A) \[\sqrt{2}\]
B) \[\frac{1}{\sqrt{2}}\]
C) 2
D) \[\sqrt{3}\]
Correct Answer: A
Solution :
| [a] \[I(\pi {{r}^{2}})=M,2M=I\pi {{(r')}^{2}}\] |
| \[r'=\sqrt{2}r\] |
| \[{{B}_{1}}=\frac{{{\mu }_{o}}I}{2r}\] |
| \[{{B}_{2}}=\frac{{{\mu }_{o}}I}{2\sqrt{2}r}\] |
| \[\therefore \frac{{{B}_{1}}}{{{B}_{2}}}=\frac{\sqrt{2}}{1}\] |
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