A) 9
B) \[1/3\]
C) \[1/4\]
D) \[1/9\]
Correct Answer: D
Solution :
: Field at the centre of a circular coil \[=\frac{{{\mu }_{0}}ni}{2R}\] In 1st case, length of wire\[(L)=\]circumference of a coil or \[L=2\pi R\] \[\therefore \] \[R=\frac{L}{2\pi }\] ...(i) In second case, wire is bent into three loops. \[\therefore \] \[L=3\times (2\pi r)\] \[\therefore \] \[\therefore r=\frac{L}{6\pi }\] ...(ii) \[\frac{{{B}_{1}}}{{{B}_{2}}}=\frac{\frac{{{\mu }_{0}}i}{2}\times \frac{2\pi }{L}}{\frac{{{\mu }_{0}}i}{2}\times \frac{3\times 6\pi }{L}}\]or\[\frac{{{B}_{1}}}{{{B}_{2}}}=\frac{2}{18}=\frac{1}{9}\]
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