Question

The segment $\text{AB}$ of wire carrying current $I_1$ is placed perpendicular to a long straight wire carrying current $I_2$ as shown in figure. The magnitude of force experienced by straight wire $\text{AB}$ is:

• A. $\frac{{\mu }_0{I}_1{I}_2}{2\pi }\left(ln3\right)$
• B. $\frac{{\mu }_0{I}_1{I}_2}{2\pi }\left(ln2\right)$
• C. $\frac{2{\mu }_0{I}_1{I}_2}{2\pi }$
• D. $\frac{{\mu }_0{I}_1{I}_2}{2\pi }$

Answer 1

The correct option is B $\frac{{\mu }_0{I}_1{I}_2}{2\pi }\left(ln2\right)$

Taking an infinitesimally small current element of length $dx$ on wire $\text{AB}$ at a distance $x$ from wire carrying current $I_2$,

The magnetic field at this element will be:

$dB=\frac{{\mu }_0{I}_2}{2\pi x}⨀$

The magnetic force experience by small element $dx$ is given by,

$dF=I_1(d\overrightarrow{l}\times d\overrightarrow{B})$

$dF=I_{1}dl\left(\frac{{\mu }_0{I}_2}{2\pi x}\right)\sin {90}^{\circ }$

$dF=\frac{{\mu }_0{I}_1{I}_2}{2\pi x}dx$

Putting the limits for $x=a$ to $x=2a$ and integrating we get net force on wire $\text{AB}$:

$F=\int dF$

$F=\frac{{\mu }_0{I}_1{I}_2}{2\pi }{\int }_{x=a}^{x=2a}\frac{dx}{x}$

$F=\frac{{\mu }_0{I}_1{I}_2}{2\pi }[lnx{]}_a^{2a}$

$F=\frac{{\mu }_0{I}_1{I}_2}{2\pi }[ln2a-lna]$

$F=\frac{{\mu }_0{I}_1{I}_2}{2\pi }\left[ln\left(\frac{2a}{a}\right)\right]$

$\therefore F=\frac{{\mu }_0{I}_1{I}_2}{2\pi }ln2$

Hence, option $\left(b\right)$ is the correct answer.