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A long wire is bent into a circular coil of one turn and then into a circular coil of smaller radius having $\mathrm{n}$ turns. If the same current passes in both the cases, the ratio of magnetic fields produced at the centre for one turn to that of $n$ turns is
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The correct answer is:
$1:n^2$
Magnetic field at the centre of coil is,
$\mathrm{B}_1=\frac{\mu_0 \mathrm{I}}{2 \mathrm{r}_1}$
$\therefore \quad$ Magnetic field of $\mathrm{n}$ turns coil at the centre:
$\begin{aligned}
& \mathrm{B}_2=\frac{\mu_0 \mathrm{nI}}{2 \mathrm{r}_2} \\
\therefore \quad \frac{\mathrm{B}_1}{\mathrm{~B}_2} & =\frac{\frac{\mu_0 \mathrm{I}}{2 \mathrm{r}_1}}{\frac{\mu_0 \mathrm{nI}}{2 \mathrm{r}_2}}=\frac{\mathrm{r}_2}{\mathrm{nr}_1}
\end{aligned}$
But, radius: $r_1=\frac{1}{2 \pi}$ and $r_2=\frac{1}{2 \pi n}$
$\begin{array}{ll}
\therefore & \frac{\mathrm{r}_2}{\mathrm{r}_1}=\frac{1}{\mathrm{n}} \\
\therefore & \frac{\mathrm{B}_1}{\mathrm{~B}_2}=\frac{1}{\mathrm{n}^2}
\end{array}$
$\mathrm{B}_1=\frac{\mu_0 \mathrm{I}}{2 \mathrm{r}_1}$
$\therefore \quad$ Magnetic field of $\mathrm{n}$ turns coil at the centre:
$\begin{aligned}
& \mathrm{B}_2=\frac{\mu_0 \mathrm{nI}}{2 \mathrm{r}_2} \\
\therefore \quad \frac{\mathrm{B}_1}{\mathrm{~B}_2} & =\frac{\frac{\mu_0 \mathrm{I}}{2 \mathrm{r}_1}}{\frac{\mu_0 \mathrm{nI}}{2 \mathrm{r}_2}}=\frac{\mathrm{r}_2}{\mathrm{nr}_1}
\end{aligned}$
But, radius: $r_1=\frac{1}{2 \pi}$ and $r_2=\frac{1}{2 \pi n}$
$\begin{array}{ll}
\therefore & \frac{\mathrm{r}_2}{\mathrm{r}_1}=\frac{1}{\mathrm{n}} \\
\therefore & \frac{\mathrm{B}_1}{\mathrm{~B}_2}=\frac{1}{\mathrm{n}^2}
\end{array}$
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