A) \[\frac{\sqrt{2}{{\mu }_{0}}I}{(4\pi R)}\]out of plane of paper
B) \[\frac{\sqrt{2}{{\mu }_{0}}I}{4\pi R}\]into the plane of paper
C) \[\frac{(\sqrt{2}-1){{\mu }_{0}}I}{4\pi R}\]out of plane of paper
D) \[\frac{(\sqrt{2}-1){{\mu }_{0}}I}{4\pi R}\]into the plane of paper
Correct Answer: C
Solution :
\[B=\frac{{{\mu }_{0}}I}{4\pi d}[\sin \alpha +\sin \beta ]\] [Magnetic field due to current carrying wire] \[=\frac{{{\mu }_{0}}I}{4\pi d(R/\sqrt{2})}\left[ \sin \left( 2\pi -\frac{\pi }{4} \right)+\sin {{90}^{o}} \right]\] \[=\frac{{{\mu }_{0}}I}{4\pi R}\times \sqrt{2}\left[ 1-\frac{1}{\sqrt{2}} \right]\] \[=\frac{{{\mu }_{0}}I}{4\pi R}\times (\sqrt{2}-1)\] (Out of plane of paper)You need to login to perform this action.
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