A neutral point is obtained at the Centre of a vertical circular coil carrying current. The angle between the plane of the coil and the magnetic meridian is

$R=\frac{\sqrt[]{2Em}}{B.q}$

$\Rightarrow R\times \frac{\sqrt[]{m}}{q}$

$\Rightarrow \frac{R_1}{R_2}=\frac{\sqrt[]{4}}{2}\times \frac{3}{\sqrt[]{16}}=\frac{3}{4}$

$\Rightarrow R_2=\frac{4R_1}{3}$

$sin\theta =\frac{d}{R}\Rightarrow \theta \alpha \frac{1}{2}\Rightarrow {\theta }_2<{\theta }_1$

Let us consider an elementary ring of radius r and thickness dr in which current is flowing.

So, No. of turns in this elementary ring

$dN=\left(\frac{N}{b-a}\right)dr$

$\therefore {\left(dB\right)}_{atcentre}=\frac{{\mu }_{0}ldN}{2r}$

$B=\frac{{\mu }_{0}lN}{2\left(b-a\right)}\mathcal{l}n\frac{b}{a}$

 $\overrightarrow{a}\perp \overrightarrow{B}$

$\Rightarrow \overrightarrow{a}.\overrightarrow{B}=0$

$\Rightarrow \left(2\alpha -12\right)=0$

= 6