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

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(C) Consider a small elemental ring of radius $r$ and thickness $dr$ within the spiral coil.The number of turns in this elemental ring is $dn = \frac{

Vedclass · Accepted Answer

$\frac{\mu_0 NI}{2(b - a)} \ln \left( \frac{b}{a} \right)$

Vedclass · Answer

(C) Consider a small elemental ring of radius $r$ and thickness $dr$ within the spiral coil.The number of turns in this elemental ring is $dn = \frac{N}{b - a} dr$.The magnetic field at the centre due to this elemental ring is $dB = \frac{\mu_0 (dn) I}{2r} = \frac{\mu_0 N I}{2(b - a)} \frac{dr}{r}$.To find the total magnetic field $B$ at the centre,we integrate $dB$ from $r = a$ to $r = b$:$B = \int_{a}^{b} \frac{\mu_0 N I}{2(b - a)} \frac{dr}{r} = \frac{\mu_0 N I}{2(b - a)} [\ln r]_{a}^{b}$.$B = \frac{\mu_0 N I}{2(b - a)} \ln \left( \frac{b}{a} \right)$.

$A$ 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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