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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