A coil having N turns is wound tightly in the | vedclass

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Question

(c) Number of turns per unit width $ = \frac{N}{{b - a}}$

 Consider an elemental ring of radius $x$ and with thickness $dx$

Number of turns

Vedclass · Accepted Answer

$\frac{{{\mu _0}NI}}{{2(b - a)}}\ln \frac{b}{a}$

Vedclass · Answer

(c) Number of turns per unit width $ = \frac{N}{{b - a}}$

 Consider an elemental ring of radius $x$ and with thickness $dx$

Number of turns in the ring $ = dN = \frac{{Ndx}}{{b - a}}$

Magnetic field at the centre due to the ring element
$dB = \frac{{{\mu _0}(dN)i}}{{2x}} = \frac{{{\mu _0}i}}{2}.\frac{{Ndx}}{{(b - a)}}.\frac{1}{x}$

$	herefore $ Field at the centre
$ = \int_{}^{} {dB = \frac{{{\mu _0}Ni}}{{2(b - a)}}} \int_a^b {\frac{{dx}}{x}} $
$ = \frac{{{\mu _0}Ni}}{{2(b - a)}}\ln \frac{b}{a}.$

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

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