A) \[\frac{1}{2}\]
B) 2
C) \[\frac{1}{4}\]
D) 1
Correct Answer: D
Solution :
(d) 1 Let the magnetic fields due to a long straight wire of radius R carrying a steady current l at a distance r from the centre of the wire are \[{{B}_{1}}=\frac{{{\mu }_{0}}lr}{2\pi {{R}^{2}}}(For\,\,r<R)\] And \[{{B}_{2}}=\frac{{{\mu }_{0}}l}{2\pi R}(For\,\,r>R)\] So, the magnetic field at \[r=\frac{R}{2}\] is \[{{B}_{1}}=\frac{{{\mu }_{0}}l}{2\pi {{R}^{2}}}\left( \frac{R}{2} \right)=\frac{{{\mu }_{0}}l}{4\pi R}\] and at \[r=2R\,is\,{{B}_{2}}=\frac{{{\mu }_{0}}l}{2\pi (2R)}=\frac{{{\mu }_{0}}l}{4\pi R}\] \[\therefore \] Their corresponding ratio is \[\frac{{{B}_{1}}}{{{B}_{2}}}=\frac{({{\mu }_{0}}l/4\pi R)}{({{\mu }_{0}}l/4\pi R)}=1\]
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