A long insulated copper wire is closely wound as a spiral of ' $N$ ' turns. The spiral has inner radius ' $a$ ' and outer radius ' $b$ '. The spiral lies in the $X-Y$ plane and a steady current ' $I$ ' flows through the wire. The $Z$-component of the magnetic field at the center of the spiral is
IIT 2011, Diffcult
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Let us consider an elementary ring of radius r and thickness dr in which current I is flowing.
Number of turns in this elementary ring $dN =\frac{ N }{ b - a } dr$
Thus magnetic field at the centre $O$ due to this ring $dB =\frac{\mu_0 IdN }{2 r }$
We get $dB =\frac{\mu_0 INdr }{2(b- a ) r }$
Net magnetic field at centre of spiral $B=\int_a^b \frac{\mu_0 I N ~ d r}{2(b-a) r}$
$\therefore B =\frac{\mu_0 IN }{2(b- a )} \int_{ a }^{ b } \frac{ dr }{ r }$
Or $B=\frac{\mu_0 I N}{2(b-a)} \times\left.\ln r\right|_a ^b$
Or $B=\frac{\mu_0 I N}{2(b-a)} \ln \frac{b}{a}$
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