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(A) The magnetic moment $M$ of a current-carrying coil is given by the formula $M = I \cdot A$,where $I$ is the current and $A$ is the area of the coi

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$1:4$

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(A) The magnetic moment $M$ of a current-carrying coil is given by the formula $M = I \cdot A$,where $I$ is the current and $A$ is the area of the coil.Since the coil is circular,the area $A = \pi r^2$,where $r$ is the radius.Therefore,the magnetic moment is $M = I \cdot \pi r^2$.For two coils with the same current $I$ and radii $r_1$ and $r_2$,the ratio of their magnetic moments is given by:$\frac{M_1}{M_2} = \frac{I \cdot \pi r_1^2}{I \cdot \pi r_2^2} = \left(\frac{r_1}{r_2}\right)^2$.Given the ratio of radii $\frac{r_1}{r_2} = \frac{1}{2}$,we substitute this into the equation:$\frac{M_1}{M_2} = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$.Thus,the ratio of their magnetic moments is $1:4$.

$2 \ A$ current is flowing through two different small circular copper coils having radii in the ratio $1:2$. The ratio of their respective magnetic moments will be

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