A square loop ABCD of side L carrying a current I₁ is placed at a distance (L/3) from a conductor coplaner with a straight conductor XY carrying current I₂ as shown in figure. The net force on the loop will be ( ₀ = magnetic permeability)

Step-by-step solution

The magnetic field produced by the straight wire XY at a distance r is given by B = ₀ I₂ 2 r , directed into the plane of the paper. The force on the side AB of the square loop, which is at a distance r = L 3 and carries current I₁ in the same direction as I₂ , is attractive: F_ AB = I₁ L B_ AB = I₁ L ( ₀ I₂ 2 (L/3) ) = 3 ₀ I₁ I₂ 2 The force on the side CD, which is at a distance r = L 3 + L = 4L 3 and carries current I₁ in the opposite direction to I₂ , is repulsive: F_ CD = I₁ L B_ CD = I₁ L ( ₀ I₂ 2 (4L/3) ) = 3