A square loop ABCD of side L, carrying a current i, is placed near and coplanar with a long straight conductor XY carrying a current I. The side nearer the conductor is at a distance L/2 from it, so the far side is at (3L)/2. The net force on the loop will be

Question

Accepted Answer

(3) (2μ₀Ii)/(3π). Given: a square loop of side L with its near side at distance L/2 from the wire and its far side at (3L)/2; the two sides perpendicular to the wire feel equal and opposite forces that cancel. Idea: the field of the wire falls as 1/d, so the near side is pulled harder than the far side is pushed. Near side: F₁=(μ₀Ii)/(2π)·L/(L/2)=(μ₀Ii)/(2π)×2, towards the wire. Far side: F₂=(μ₀Ii)/(2π)·L/(3L/2)=(μ₀Ii)/(2π)×2/3, away from the wire. Net: F=(μ₀Ii)/(2π)(2-2/3)=(μ₀Ii)/(2π)×4/3. F=(2μ₀Ii)/(3π) — with no L left, as it must be.

A square loop ABCD of side L, carrying a current i, is placed near and coplanar with a long straight conductor XY carrying a current I. The side nearer the conductor is at a distance L/2 from it, so the far side is at (3L)/2. The net force on the loop will be

Step-by-step solution

Why the other options are wrong