Question

Two long conductors, separated by a distance d carry currents $ {{I}_{1}} $ and $ {{I}_{2}} $ in the same direction. They exert a force F on each other. Now the current in one of them is increased to two times and its direction is reversed. The distance is also increased to 3 d. The new value of the force between them is

• A. $ -2F $
• B. $ F/3 $
• C. $ -2F/3 $
• D. $ -F/3 $

Answer 1

Answer: (C)

Force acting between two current carrying conductors $ F=\frac{{{\mu }_{0}}}{2\pi }\frac{{{I}_{1}}{{I}_{2}}}{d}l $ ?(i) where $ d= $ distance between the conductors, $ l= $ length of each conductor. Again $ F=\frac{{{\mu }_{0}}}{2\pi }\frac{(-2{{I}_{1}})({{I}_{2}})}{(3d)}.l $ $ =-\frac{{{\mu }_{0}}}{2\pi }\frac{2{{I}_{1}}{{I}_{2}}}{3d}.l $ ...(ii) Thus, from Eqs. (i) and (ii) $ \frac{F}{F}=-\frac{2}{3} $ $ \Rightarrow $ $ F=-\frac{2}{3}F $