A) \[\frac{{{H}_{o}}I}{2}\left( \frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}} \right)\]
B) \[\frac{{{H}_{o}}I}{4}\left( \frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}} \right)\]
C) \[\frac{{{H}_{o}}I}{2}\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\]
D) \[\frac{{{H}_{o}}I}{4}\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\]
Correct Answer: D
Solution :
The magnitude of magnetic field due to circular loop which radius \[{{R}_{1}}\] at centre C is \[{{B}_{1}}=-\frac{{{\mu }_{0}}}{4},\frac{I}{{{R}_{1}}}\] Similarly, for radius \[{{R}_{2}}\] \[{{B}_{2}}=\frac{{{\mu }_{0}}}{4},\frac{I}{{{R}_{2}}}\] (direction of current is opposite). Now the resultant magnitude of magnetic field due to circular loop at the centre C is: \[B={{B}_{1}}+{{B}_{2}}\] (as magnetic field at C due to straight part of the conductor, will be zero). \[\Rightarrow \] \[B=\frac{{{\mu }_{0}}}{4},\frac{I}{{{R}_{1}}}-\frac{{{\mu }_{0}}}{4}-\frac{I}{{{R}_{2}}}\] \[B=\frac{{{\mu }_{0}}I}{4}\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)\]
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