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

• A. $-2F$
• B. $\frac{F}{3}$
• C. $\frac{-2F}{3}$
• D. $\frac{-F}{3}$

Answer 1

The correct option is C $\frac{-2F}{3}$

$\text{Magnetic field due to a wire at a distance}d\text{is}B=\frac{{\mu }_0{I}_1}{2\pi d}$

$\text{Force on other wire due to this magnetic field of length}l\text{is}F=B{I}_{2}l=\frac{{\mu }_0{I}_1}{2\pi d}{I}_{2}l$

$\text{Therefore force per unit length}=F/l=\frac{{\mu }_0{I}_1{I}_2}{2\pi d}$

$\Rightarrow F/l\propto \frac{I_1{I}_2}{d}$

$\text{When current becomes two times and distance is increased to three times,}$

$F_1=\frac{2F}{3}$

$\text{Since the direction of current is also reversed in one of them, the direction of force also reverses.}$

$\Rightarrow F^{{}^{\prime }}=\frac{-2F}{3}$