Question

An arrangement of three infinitely long straight wires placed perpendicular to plane of paper carrying same current $I$ along the same direction is shown in the figure. The magnitude of force per unit length on wire $\text{B}$ is

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

Answer 1

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

Clearly, the magnetic field due to wire $\text{C}$ ie ${\text{B}}_C$ is pointing downwards as shown in the figure.

Hence, the force on wire $\text{B}$ due to wire $\text{C}$ ie ${\text{F}}_C$ is pointing rightwards as shown below.

Similarly, the magnetic field due to wire $\text{A}$ ie ${\text{B}}_A$ is pointing leftwards as shown in the figure.

Hence, the force on wire $\text{B}$ due to wire $\text{A}$ ie ${\text{F}}_A$ is pointing downwards as shown in the figure.


$\frac{|\overrightarrow{F_A}|}{l}=\frac{|\overrightarrow{F_C}|}{l}=\frac{{\mu }_0{I}^2}{2\pi d}=f\left(say\right)$

$\frac{F_{net}}{l}=\sqrt{f^2+f^2+2f.f.\cos {90}^{\circ }}=f\sqrt{2}$

$\because \frac{F_{net}}{l}=\frac{{\mu }_0{I}^2\sqrt{2}}{2\pi d}=\frac{{\mu }_0{I}^2}{\sqrt{2}.\pi d}$

Why this Question? Note: The magnitude of force per unit length between two parallel current carrying infinite wires having currents $I_1$ and $I_2$ is given by, $F=\frac{{\mu }_0{I}_1{I}_2}{2\pi d}$ Where, $d$ = seperation between the two parallel wires.