Question

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

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

Answer 1

Answer: (C)

An element is assumed at distance $x$ from center whose width is $dx$ . No. of turns in width $b-a=N$
$\therefore$ No. of turns is width $dx=n=\left(\frac{N}{b-a}\right)dx$
$dB=\frac{{\mu }_{0}ni}{2x}=\frac{{\mu }_{0}iNdx}{2(b-a{)}^x}$
$\therefore B=\frac{{\mu }_{0}iN}{2(b-a)}{\int }_a^b\frac{dx}{X}=\frac{{\mu }_{0}Ni}{2(b-a)}{\log }_e\left(\frac{b}{a}\right)$