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. \[\Rightarrow \]\[\frac{{{B}_{1}}}{2}={{B}_{1}}-{{B}_{2}}\left[ \because B=\frac{{{B}_{1}}}{2} \right]\] \[\Rightarrow \]\[{{B}_{2}}=\frac{{{B}_{1}}}{2}\]\[\Rightarrow \]\[\frac{{{B}_{2}}}{{{B}_{1}}}=\frac{1}{2}\] \[\frac{\frac{\frac{{{\mu }_{0}}{{l}_{2}}}{2{{R}_{2}}}}{{{\mu }_{0}}{{l}_{1}}}}{2{{R}_{1}}}=\frac{{{l}_{2}}}{{{l}_{1}}}\times \frac{{{R}_{1}}}{{{R}_{2}}}=\frac{1}{2}\Rightarrow \frac{{{l}_{2}}}{{{l}_{1}}}=1\]You need to login to perform this action.
You will be redirected in
3 sec