question_answer

Two concentric coplanar circular loops of radii \[{{R}_{1}}\] and \[{{R}_{2}}\] carry currents \[{{l}_{1}}\] and \[{{l}_{2}}\] respectively in opposite directions. The magnetic field at the center of the loops is half that due to\[{{l}_{1}}\] at the center. Lf \[{{R}_{2}}=2{{R}_{1}},\]the value of \[{{I}_{2}}/{{I}_{1}}\] is

A)  1                            

B)  2         

C)  3                            

D)  4

Correct Answer: A

Solution :

Resultant magnetic field at O is \[B={{B}_{1}}-{{B}_{2}}\] where \[{{B}_{1}}\And {{B}_{2}}\] are magnetic field due to inner and outer circle at O. \[\Rightarrow \]\[\frac{{{B}_{1}}}{2}={{B}_{1}}-{{B}_{2}}\left[ \because B=\frac{{{B}_{1}}}{2} \right]\] \[\Rightarrow \]\[{{B}_{2}}=\frac{{{B}_{1}}}{2}\]\[\Rightarrow \]\[\frac{{{B}_{2}}}{{{B}_{1}}}=\frac{1}{2}\] \[\frac{\frac{\frac{{{\mu }_{0}}{{l}_{2}}}{2{{R}_{2}}}}{{{\mu }_{0}}{{l}_{1}}}}{2{{R}_{1}}}=\frac{{{l}_{2}}}{{{l}_{1}}}\times \frac{{{R}_{1}}}{{{R}_{2}}}=\frac{1}{2}\Rightarrow \frac{{{l}_{2}}}{{{l}_{1}}}=1\]