A) 1 : 1
B) 1 : 3
C) 1 : 9
D) 9 : 1
Correct Answer: B
Solution :
[b] For loop \[B=\frac{{{\mu }_{0}}nI}{2a}\] where, a is the radius of loop. Then, \[{{B}_{1}}=\frac{{{\mu }_{0}}I}{2a}\] Now, for coil \[B=\frac{{{\mu }_{0}}I}{4\pi }.\frac{2n\,A}{{{x}^{3}}}\] at the centre x = radius of loop \[{{B}_{2}}=\frac{{{\mu }_{0}}}{4\pi }.\frac{2\times 3\times (I/3)\times \pi {{(a/3)}^{2}}}{{{(a/3)}^{3}}}=\frac{{{\mu }_{0}}.3I}{2a}\] \[\therefore \,\,\frac{{{B}_{1}}}{{{B}_{2}}}=\frac{{{\mu }_{0}}I/2a}{{{\mu }_{0}}.3I/2a}\] \[{{B}_{1}}:{{B}_{2}}=1:3\]
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