A) 1/6
B) ¼
C) 1/2
D) 1/3
Correct Answer: A
Solution :
Magnetic field produced at the centre by loops, \[{{B}_{1}}=\frac{{{\mu }_{0}}{{I}_{1}}}{2{{r}_{1}}},\] s\[Bl2=\frac{{{\mu }_{0}}{{I}_{2}}}{2{{r}_{2}}}\] \[\frac{{{B}_{1}}}{{{B}_{2}}}=\frac{{{I}_{1}}}{{{I}_{2}}}\times \frac{{{r}_{2}}}{{{r}_{1}}}\] But given \[\frac{{{B}_{1}}}{{{B}_{2}}}=\frac{1}{3}\] \[\therefore \] \[\frac{1}{3}=\frac{{{I}_{1}}}{{{I}_{2}}}\times \frac{2}{1}\] \[\therefore \] \[\frac{{{I}_{1}}}{{{I}_{2}}}=\frac{1}{6}\]You need to login to perform this action.
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