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

Question

Dr. Abubacker · Accepted Answer

(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, $F^{'} = \frac{\mu_{0}}{2\pi}\frac{\left( - 2I_{1} \right)(I_{2})}{(3d)}.l$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \  = - \frac{\mu_{0}}{2\pi}\frac{2I_{1}I_{2}}{d}.l\ \ldots(ii)$
Thus, from Eqs.($i$) and $ii$
$\frac{F^{'}}{F} = - \frac{2}{3} \Longrightarrow F^{'} = - \frac{2}{3}F$

Explanation

Reviewed by Dr. Abubacker

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