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Two circular coils made from same wire but radius of $1^{\text {st }}$ coil is twice that of $2^{\text {nd }}$ coil. If magnetic field at their centres is same then ratio of potential difference applied across them is $\left(1^{\text {st }}\right.$ to $2^{\text {nd }}$ coil)
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The correct answer is:
$4$
Magnetic fields at the centre of the coils are equal.
$$
\therefore \quad \frac{\mu_0 I_1}{2 r_1}=\frac{\mu_0 I_2}{2 r_2}
$$
$\therefore \quad \frac{\mathrm{I}_1}{\mathrm{I}_2}=\frac{\mathrm{r}_1}{\mathrm{r}_2}=\frac{2 \mathrm{r}}{\mathrm{r}}$....(given $r_1=2 r_2$ )
$\therefore \quad \frac{\mathrm{I}_1}{\mathrm{I}_2}=2$...(ii)
The resistance through the coil, $\mathrm{R} \propto l$
Here, $l=2 \pi \mathrm{r}$
$\therefore \quad \frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{2 \pi \mathrm{r}_1}{2 \pi \mathrm{r}_2}=2$...(ii)
$$
\therefore \quad \frac{\mathrm{V}_1}{\mathrm{~V}_2}=\frac{\mathrm{I}_1 \mathrm{R}_1}{\mathrm{I}_2 \mathrm{R}_2}=2 \times 2=4
$$
[From (i) and (ii)]
$$
\therefore \quad \frac{\mu_0 I_1}{2 r_1}=\frac{\mu_0 I_2}{2 r_2}
$$
$\therefore \quad \frac{\mathrm{I}_1}{\mathrm{I}_2}=\frac{\mathrm{r}_1}{\mathrm{r}_2}=\frac{2 \mathrm{r}}{\mathrm{r}}$....(given $r_1=2 r_2$ )
$\therefore \quad \frac{\mathrm{I}_1}{\mathrm{I}_2}=2$...(ii)
The resistance through the coil, $\mathrm{R} \propto l$
Here, $l=2 \pi \mathrm{r}$
$\therefore \quad \frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{2 \pi \mathrm{r}_1}{2 \pi \mathrm{r}_2}=2$...(ii)
$$
\therefore \quad \frac{\mathrm{V}_1}{\mathrm{~V}_2}=\frac{\mathrm{I}_1 \mathrm{R}_1}{\mathrm{I}_2 \mathrm{R}_2}=2 \times 2=4
$$
[From (i) and (ii)]
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