Q. 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) $- 2\ F$
B) $F/3$
C) $- 2F/3$
D) $- F/3$
Explanation
(c) Force acting between two current carrying conductors
$F = \frac{\mu_{0}}{2\pi}\frac{I_{1}I_{2}}{d}\ l\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
Where, $d =$ distance between the conductors,
$\ \ l =$ length of each conductor.
Again, $F^{'} = \frac{\mu_{0}}{2\pi}\frac{\left( - 2I_{1} \right)\left( I_{2} \right)}{(3d)}\ .l$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = - \frac{\mu_{0}}{2\pi}\frac{2I_{1}I_{2}}{3d}\ .l\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(ii)$
Thus, from Eqs. (i) and (ii)
$\frac{F^{'}}{F} = - \frac{2}{3}$
$\Rightarrow \ \ F^{'} = - \frac{2}{3}\ F$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.