A square coil ABCD of side ' L ' is carrying a current in clockwise direction. A straight conductor carrying current I ₂ (upward direction) is kept parallel to side A B at a distance L 3 in the plane of A B C D . The net force on the coil ABCD is ( ₀= magnetic permeability)

Step-by-step solution

Magnetic force on a square coil near a straight current-carrying conductor The straight conductor along the y-axis carrying current I₂ produces a magnetic field at perpendicular distance x given by B = ₀ I₂ 2 x , directed into the page ( - k ) for x > 0 when I₂ flows upward. For side AB, located at distance L/3 from the conductor with current I₁ flowing downward ( L _ AB = -L j ), the magnetic field is B _ AB = - 3 ₀ I₂ 2 L k . The force is F _ AB = I₁ ( L _ AB B _ AB ) = 3 ₀ I₁ I₂ 2 i . For side CD, at distance 4L