Question

A coil having $N$ turns is wound tightly in the form of 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. $\frac{\mu _{0} N I}{b - a}$
• B. $\frac{\mu_0 N I}{2 \pi(b-a)}$
• C. $\frac{\mu_0 N I}{2(b-a)} \ln \frac{b}{a}$
• D. $\frac{\mu _{0} N I}{b - a}ln\frac{b}{a}$

Answer 1

Answer: (C)

Let $d N$ be the number of turns in width $d r$.
$ \begin{array}{l} d N=\frac{N}{b-a} d r \\ d B=\frac{\mu_0 d N i}{2 r}=\frac{\mu_0 N i}{2(b-a)} d r \\ B=\int_a^b \frac{\mu_0 N i}{2(b-a)} \frac{d r}{r}=\frac{\mu_0 N i}{2(b-a)} \ln \frac{b}{a} \end{array} $