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

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(A) The number of turns per unit radial width is $n = \frac{N}{b-a}$.Consider a small elemental ring of radius $x$ and width $dx$. The number of turns

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The number of turns per unit radial width is $n = \frac{N}{b-a}$.Consider a small elemental ring of radius $x$ and width $dx$. The number of turns in this element is $dN = n \cdot dx = \frac{N}{b-a} dx$.The magnetic field at the centre due to this elemental ring is $dB = \frac{\mu_{0} (dN) I}{2x} = \frac{\mu_{0} I}{2x} \left( \frac{N}{b-a} \right) dx$.Integrating from $x = a$ to $x = b$:$B = \int_{a}^{b} \frac{\mu_{0} I N}{2(b-a)} \frac{dx}{x} = \frac{\mu_{0} I N}{2(b-a)} [\ln x]_{a}^{b} = \frac{\mu_{0} I N}{2(b-a)} \ln \left( \frac{b}{a} \right)$.

$A$ coil having $N$ turns is wound tightly in the form of a spiral with inner and outer radii $a$ and $b$ respectively. Find the magnetic field at the centre,when a current $I$ passes through the coil.

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