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) \[\frac{{{\mu }_{0}}}{4\pi r}{{i}_{1}}\,{{i}_{2}}\,dl\,\sin \theta \]
Correct Answer: C
Solution :
Length of the component dl which is parallel to wire (1) is \[dl\cos \theta \], so force on it. \[F=\frac{{{\mu }_{0}}}{4\pi }.\frac{2{{i}_{1}}{{i}_{2}}}{r}(dl\cos \theta )=\frac{{{\mu }_{0}}{{i}_{1}}{{i}_{2}}dl\cos \theta }{2\pi r}\]
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