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