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:

We want the magnetic field at the centre of a flat spiral coil that starts from an inner radius $$a$$ and ends at an outer radius $$b$$. The coil carries a steady current $$I$$ and has a total of $$N$$ closely packed turns.

For a single circular loop of radius $$r$$ carrying current $$I$$, the magnetic field at its centre is given by the well-known formula

$$B_{\text{single}}=\dfrac{\mu_0 I}{2r}.$$

In the spiral, the turns are so tightly wound that as we move radially outwards by a small distance $$dr$$ we encounter a small fractional turn $$dN$$. Because the winding is uniform, the number of turns per unit radial length is the same everywhere. The total radial width of the spiral is $$b-a$$, so

$$\text{Turns per unit length}= \dfrac{N}{\,b-a\,},\qquad\text{hence}\qquad dN=\dfrac{N}{\,b-a\,}\,dr.$$

Every elementary circular ring between radii $$r$$ and $$r+dr$$ behaves like $$dN$$ separate loops, each contributing the field $$\dfrac{\mu_0 I}{2r}$$ at the centre. Therefore the differential contribution to the field is

$$dB = \left(\dfrac{\mu_0 I}{2r}\right)\,dN =\left(\dfrac{\mu_0 I}{2r}\right)\left(\dfrac{N}{\,b-a\,}\,dr\right) =\dfrac{\mu_0 I N}{2(b-a)}\;\dfrac{dr}{r}.$$

All these elemental fields point in the same direction (perpendicular to the plane of the coil), so we add them directly. Integrating $$dB$$ from the inner radius $$a$$ to the outer radius $$b$$ gives the total magnetic field at the centre:

$$ \begin{aligned} B &= \int_{a}^{b} dB = \dfrac{\mu_0 I N}{2(b-a)}\int_{a}^{b}\dfrac{dr}{r} \\ &= \dfrac{\mu_0 I N}{2(b-a)}\Bigl[\ln r\Bigr]_{a}^{b} \\ &= \dfrac{\mu_0 I N}{2(b-a)}\;\ln\!\left(\dfrac{b}{a}\right). \end{aligned} $$

This expression matches option 4 exactly.

Hence, the correct answer is Option 4.