Q 12-04-164JEE MainJEE Main 2021 (31 Aug, Shift 1)Medium
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 centre, when a current $I$ passes through coil:
Answer: (D) $\dfrac{\mu_0IN}{2(b - a)}\log_e\left(\dfrac ba\right)$
Turns per unit radial width: $\dfrac{N}{b - a}$. A ring of radius $r$ and width $dr$ has $dN = \dfrac{N\,dr}{b - a}$ turns.
$$B = \int_a^b\frac{\mu_0I}{2r}\cdot\frac{N\,dr}{b - a} = \frac{\mu_0NI}{2(b - a)}\ln\frac ba$$
Solution by Sreeraj P, M.Sc Physics