A) \[\frac{{{\mu }_{o}}I\,(y\hat{i}-x\hat{j})}{2\pi ({{x}^{2}}+{{y}^{2}})}\]
B) \[\frac{{{\mu }_{o}}I\,(x\hat{i}+y\hat{j})}{2\pi ({{x}^{2}}+{{y}^{2}})}\]
C) \[\frac{{{\mu }_{o}}I\,(x\hat{j}-y\hat{i})}{2\pi ({{x}^{2}}+{{y}^{2}})}\]
D) \[\frac{{{\mu }_{o}}I\,(x\hat{i}-y\hat{j})}{2\pi ({{x}^{2}}+{{y}^{2}})}\]
Correct Answer: A
Solution :
Magnetic field at P is \[\overrightarrow{B}\], perpendicular to OP in the direction shown in figure. So, \[\overrightarrow{B}=B\sin \theta \,\hat{i}-B\cos \theta \,\hat{j}\] Here \[B=\frac{{{\mu }_{0}}}{2\pi }\frac{I}{r}\] \[\sin \theta =\frac{y}{r}\] and \[\cos \theta =\frac{x}{r}\] \ \[\overrightarrow{B}=\frac{{{\mu }_{0}}I}{2\pi }\cdot \frac{1}{{{r}^{2}}}(y\hat{i}-x\hat{j})=\frac{{{\mu }_{0}}I(y\hat{i}-x\hat{j})}{2\pi ({{x}^{2}}+{{y}^{2}})}\](as \[{{r}^{2}}={{x}^{2}}+{{y}^{2}}\])You need to login to perform this action.
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