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Q.

A thin insulated wire forms a plane spiral of N turns carrying a current I. The radii of inside and outside turns are equal to a and b. The magnetic induction at the centre of spiral

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a

$\frac{μ_{0} N I}{2 (b - a)}$

b

$\frac{μ_{0} N I}{2 (b - a)} l n (\frac{a}{b})$

c

$\frac{μ_{0} N I}{(b - a)} l n (\frac{b}{a})$

d

$\frac{μ_{0} N I}{2 (b - a)} l n (\frac{b}{a})$

answer is D.

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Detailed Solution

Question Image
Number of turn per unit length $= \frac{N}{b - a}$
Consider a ring of radius r and thickness dr.
Number of turns in $d r, N' = \frac{N}{b - a} d r$
$d B = \frac{μ_{0} N' I}{2 r}$
$B_{0} = \frac{μ_{0} N I}{2 (b - a)} \int_{a}^{b} \frac{d r}{r} = \frac{μ_{0} N I}{2 (b - a)} \ln (\frac{b}{a})$

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