Question

The length of conductor ab carrying current $I_2$ is $l$. The force acting on it due to a long current carrying conductor as shown in figure. The midpoint of wire ab is distance $x$ apart from long wire. The magnitude of the force on the wire is :

• A. $\frac{{\mu }_{0}I}{2\pi }lo{g}_e\frac{x+l/2}{x-l/2}$
• B. $\frac{{\mu }_{0}I}{2\pi x}$
• C. $\frac{{\mu }_{0}I}{2\pi }lo{g}_e\frac{x-l/2}{x+l/2}$
• D. ${\mu }_{0}Ilo{g}_e\frac{x-l/2}{x+l/2}$

Answer 1

The correct option is A $\frac{{\mu }_{0}I}{2\pi }lo{g}_e\frac{x+l/2}{x-l/2}$

Force on a small section of wire of length dx is given by:

$dF=B{I}_{2}dx$

$dF=\frac{{\mu }_o{I}_1}{2\pi x}{I}_{2}dx$

Integrating the force over length:

$\int dF={\int }_{x-\frac{l}{2}}^{x+\frac{l}{2}}\frac{{\mu }_0{I}_1}{2\pi x}{I}_{2}dx$

$F=\frac{{\mu }_0{I}_1{I}_2}{2\pi x}ln\left(\frac{x+\frac{l}{2}}{x-\frac{l}{2}}\right)$
$F=\frac{{\mu }_0{I}_1{I}_2}{2\pi r}ln\left(\frac{x+\frac{l}{2}}{x-\frac{l}{2}}\right)$