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 \[B=\frac{{{\mu }_{0}}ni}{2r}\] 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, \[n=10,{{i}_{1}}=0.2A,{{r}_{1}}=20cm=0.20\,m\] \[{{i}_{2}}=0.3A,\,{{r}_{2}}=40\,cm\,=0.40\,m\] \[\therefore \,B=\mu \,\left[ \frac{10\times 0.2}{2\times 0.20}-\frac{10\times 0.3}{2\times 0.40} \right]\,\,=\frac{5}{4}{{\mu }_{0}}\]
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