Two long conductors separated by a distance 'd' carry currents ' I ₁ ' and ' I ₂ ' in the same directions. They exert a force ' F ' on each other. The distance between them is increased to ' 3 d '. If new repulsive force of magnitude '2/3 F' is found between these conductors, the required change in the magnitude and direction of one of the currents in the conductor is respectively [length of the conductors is constan

Step-by-step solution

The force per unit length between parallel current-carrying conductors is F = ₀ I₁ I₂ L 2 d , where currents in the same direction produce attraction. After increasing the separation to d' = 3d and reversing one current's direction to create repulsion, the new force magnitude becomes F' = 2 3 F . With I₁ unchanged and I₂' as the modified current, the force equation becomes: F' = ₀ I₁ I₂' L 2 (3d) = 2 3 ( ₀ I₁ I₂ L 2 d ) Canceling common terms yields I₂' 3 = 2 3 I₂ , so I₂' = 2I₂ . The current magnitude must double