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Three infinitely-long conductors carrying currents $I_{1}, I_{2} a n d I_{3}$ lie perpendicular to the plane of the paper as shown below. If the value of integral $\oint_{C} \vec{B} . d \vec{l}$ for the loops $C_{1} {,C}_{2} {and C}_{3}$ are ${2μ}_{0} {,4μ}_{0} {and μ}_{0}$ in SI units, respectively, then

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$I_{3} =0$

$I_{2} =5A$ out of the paper

$I_{2}$ is 4 Amp out of the paper

$I_{1} =3A$ into the paper

answer is A, B.

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Detailed Solution

$I_{1} {+I}_{2} =2 I_{2} {+I}_{3} =4 I_{1} {+I}_{2} {+I}_{3} = 1 I_{2} = 5A out of the paper I_{3} = 1A into paper I_{1} = 3A into the paper$

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