Two circular coils are made from the same wir | vedclass

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(C) The magnetic field at the center of a circular coil is given by $B = \frac{\mu_0 I}{2r}$.Given that the magnetic fields at the centers are equal,w

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$4$

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(C) The magnetic field at the center of a circular coil is given by $B = \frac{\mu_0 I}{2r}$.Given that the magnetic fields at the centers are equal,we have $\frac{\mu_0 I_1}{2 r_1} = \frac{\mu_0 I_2}{2 r_2}$.This implies $\frac{I_1}{I_2} = \frac{r_1}{r_2}$. Since $r_1 = 2r_2$,we get $\frac{I_1}{I_2} = 2$ (Equation $i$).The resistance of a wire is given by $R = ho \frac{l}{A}$. Since the coils are made from the same wire,$ho$ and $A$ are constant,so $R \propto l$.The length of the wire for a circular coil is $l = 2\pi r$,so $R \propto r$.Therefore,$\frac{R_1}{R_2} = \frac{r_1}{r_2} = 2$ (Equation $ii$).The potential difference is $V = IR$. The ratio of potential differences is $\frac{V_1}{V_2} = \frac{I_1 R_1}{I_2 R_2}$.Substituting the ratios from (Equation $i$) and (Equation $ii$),we get $\frac{V_1}{V_2} = 2 	imes 2 = 4$.

Two circular coils are made from the same wire,but the radius of the $1^{\text{st}}$ coil is twice that of the $2^{\text{nd}}$ coil. If the magnetic field at their centers is the same,then the ratio of the potential difference applied across them is ($1^{\text{st}}$ to $2^{\text{nd}}$ coil).

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