A long insulated copper wire is closely wound as a spiral of N...

If we take a small strip of $d r$ at distancer from centre, then number of turns in this strip would be d N=$\frac{N}{b-a}$d r Magnetic field due to this element at the centre of the coil will be $$\begin{aligned}& d B=\frac{\mu_0(d N) I}{2 r}=\frac{\mu_0 N I}{2(b-a)} \frac{d r}{r}\\ \therefore \quad & B=\int_{r=a}^{r=b} d B=\frac{\mu_0 N I}{2(b-a)} \ln \frac{b}{a} \end{aligned}$$\therefore$Correct answer is (a). Analysis of Question (i) If we see this problem independently, then I will rate this question moderately difficult. But the idea of this question is taken from question number$3.245$ of IE Irodov. (ii) Interestingly the same question was asked in IIT-JEE 2001 also.