Two long straight wires A and B carrying equal current 'I' were kept parallel to each other at distance ' d ' apart. Magnitude of magnetic force experienced by length L of wire A is ' F '. If the distance between the wires is made half and currents are doubled, force F₂ on length L of wire A will be

Step-by-step solution

The magnetic force on a length L of wire A due to wire B is given by F = ₀ I_A I_B L 2 d . Initially, with currents I_A = I_B = I and separation d₁ = d , the force is F₁ = ₀ I^2 L 2 d . When both currents are doubled to 2I and the separation is halved to d/2 , the new force becomes F₂ = ₀ (2I)^2 L 2 (d/2) = 4 ₀ I^2 L d . Substituting the expression for F₁ , we find F₂ = 8 ( ₀ I^2 L 2 d ) = 8F₁ . The final force on length L of wire A is 8F .