A) \[\text{35}{{}_{\text{0}}}\text{/4}\]
B) \[{{}_{\text{0}}}\text{/80}\]
C) \[\text{7}{{}_{\text{0}}}\text{/80}\]
D) \[\text{5}{{}_{\text{0}}}\text{/4}\]
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,\[n=10,\,\,{{i}_{1}}=0.2\,\,A,\,\,{{r}_{1}}=20\,\,cm=0.20\,\,m\], \[{{i}_{2}}=0.3\,\,A,\,\,{{r}_{2}}=40\,\,cm=0.40\,\,m\] \[\therefore \] \[B={{\mu }_{0}}\left[ \frac{10\times 0.2}{2\times 0.20}-\frac{10\times 0.3}{2\times 0.40} \right]=\frac{5}{4}{{\mu }_{0}}\]
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