A coil of n number of turns is wound tightly | vedclass

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(D) Consider an element of thickness $dr$ at a distance $r$ from the centre of the spiral coil.Total number of turns $= n$.The radial width of the coi

Vedclass · Accepted Answer

$\frac{\mu_{0} n I}{2(b-a)} \log _{e} \frac{b}{a}$

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(D) Consider an element of thickness $dr$ at a distance $r$ from the centre of the spiral coil.Total number of turns $= n$.The radial width of the coil is $(b - a)$.Number of turns per unit radial length $= \frac{n}{b - a}$.Number of turns in the element of thickness $dr$ is $dn = \frac{n}{b - a} dr$.The magnetic field at the centre due to this circular element of radius $r$ is $dB = \frac{\mu_{0} I dn}{2r}$.Substituting $dn$,we get $dB = \frac{\mu_{0} I}{2r} \left( \frac{n}{b - a} \right) dr$.To find the total magnetic field $B$,we integrate from $r = a$ to $r = b$:$B = \int_{a}^{b} \frac{\mu_{0} n I}{2(b - a)} \frac{dr}{r} = \frac{\mu_{0} n I}{2(b - a)} [\log_{e} r]_{a}^{b}$.$B = \frac{\mu_{0} n I}{2(b - a)} \log_{e} \left( \frac{b}{a} \right)$.

$A$ coil of $n$ number of turns is wound tightly in the form of a spiral with inner and outer radii $a$ and $b$ respectively. When a current of strength $I$ is passed through the coil,the magnetic field at its centre is

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