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(B) The magnetic field at the center of a circular coil is given by $B = \frac{\mu_0 i}{2r}$.Given that $B_1 = B_2$,we have $\frac{i_1}{r_1} = \frac{i

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$4: 1$

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(B) The magnetic field at the center of a circular coil is given by $B = \frac{\mu_0 i}{2r}$.Given that $B_1 = B_2$,we have $\frac{i_1}{r_1} = \frac{i_2}{r_2}$.Since $r_1 = 2r_2$,we get $\frac{i_1}{2r_2} = \frac{i_2}{r_2}$,which implies $i_1 = 2i_2$.The resistance of a wire is $R = \rho \frac{L}{A}$. Since both coils are made of the same wire,$\rho$ and $A$ are constant. The length $L = 2\pi r$.Thus,$R_1 = 2\pi r_1$ and $R_2 = 2\pi r_2$. Since $r_1 = 2r_2$,$R_1 = 2R_2$.The potential difference is $V = iR$. Therefore,$\frac{V_1}{V_2} = \frac{i_1 R_1}{i_2 R_2}$.Substituting the values: $\frac{V_1}{V_2} = \frac{(2i_2)(2R_2)}{i_2 R_2} = 4$.Thus,the ratio is $4: 1$.

Two circular coils $1$ and $2$ are made from the same wire,but the radius of the first coil is twice that of the second coil. What is the ratio of the potential difference applied across them so that the magnetic field at their centres is the same?

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