Question

A coil having $N$ turns is wound tightly in the form of a spiral with inner and outer radii $a$ and $b$ respectively. When a current $I$ passes through the coil, the magnetic field at the centre is

• A. $\frac{\mu _{0} NI}{b}$
• B. $\frac{2 \mu _{0} N I}{a}$
• C. $\frac{\left(\mu \right)_{0} N I}{2 \left(b - a\right)}\left(log\right)_{e}\left(\frac{b}{a}\right)$
• D. $\frac{\left(\mu \right)_{0} I N}{2 \left(b - a\right)}\left(log\right)_{e}\left(\frac{a}{b}\right)$

Answer 1

Answer: (C)

An element is assumed at distance x from centre whose width is dx. No. of turns in width $b-a$ is $N$
$\therefore $ No. of turns is width $dx=n=\left(\frac{N}{b - a}\right)dx$
$dB=\frac{\left(\mu \right)_{0} n i}{2 x}=\frac{\left(\mu \right)_{0} i N}{2 \left(b - a\right)}\frac{d x}{x}$
$\therefore B=\frac{\left(\mu \right)_{0} i n}{2 \left(b - a\right)}\displaystyle \int _{a}^{b} \frac{d x}{x}=\frac{\left(\mu \right)_{0} N i}{2 \left(b - a\right)}\left(log\right)_{e} \left(\frac{b}{a}\right)$