A) R
B)
C) 2R
D)
Correct Answer: D
Solution :
If a coil of radius R is carrying current I, then magnetic field on its axis at a distance x from its centre is given by \[{{B}_{axis}}=\frac{{{\mu }_{0}}\,2\pi \,NI{{R}^{2}}}{4\pi {{({{x}^{2}}+{{R}^{2}})}^{3/2}}}\] At centre \[x=0\] \[\Rightarrow \] \[{{B}_{centre}}=\frac{{{\mu }_{0}}NI}{2R}\] Given, \[\frac{{{B}_{axis}}}{{{B}_{centre}}}=\frac{1}{8}\] \[\Rightarrow \] \[\frac{{{\mu }_{0}}}{4\pi }\frac{2\pi NI{{R}^{2}}}{{{({{x}^{2}}+{{R}^{2}})}^{3/2}}}=\frac{1}{8}\frac{{{\mu }_{0}}NI}{2R}\] or \[8{{R}^{3}}={{({{x}^{2}}+{{R}^{2}})}^{3/2}}\] or \[3{{R}^{2}}={{x}^{2}}\] or \[x=\sqrt{3}R\]
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