Question

A conductor $\text{PQ}$ of length $b$ carries a current $i_2$ is placed perpendicular to a long conductor $\text{XY}$ carrying a current $i_1$ as shown in figure. The magnitude and direction of the magnetic force acting on $\text{PQ}$ is

Both the conductors are in $xy-$plane.

Answer 1

Let us consider, a current element of size $dx$ on $\text{PQ}$ at a distance $x$ from the current carrying conductor $\text{PQ}$ as shown in figure.


Force on the current element,

$\Rightarrow \overrightarrow{dF}=i_2(\overrightarrow{dx}\times \overrightarrow{B})$

$=i_2[dx\hat{i}\times \frac{{\mu }_0{i}_1}{2\pi x}(-\hat{k}\left)\right][B:\text{is the field due to XY}]$

$=\frac{{\mu }_0{i}_1{i}_{2}dx}{2\pi x}\hat{j}$

Integrating, both sides with proper limit,

${\int }_0^{F}dF=\frac{{\mu }_0{i}_1{i}_2}{2\pi }{\int }_a^{a+b}\frac{1}{x}dx$

$\Rightarrow F=\frac{{\mu }_0{i}_1{i}_2}{2\pi }\ln \left(\frac{a+b}{a}\right)$

Since, the direction of the force is along $+\hat{j}$ direction,

$\therefore \overrightarrow{F}=\frac{{\mu }_0{i}_1{i}_2}{2\pi }\ln \left(\frac{a+b}{a}\right)\hat{j}$

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

Why this question? To understand the concept of force between two straight current carrying conductors.