A tangent galvanometer is directly connected to an ideal battery....

The magnetic field at the center of the coil of a tangent galvanometer is given by $B = \dfrac{\mu_0 N I}{2r}$.

The current through the coil is $I = \dfrac{V}{R}$, where $V$ is the voltage of the ideal battery and $R$ is the resistance of the coil.

The resistance of the coil is proportional to the total length of the wire, which is proportional to the number of turns $N$ (assuming the radius $r$ of the coil is constant). Let $R = c N$, where $c$ is a constant.

Substituting $I$ in the expression for $B$, we get:

$B = \dfrac{\mu_0 N}{2r} \left( \dfrac{V}{c N} \right) = \dfrac{\mu_0 V}{2 r c}$

This shows that the magnetic field $B$ is independent of the number of turns $N$.

The deflection $\theta$ in a tangent galvanometer is given by $B = B_H \tan \theta$, where $B_H$ is the horizontal component of the Earth's magnetic field.

Since $B$ remains unchanged, the deflection $\theta$ will also remain unchanged.