A) 1 : 1
B) 1 : 2
C) 2 : 1
D) 1 : 4
Correct Answer: A
Solution :
(a) 1 : 1. We know that, \[{{B}_{out}}=\frac{{{\mu }_{0}}l}{2\pi r}(r>R)\] Here, \[r=2R\] \[\therefore {{B}_{out}}=\frac{{{\mu }_{0}}l}{2\pi \times 2R}=\frac{{{\mu }_{0}}l}{4\pi R}\] (i) Also. \[{{B}_{in}}=\frac{{{\mu }_{0}}lr}{2\pi {{R}^{2}}}(r<R)\] Here, \[r=R/2\] \[\therefore {{B}_{in}}=\frac{{{\mu }_{0}}lR/2}{2\pi {{R}^{2}}}=\frac{{{\mu }_{0}}l}{4\pi R}\] (ii) Required ratio, \[\frac{{{B}_{in}}}{{{B}_{out}}}=\frac{1}{1}.\]
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