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(C) The number of turns per unit width is $n = \frac{N}{b - a}$.Consider an elemental ring of radius $x$ and thickness $dx$.The number of turns in thi

Vedclass · Accepted Answer

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

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(C) The number of turns per unit width is $n = \frac{N}{b - a}$.Consider an elemental ring of radius $x$ and thickness $dx$.The number of turns in this elemental ring is $dN = n \cdot dx = \frac{N}{b - a} dx$.The magnetic field at the centre due to this elemental ring is given by $dB = \frac{\mu_0 (dN) I}{2x}$.Substituting $dN$,we get $dB = \frac{\mu_0 I}{2x} \cdot \frac{N}{b - a} dx = \frac{\mu_0 NI}{2(b - a)} \cdot \frac{dx}{x}$.To find the total magnetic field $B$ at the centre,we integrate $dB$ from $x = a$ to $x = b$:$B = \int_a^b \frac{\mu_0 NI}{2(b - a)} \frac{dx}{x} = \frac{\mu_0 NI}{2(b - a)} \int_a^b \frac{dx}{x}$.$B = \frac{\mu_0 NI}{2(b - a)} [\ln x]_a^b = \frac{\mu_0 NI}{2(b - a)} \ln \frac{b}{a}$.

$A$ 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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