A) 1 : 1
B) 1 : 3
C) 1 : 9
D) 9 : 1
Correct Answer: B
Solution :
| [b] For loop 0nI\[B=\frac{{{\mu }_{0}}nI}{2a}\]where, a is the radius of loop. |
| Then, \[{{B}_{1}}=\frac{{{\mu }_{0}}I}{2a}\] |
| Now, for coil \[B=\frac{{{\mu }_{0}}I}{4\pi }.\frac{2nA}{{{x}^{3}}}\] |
| at the centre x = radius of loop |
| \[{{B}_{2}}=\frac{{{\mu }_{0}}}{4\pi }.\frac{2\times 3\times \left( I/3 \right)\times \pi {{\left( a/3 \right)}^{2}}}{{{\left( a/3 \right)}^{3}}}\] |
| \[=\frac{{{\mu }_{0}}.3I}{2a}\] |
| \[\therefore \]\[\frac{{{B}_{1}}}{{{B}_{2}}}=\frac{{{\mu }_{0}}I/2a}{{{\mu }_{0}}.3I/2a}\]\[{{B}_{1}}:{{B}_{2}}=1:3\] |
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