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Six wires of current $I_{1} = 1 A, I_{2} = 2 A, I_{3} = 3 A, I_{4} = 1 A, I_{5} = 5 A$ and I₆ = 4A cut the page perpendicular at the points 1, 2, 3, 4, 5 and 6 respectively as shown in the figure. Find value of the integral $∮ \vec{B} \cdot d \vec{l}$ around the closed path.

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$2 μ_{0}$

$3 μ_{0}$

$4 μ_{0}$

$μ_{0}$

answer is A.

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Detailed Solution

$μ_{0} i_{net} = ∮ \vec{B} \cdot d \vec{l}$

$i_{n e t} = 1 A; ∮ \vec{B} \cdot d \vec{l} = μ_{0}$

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