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 0, in a direction perpendicular to the plane of the wires AOB and COD, will be given by                                                       [AIEEE 2007]

A) \[f:R/\{0\}\to R\]

B) \[f(x)=\frac{1}{x}-\frac{2}{{{e}^{2x}}-1}\]

C) \[x=0\]

D) \[x\]

Correct Answer: A

Solution :

[b] 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, \[\frac{1}{2}\log \tan \left( \frac{x}{2}-\frac{\pi }{12} \right)+C\]

\[={{\left[ {{\left( \frac{{{\mu }_{0}}}{4\pi }\,\frac{2{{l}_{1}}}{d} \right)}^{2}}\,+{{\left( \frac{{{\mu }_{0}}}{4\pi }\,\frac{2{{l}_{2}}}{d} \right)}^{2}} \right]}^{1/2}}\]

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