AIEEE Solved Paper-2002
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) \[\frac{{{\mu }_{0}}}{2\pi r}{{i}_{1}}{{i}_{2}}\,dl\sin \theta \]
Correct Answer: C
Solution :
The component \[d\,l\cos \theta \] of element dl is parallel to the length of the wire 1. Hence, force on this elemental component \[F=\frac{{{\mu }_{0}}}{4\pi }.\frac{2{{i}_{1}}{{i}_{2}}}{r}(d\,l\cos \theta )\] \[=\frac{{{\mu }_{0}}\,{{i}_{1}}\,{{i}_{2}}d\,l\cos \theta }{2\pi r}\]
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