Question

Wires 1 and 2 carrying currents $i_1$ 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. $\frac{{\mu }_0}{2\pi r}{i}_1{i}_{2}dltan\theta$
  
• B. $\frac{{\mu }_0}{2\pi r}{i}_1{i}_{2}dlsin\theta$
  
• C. $\frac{{\mu }_0}{2\pi r}{i}_1{i}_{2}dlcos\theta$
  
• D. $\frac{{\mu }_0}{4\pi r}{i}_1{i}_{2}dlsin\theta$
  

Answer 1

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


Length of the component dl which is parallel to wire (1) is $dlcos\theta$, so force on it.
$F=\frac{{\mu }_0}{4\pi }.\frac{2i_1{i}_2}{r}\left(dlcos\theta \right)=\frac{{\mu }_0{i}_1{i}_{2}dlcos\theta }{2\pi r}$