A square loop of length L is placed with its edges parallel to the X Y -axes. The loop is carrying the current I . If the magnetic field in the region varies as B =B₀ (1+ X y L^2 ) k , then the magnitude of the force on the loop will be

Step-by-step solution

We take one vertex of square loop at origin, for wire A B , magnetic field A B=B₀(1+0)=B₀ . n=0 For wire C D , magnetic field =B₀ (1+ y L ) , x=L For wire B C , magnetic field =B₀ (1+ L ) , y=L and for wire A D , magnetic field =B₀ y=0 Now me calculate forces F₁, F₂, F₃ and F₄ , aligned F₁ & = ₀^L I B₀ dy =I B₀ L F₂ & = ₀^L I B₀ (1+ x L ) d x & = 3 2 I B₀ L F₃ & = ₀^I I B₀ (1+ y L ) d x & = 3 2 I B₀ L F₄ & = ₀^L I B₀ d x=I B₀ L aligned