Question

The segment 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 the straight wire AB is

• A. $\frac{{\mu }_o{I}_1{I}_2}{2\pi }1n3$
• B. $\frac{{\mu }_o{I}_1{I}_2}{2\pi }1n2$
• C. $\frac{2{\mu }_o{I}_1{I}_2}{2\pi }$
• D. $\frac{{\mu }_o{I}_1{I}_2}{2\pi }$

Answer 1

The correct option is B $\frac{{\mu }_o{I}_1{I}_2}{2\pi }1n2$

The magnetic filed at distance $x$ due to long wire is $B=\frac{{\mu }_0{I}_2}{2\pi x}$

Now the force experienced by small element $dx$ of wire AB is

$dF=I_{1}dxB\sin 90$

or $F=\frac{{\mu }_0{I}_1{I}_2}{2\pi }{\int }_a^{2a}\frac{dx}{x}=\frac{{\mu }_0{I}_1{I}_2}{2\pi }ln2$