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(A) The magnetic field $B$ at the center of a circular current-carrying loop is given by the formula: $B = \frac{\mu_0 i}{2r}$.Given the ratio of radi

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$1/6$

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(A) The magnetic field $B$ at the center of a circular current-carrying loop is given by the formula: $B = \frac{\mu_0 i}{2r}$.Given the ratio of radii $r_1 : r_2 = 1 : 2$ and the ratio of magnetic flux densities $B_1 : B_2 = 1 : 3$.Using the formula for both loops:$B_1 = \frac{\mu_0 i_1}{2r_1}$ and $B_2 = \frac{\mu_0 i_2}{2r_2}$.Taking the ratio:$\frac{B_1}{B_2} = \frac{i_1}{r_1} 	imes \frac{r_2}{i_2} = \frac{i_1}{i_2} 	imes \frac{r_2}{r_1}$.Substituting the given values:$\frac{1}{3} = \frac{i_1}{i_2} 	imes \frac{2}{1}$.Therefore,$\frac{i_1}{i_2} = \frac{1}{3} 	imes \frac{1}{2} = \frac{1}{6}$.

$A$ and $B$ are two concentric circular conductors with center $O$,carrying currents $i_1$ and $i_2$ as shown in the figure. If the ratio of their radii is $1:2$ and the ratio of the magnetic flux densities at $O$ due to $A$ and $B$ is $1:3$,then the value of $i_1/i_2$ is:

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