Two long parallel wires A and B , made of the same material and having equal lengths, are placed at a fixed distance apart. The cross-sectional area of wire B is twice that of wire A . When they are connected in parallel across an ideal battery, the magnetic force per unit length between them is f₁ . When they are connected in series across the same battery, the magnetic force per unit length between them becomes f₂

Step-by-step solution

Let the resistance of wire B be R . Since wire A has half the cross-sectional area of wire B and the same length and material, its resistance is 2R . When connected in parallel across a battery of voltage V , the potential difference across each wire is V . Current in wire A , I_A = V 2R Current in wire B , I_B = V R The magnetic force per unit length between them is f₁ I_A I_B = ( V 2R ) ( V R ) = V^2 2R^2 . When connected in series across the same battery, the equivalent resistance is R_ eq = 2R + R = 3R . The cu