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 increases to $3d$. The new value of the force between them is:

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

Answer 1

The correct option is C $-2F/3$

Two conductors, separated by a distance $=d$

current $=I_1$ and $I_2$

They exert a force $=F$

Force acting between two current carrying conductors,

$F=\frac{{\mu }_0}{4\pi }\frac{I_1{I}_2}{d}⟶\left(i\right)$

where, $d=$ distance between the conductors

$I=$ length of each conductor

$F^1=\frac{{\mu }_c}{2\pi }\frac{\left({-2I}_1\right)\left(I_2\right)}{3d}-1$

$=-\frac{{\mu }_0}{2\pi }\frac{\left({-2I}_1\right)\left(I_2\right)}{3d}-1$

$=-\frac{{\mu }_0}{2\pi }\frac{2I_1{I}_2}{3d}-1⟶\left(ii\right)$

thus, from equation $\left(i\right)$ and $\left(ii\right)$

$\frac{F^1}{F}=-\frac{2}{3}$

$F^1=-\frac{2F}{3}$