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(C) The magnetic field $B$ at the center of a circular coil with $n$ turns,radius $r$,and current $i$ is given by $B = \frac{\mu_0 n i}{2r}$.Given tha

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$8: 15$

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(C) The magnetic field $B$ at the center of a circular coil with $n$ turns,radius $r$,and current $i$ is given by $B = \frac{\mu_0 n i}{2r}$.Given that the current $i$ is the same for both coils and the magnetic fields $B_1$ and $B_2$ are equal,we have $B_1 = B_2$.Substituting the formula,we get $\frac{\mu_0 n_1 i}{2 r_1} = \frac{\mu_0 n_2 i}{2 r_2}$.Simplifying this,we get $\frac{n_1}{r_1} = \frac{n_2}{r_2}$,which implies $\frac{r_1}{r_2} = \frac{n_1}{n_2}$.Given the ratio of the number of turns is $\frac{n_1}{n_2} = \frac{8}{15}$,therefore the ratio of their radii is $\frac{r_1}{r_2} = \frac{8}{15}$.

When equal current is passed through two coils,equal magnetic field is produced at their centres. If the ratio of the number of turns in the coils is $8: 15$,then the ratio of their radii will be

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