Physics - Magnetism Question with Solution | TestHub

An element is assumed at distance x from center whose width is dx. No. of turns in width$\mathrm{b}–\mathrm{a}\mathrm{ }=\mathrm{ }\mathrm{N}$

$\therefore$No. of turns is width$\mathrm{d}\mathrm{x}=\mathrm{n}=\left(\frac{\mathrm{N}}{\mathrm{b}-\mathrm{a}}\right)\mathrm{d}\mathrm{x}$

$\mathrm{d}\mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{n}\mathrm{i}}{2\mathrm{x}}=\frac{{\mathrm{\mu }}_0\mathrm{i}\mathrm{N}}{2\left(\mathrm{b}-\mathrm{a}\right)}\frac{\mathrm{d}\mathrm{x}}{\mathrm{x}}$

$\therefore \mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{i}\mathrm{n}}{2\left(\mathrm{b}-\mathrm{a}\right)}{\int }_{\mathrm{a}}^{\mathrm{b}}\frac{\mathrm{d}\mathrm{x}}{\mathrm{x}}=\frac{{\mathrm{\mu }}_0\mathrm{N}\mathrm{i}}{2\left(\mathrm{b}-\mathrm{a}\right)}{\log }_{\mathrm{e}}\left(\frac{\mathrm{b}}{\mathrm{a}}\right)$