A) \[35{{\mu }_{0}}/4\]
B) \[{{\mu }_{0}}/80\]
C) \[7{{\mu }_{0}}/80\]
D) \[5{{\mu }_{0}}/4\]
Correct Answer: D
Solution :
Magnetic field at the centre of circular coil of n turns and radius r is \[B=\frac{{{\mu }_{0}}ni}{2\,r}\] For first coil; \[{{B}_{1}}=\frac{{{\mu }_{0}}n{{i}_{1}}}{2{{r}_{1}}}\] For second coil; \[{{B}_{2}}=\frac{{{\mu }_{0}}n{{i}_{2}}}{2{{r}_{2}}}\] Hence, resultant magnetic field at the centre of concentric loop is \[B=\frac{{{\mu }_{0}}n{{i}_{1}}}{2{{r}_{1}}}-\frac{{{\mu }_{0}}n{{i}_{2}}}{2{{r}_{2}}}\] Given, \[n=10,\,{{i}_{1}}=0.2A,\,\,{{r}_{1}}=20\,cm=0.20\,m\] \[{{i}_{2}}=0.3\,A,\,{{r}_{2}}=40\,cm=0.40\,m\]. \[\therefore \]\[B={{\mu }_{0}}\left[ \frac{10\times 0.2}{2\times 0.20}-\frac{10\times 0.3}{2\times -0.40} \right]=\frac{5}{4}{{\mu }_{0}}\]
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