question_answer

Two identical conducting wires AOB and COD are placed at right angles to each other. The wire AOB carries an electric current\[f(x)\]and COD carries a current\[x=0\]. The magnetic field on a point lying at a distance d from O, in a direction perpendicular to the plane of the wires AOB and COD, will be given by       AIEEE  Solved  Paper-2007

A)  \[\frac{{{\mu }_{0}}}{2\pi }\,{{\left( \frac{({{l}_{1}}+{{l}_{2}})}{d} \right)}^{1/2}}\]

B)         \[\frac{{{\mu }_{0}}}{2\pi d}\,{{(l_{1}^{2}+l_{2}^{2})}^{1/2}}\]

C)                         \[\frac{{{\mu }_{0}}}{2\pi d}\,(l_{1}^{{}}+l_{2}^{{}})\]

D)                         \[\frac{{{\mu }_{0}}}{2\pi d}\,(l_{1}^{2}+l_{2}^{2})\]

Correct Answer: B

Solution :

The magnetic field induction at a point P, at a distance d from O in a direction perpendicular to the plane ABCD due to currents through AOB and COD are perpendicular to each other, is                                 Hence, \[\frac{1}{2}\log \tan \left( \frac{x}{2}-\frac{\pi }{12} \right)+C\] \[={{\left[ {{\left( \frac{{{\mu }_{0}}}{4\pi }\,\frac{2{{l}_{1}}}{d} \right)}^{2}}\,+{{\left( \frac{{{\mu }_{0}}}{4\pi }\,\frac{2{{l}_{2}}}{d} \right)}^{2}} \right]}^{1/2}}\] \[=\frac{{{\mu }_{0}}}{2\pi d}\,\sqrt{(l_{1}^{2}+l_{2}^{2})}\]