question_answer

A long straight wire along the z-axis carries a current I in the negative z direction. The magnetic vector field \[\overset{\to }{\mathop{B}}\,\] at a point having coordinates (x, y)  in the z = 0 plane is  [IIT-JEE (Screening) 2002]

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 \[{{r}^{2}}={{x}^{2}}+{{y}^{2}}\])