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Question:

A square loop ABCD, carrying a current I1, is placed near and coplanar with a long straight conductor XY carrying a current I1, as shown in figure. The net force on the loop will be

2μ₀I₁I₂L/2π

μ₀I₁I₂/2π

μ₀I₁I₂L/2π

2μ₀I₁I₂/3π

Solution:

Force on arm AB due to current in conductor XY is F₁ = μ₀/4π * 2I₁I₂L(L/2) = μ₀I₁I₂/2π acting towards the wire in the plane of loop.
Force on arm CD due to current in conductor XY is F₂ = μ₀/4π * 2I₁I₂L(3L/2) = 3μ₀I₁I₂/2π away from the wire the plane of loop.
∴Net force on the loop = F₁ - F₂ = μ₀I₁I₂/2π[1 - 3] = -μ₀I₁I₂/π.