Question

A coil of ${}^{\prime }{n}^{\prime }$ number of turns is wound tightly in the form of a spiral with inner and outer radii ${}^{\prime }{a}^{\prime }$ and ${}^{\prime }{b}^{\prime }$ respectively. When a current of strength $I$ is passed through the coil, the magnetic field at its centre is

• A. $\frac{{\mu }_{0}nI}{2(b-a)}$
• B. $\frac{2{\mu }_{0}nI}{b}$
• C. $\frac{{\mu }_{0}nI}{2(b-a)}lo{g}_e\frac{b}{a}$
• D. $\frac{{\mu }_{0}nI}{(b-a)}lo{g}_e\frac{a}{b}$

Answer 1

Answer: (C)

Consider an element of thickness $dr$ 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 $dr=dn$
Number of turns per unit length in element $dr$
$=\frac{ndr}{b-a}$
ie, $dn=\frac{ndr}{b-a}$
Magnetic field at its centre due to element $dr$ is
$dB=\frac{{\mu }_{0}Idn}{2r}=\frac{{\mu }_{0}I}{2}\frac{n}{(b-a)}\frac{dr}{r}$
$\therefore B={\int }_a^b\frac{{\mu }_{0}Indr}{2(b-a)r}=\frac{{\mu }_{0}In}{2(b-a)}{\int }_a^b\frac{dr}{r}$
$=\frac{{\mu }_{0}In}{2(b-a)}{\log }_e\left(\frac{b}{a}\right)$