Question

An arrangement of three parallel straight wires placed perpendicular to plane of paper carrying same current '$I$' along the same direction is shown in Fig. Magnitude of force per unit length on the middle wire '$B$' is given by:

• A. $\frac{{\mu }_o{i}^2}{\sqrt{2}\pi d}$
• B. $\frac{{\mu }_o{i}^2}{2\pi d}$
• C. $\frac{2{\mu }_o{i}^2}{\pi d}$
• D. $\frac{\sqrt{2}{\mu }_o{i}^2}{\pi d}$

Answer 1

The correct option is A $\frac{{\mu }_o{i}^2}{\sqrt{2}\pi d}$

Magnetic field due to current

carrying wire $=\frac{{\mu }_{0}I}{2\pi d}$

force due to magnetic

per field on current carrying wire per unit length

$\begin{array}{c}\left(\frac{{\mu }_{o}I}{2\pi d}\right)Il=\frac{{\mu }_0{I}_2}{2\pi d}\\ \frac{{\mu }_o{I}^2}{2\pi d}\end{array}$

Resultant = $\sqrt{{\left(\frac{{\mu }_o{I}^2}{2\pi d}\right)}^2}=\frac{{\mu }_o{I}^2}{\sqrt{2}\pi d}$