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

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(C) Consider an element of thickness $dr$ at a distance $r$ from the centre. The number of turns in this element is given by $dN = \left(\frac{N}{b -

Vedclass · Accepted Answer

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

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(C) Consider an element of thickness $dr$ at a distance $r$ from the centre. The number of turns in this element is given by $dN = \left(\frac{N}{b - a}\right) dr$.The magnetic field due to this element at the centre of the coil is given by $dB = \frac{\mu_0 (dN) i}{2r} = \frac{\mu_0 i}{2} \left(\frac{N}{b - a}\right) \frac{dr}{r}$.To find the total magnetic field $B$,we integrate $dB$ from $r = a$ to $r = b$:$B = \int_{a}^{b} dB = \frac{\mu_0 Ni}{2(b - a)} \int_{a}^{b} \frac{dr}{r}$.$B = \frac{\mu_0 Ni}{2(b - a)} [\ln r]_{a}^{b} = \frac{\mu_0 Ni}{2(b - a)} \ln\left(\frac{b}{a}\right)$.

$A$ 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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