Magnetic Effects Of Current Question 154
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 wire1 [AIEEE 2002]
Options:
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 $
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Answer:
Correct Answer: C
Solution:
Length of the component dl which is parallel to wire (1) is $ dl\cos \theta $ , so force on it. $ F=\frac{{\mu_{0}}}{4\pi }.\frac{2i_{1}i_{2}}{r}(dl\cos \theta )=\frac{{\mu_{0}}i_{1}i_{2}dl\cos \theta }{2\pi r} $
BETA
