector of ABC is along the right direction, while ACD is along the upward direction.Now currents passing through ABC are I1,I3 along the area directions.Now through ACD, I2,I3;I2 along the area vector while I3 is opposite to it.Thus, μ0(I1+I3+I2−I3)=∮¯B.¯dl=μ0(I1+I2)

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