question_answer

Two identical long conducting wires AOB and COD are placed at right angle to each other, with one above other such that O is their common point for the two. The wires carry \[{{I}_{1}}\] and \[{{I}_{2}}\] currents, respectively. Point P is lying at distance d from O along a direction perpendicular to the plane containing the wires. The magnetic field at the point P will be                                                                                                                     [NEET 2014]

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

B)       \[\frac{{{\mu }_{0}}}{2\pi d}({{l}_{1}}+{{l}_{2}})\]

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

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

Correct Answer: D

Solution :

[d]

The point P is lying at a distance d along the z axis.

\[\left| {{\mathbf{B}}_{1}} \right|=\frac{{{\mu }_{0}}}{2\pi }\frac{{{l}_{1}}}{d}\]

and       \[\left| {{\mathbf{B}}_{2}} \right|=\frac{{{\mu }_{0}}}{2\pi }\frac{{{l}_{2}}}{d}\]

\[{{B}_{net}}=\sqrt{B_{1}^{2}+B_{2}^{2}},\]

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