Question

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

• A. $\frac {\mu_0 n I}{2(b-a)}$
• B. $\frac {2\mu_0 n I}{b}$
• C. $\frac {\mu_0 n I}{2(b-a)}log_e \frac{b}{a}$
• D. $\frac {\mu_0 n I}{(b-a)}log_e\frac{a}{b}$

Answer 1

Answer: (C)

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 $d r$
$=\frac{n d r}{b-a}$
ie, $d n=\frac{n d r}{b-a}$
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 \,\,\,\,B=\displaystyle\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} $
$=\frac{\mu_{0} I n}{2(b-a)} \log _{e}\left(\frac{b}{a}\right)$