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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 $b$ 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:
$\frac{\mu_o n I}{2(b-a)} \log _e \frac{b}{a}$
Consider an element of thickness $d r$ at a distance $r$ from the centre of spiral coil.
Number of turns in coil $=n$
Number of turns per unit length
$=\frac{n}{b-a}$
Number of turns in element $d r=d n$
Number of turns per unit length in element $d r$
i.e., $d n=\frac{n d r}{b-a}$
Magnetic field at its centre due to element $d r$ is
$d B=\frac{\mu_0 I d n}{2 r}=\frac{\mu_0 I}{2} \frac{n}{(b-a)} \frac{d r}{r}$
$\therefore \quad B=\int_a^b \frac{\mu_0 I n d r}{2(b-a) r}=\frac{\mu_0 I n}{2(b-a)} \int_a^b \frac{d r}{r}$
$=\frac{\mu_0 I n}{2(b-a)} \log _e\left(\frac{b}{a}\right)$
Number of turns in coil $=n$
Number of turns per unit length
$=\frac{n}{b-a}$
Number of turns in element $d r=d n$
Number of turns per unit length in element $d r$
i.e., $d n=\frac{n d r}{b-a}$
Magnetic field at its centre due to element $d r$ is
$d B=\frac{\mu_0 I d n}{2 r}=\frac{\mu_0 I}{2} \frac{n}{(b-a)} \frac{d r}{r}$
$\therefore \quad B=\int_a^b \frac{\mu_0 I n d r}{2(b-a) r}=\frac{\mu_0 I n}{2(b-a)} \int_a^b \frac{d r}{r}$
$=\frac{\mu_0 I n}{2(b-a)} \log _e\left(\frac{b}{a}\right)$
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