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(C) Let $r_{1}$ and $r_{2}$ be the radii of coil $1$ and $2$ respectively. Given $r_{1} = 2r_{2}$.The magnetic field at the center of a circular coil

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$4$

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(C) Let $r_{1}$ and $r_{2}$ be the radii of coil $1$ and $2$ respectively. Given $r_{1} = 2r_{2}$.The magnetic field at the center of a circular coil is given by $B = \frac{\mu_{0} I}{2r}$.For the fields to be equal,$B_{1} = B_{2} \implies \frac{\mu_{0} I_{1}}{2r_{1}} = \frac{\mu_{0} I_{2}}{2r_{2}}$.Substituting $r_{1} = 2r_{2}$,we get $\frac{I_{1}}{2r_{2}} = \frac{I_{2}}{r_{2}}$,which simplifies to $I_{1} = 2I_{2}$.The resistance of a wire is proportional to its length,$R \propto L = 2\pi r$. Since both coils are made of the same wire,$R_{1} = 2R_{2}$.The potential difference is $V = IR$. Therefore,the ratio of potential differences is $\frac{V_{1}}{V_{2}} = \frac{I_{1} R_{1}}{I_{2} R_{2}}$.Substituting the values,$\frac{V_{1}}{V_{2}} = \frac{(2I_{2})(2R_{2})}{I_{2} R_{2}} = 4$.

Two circular coils $1$ and $2$ are made from the same wire,but the radius of the $1^{st}$ coil is twice that of the $2^{nd}$ coil. What is the ratio of potential difference in volts that should be applied across them so that the magnetic field at their centres is the same?

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