Question

An arrangement of three parallel straight wires placed perpendicular to the plane of paper carrying the same current $I$ in the same direction as shown in figure. Find the magnitude of force per unit length on the wire $B$ is

• A. $\frac{{\mu }_0{I}^2}{2\pi d}$
• B. $\frac{{\mu }_0{I}^2}{\sqrt{2}\pi d}$
• C. $\frac{2{\mu }_0{I}^2}{\pi d}$
• D. $\frac{2{\mu }_0{I}^2}{3\pi d}$
  

Answer 1

The correct option is B $\frac{{\mu }_0{I}^2}{\sqrt{2}\pi d}$




Force between two long parallel current carrying conductors is,

$F=\frac{{\mu }_0{I}_1{I}_{2}L}{2\pi d}=\frac{{\mu }_0{I}^{2}L}{2\pi d}[\text{If}{I}_1=I_2=I]$

$\Rightarrow F_{BA}=\frac{{\mu }_0{I}^{2}L}{2\pi d}$

$\Rightarrow F_{BC}=\frac{{\mu }_0{I}^{2}L}{2\pi d}$

$\Rightarrow F_{BA}=F_{BC}=F$ and their directions are shown in figure

$\therefore F_{net}=\sqrt{2}F=\sqrt{2}\times \frac{{\mu }_0{I}^{2}L}{2\pi d}$

Now, force per unit length is,

$\frac{F_{net}}{L}=\sqrt{2}\times \frac{{\mu }_0{I}^2}{2\pi d}=\frac{{\mu }_0{I}^2}{\sqrt{2}\pi d}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.