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. $\frac{2{\mu }_{0}NI}{a}$
• C. $\frac{{\left(\mu \right)}_{0}NI}{2\left(b-a\right)}{\left(log\right)}_e\left(\frac{b}{a}\right)$
• D. $\frac{{\left(\mu \right)}_{0}IN}{2\left(b-a\right)}{\left(log\right)}_e\left(\frac{a}{b}\right)$

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 $dx=n=\left(\frac{N}{b-a}\right)dx$
$dB=\frac{{\left(\mu \right)}_{0}ni}{2x}=\frac{{\left(\mu \right)}_{0}iN}{2\left(b-a\right)}\frac{dx}{x}$
$\therefore B=\frac{{\left(\mu \right)}_{0}in}{2\left(b-a\right)}{\int }_a^b\frac{dx}{x}=\frac{{\left(\mu \right)}_{0}Ni}{2\left(b-a\right)}{\left(log\right)}_e\left(\frac{b}{a}\right)$