A) \[1:1\]
B) \[1:2\]
C) \[2:1\]
D) \[8:1\]
Correct Answer: D
Solution :
The magnetic field \[(B)\] at a distance \[x\] from the circular coil is given by \[B=\frac{{{\mu }_{0}}ni{{R}^{2}}}{2{{({{R}^{2}}+{{x}^{2}})}^{3/2}}}\] where \[R\] is radius of the coil. At centre,\[x=0;\,\,\,\,\,\,\,\,{{B}_{0}}=\frac{{{\mu }_{0}}ni}{2R}\] At\[x=\sqrt{3}R\] \[\therefore \]\[{{B}_{a}}=\frac{{{\mu }_{0}}ni{{R}^{2}}}{2{{({{R}^{2}}+3{{R}^{2}})}^{3/2}}}=\frac{{{\mu }_{0}}ni}{16R}\] Hence, \[\frac{{{B}_{0}}}{{{B}_{a}}}=\frac{8}{1}\]
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