A) 10 : 1
B) 100 : 1
C) 1000 : 1
D) 2000 : 1
Correct Answer: C
Solution :
Field at centre\[=\frac{{{\mu }_{0}}i}{2a}\] Field on the axis at a distance>>radius of ring is given by \[B=\frac{{{\mu }_{0}}}{4\pi }.\frac{2m}{{{d}^{3}}}\] Here, m is magnetic moment =\[i\times \] area of ring \[\Rightarrow \]\[m=\pi {{a}^{2}}.i\] So, \[{{B}_{at}}_{P}=\frac{{{\mu }_{0}}}{4\pi }\times \frac{2\times \pi {{a}^{2}}\times i}{{{(10a)}^{3}}}\] \[=\frac{{{\mu }_{0}}i}{2\times 1000a}\] \[\therefore \]Ratio is\[\frac{{{B}_{0}}}{{{B}_{P}}}=\frac{{{\mu }_{0}}i}{2a}\times \frac{2\times 1000a}{{{\mu }_{0}}i}\] \[=1000:1\]
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