Question

Wires $1$ and $2$ carrying $i_1$ and $i_2$, respectively are inclined at an angle $\theta$ to each other. What is the force on a small element $dl$ of wire $2$ at a distance of $r$ from wire $1$(as shown in figure) due to the magnetic field of wire $1$?

• A. $\frac{{\mu }_0}{2\pi r}{i}_1{i}_{2}dl\tan \theta$
• B. $\frac{{\mu }_0}{2\pi r}{i}_1{i}_{2}dl\sin \theta$
• C. $\frac{{\mu }_0}{2\pi r}{i}_1{i}_{2}dl\cos \theta$
• D. $\frac{{\mu }_0}{4\pi r}{i}_1{i}_{2}dl\sin \theta$

Answer 1

The correct option is C $\frac{{\mu }_0}{2\pi r}{i}_1{i}_{2}dl\cos \theta$

Let us calculate,magnetic field field due to $I_1$ at P

$B=\frac{{\mu }_0}{4\pi }\frac{I}{r}[\cos \theta +\cos \theta ]$

$=\frac{{\mu }_{0}I}{2\pi r}\cos \theta$ [direct perpendicular to plane]

Force due to $I_{2}dl$ at wire 1 is

$dF=I_{2}dlB\sin 90°$

$=\frac{I_{2}dl}{r}\frac{{\mu }_0}{2\pi }{I}_1\cos \theta$

$⟹dF=\frac{{\mu }_0}{2\pi }\frac{I_1{I}_2}{r}dl\cos \theta$