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A coil of $n$ number of turns is wound tightly in the form of a spiral with inner and outer radii a and $\mathrm{b}$ respectively. When a current of strength $I$ is passed through the coil, the magnetic field at its centre is
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Verified Answer
The correct answer is:
$\frac{\mu_{0} n I}{2(b-a)} \log _{e} \frac{b}{a}$
Consider an element of thickness dr at a distance $r$ from the centre of spiral coil.
Number of turns in coil $=\mathrm{n}$
Number of turns per unit length
$$
=\frac{\mathrm{n}}{\mathrm{b}-\mathrm{a}}
$$
Number of turns in element $d r=d n$
Number of turns per unit length in element $d r$
$$
=\frac{\mathrm{ndr}}{\mathrm{b}-\mathrm{a}}
$$
ie, $\quad \mathrm{dn}=\frac{\mathrm{ndr}}{\mathrm{b}-\mathrm{a}}$
Magnetic field at its centre due to element $d r$ is
$$
\begin{gathered}
\mathrm{dB}=\frac{\mu_{0} \mathrm{Idn}}{2 \mathrm{r}}=\frac{\mu_{0} \mathrm{I}}{2} \frac{\mathrm{n}}{(\mathrm{b}-\mathrm{a})} \frac{\mathrm{dr}}{\mathrm{r}} \\
\therefore \quad \mathrm{B}=\int_{\mathrm{a}}^{\mathrm{b}} \frac{\mu_{0} \mathrm{Indr}}{2(\mathrm{~b}-\mathrm{a}) \mathrm{r}}=\frac{\mu_{0} \mathrm{In}}{2(\mathrm{~b}-\mathrm{a})} \int_{\mathrm{a}}^{\mathrm{b}} \frac{\mathrm{dr}}{\mathrm{r}} \\
=\frac{\mu_{0} \mathrm{In}}{2(\mathrm{~b}-\mathrm{a})} \log _{\mathrm{e}}\left(\frac{\mathrm{b}}{\mathrm{a}}\right)
\end{gathered}
$$
Number of turns in coil $=\mathrm{n}$
Number of turns per unit length
$$
=\frac{\mathrm{n}}{\mathrm{b}-\mathrm{a}}
$$
Number of turns in element $d r=d n$
Number of turns per unit length in element $d r$
$$
=\frac{\mathrm{ndr}}{\mathrm{b}-\mathrm{a}}
$$
ie, $\quad \mathrm{dn}=\frac{\mathrm{ndr}}{\mathrm{b}-\mathrm{a}}$
Magnetic field at its centre due to element $d r$ is
$$
\begin{gathered}
\mathrm{dB}=\frac{\mu_{0} \mathrm{Idn}}{2 \mathrm{r}}=\frac{\mu_{0} \mathrm{I}}{2} \frac{\mathrm{n}}{(\mathrm{b}-\mathrm{a})} \frac{\mathrm{dr}}{\mathrm{r}} \\
\therefore \quad \mathrm{B}=\int_{\mathrm{a}}^{\mathrm{b}} \frac{\mu_{0} \mathrm{Indr}}{2(\mathrm{~b}-\mathrm{a}) \mathrm{r}}=\frac{\mu_{0} \mathrm{In}}{2(\mathrm{~b}-\mathrm{a})} \int_{\mathrm{a}}^{\mathrm{b}} \frac{\mathrm{dr}}{\mathrm{r}} \\
=\frac{\mu_{0} \mathrm{In}}{2(\mathrm{~b}-\mathrm{a})} \log _{\mathrm{e}}\left(\frac{\mathrm{b}}{\mathrm{a}}\right)
\end{gathered}
$$
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