Q. 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) $\frac{\mu_{0}}{2\pi r}i_{1}i_{2}\ dl\tan\theta$
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) $\frac{\mu_{0}}{4\pi r}i_{1}i_{2}\ dl\sin\theta$
Explanation
(c) Length of the component $dl$ which is parallel to wire (1) is $dl\cos\theta,$ so force on it
$F\text{=}\frac{\mu_{0}}{4\pi}.\frac{2i_{1}i_{2}}{r}\left( dl\cos\theta \right) = \frac{\mu_{0}i_{1}i_{2}dl\cos\theta}{2\pi r}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.