Six perpendicular wires carry currents I₁=1 A, I₂=2 A, I₃=3 A, I₄=1 A, I₅=5 A and I₆=4 A as shown. If ∮B⃗·dℓ⃗=αμ₀/2, find α.

Question

G1 Physics · Accepted Answer

Ampere's law gives ∮B⃗·dℓ⃗=μ₀I enclosed. From the figure, signed enclosed current is (1+2+3)-(1+4)=1 A. Thus μ₀=αμ₀/2, so α=2.