The magnitude of the current has no effect on the (relative) uniformity of the field. This is a consequence of the Biot-Savart Law,
$$\vec{B}(\vec{r})=\frac{\mu_{0}}{4\pi}\int_{C}\frac{I\,d\vec{\ell}\,'\times(\vec{r}-\vec{r}\,')}{|\vec{r}-\vec{r}\,'|^{3}}.$$
(The integration path $C$ follows the circuital path, in this case encompassing both loops.) The key thing to observe about the expression for the magnetic field is that it depends on the current only through an overall factor of $I$; everything else is composed of fundamental constants and purely geometrical expressions. Therefore, changing the current has no effect on the three-dimensional shape of the magnetic field; it just scales the magnitude of the field vector $\vec{B}(\vec{r})$, while leaving the ratio of field strengths at different points $|\vec{B}(\vec{r}_{1})|/|\vec{B}(\vec{r}_{2})|$ unchanged.
Whenever you have two idential, parallel, coaxial (normal to the $z$-axis) coils, each carrying the same current, there will be a region in the center of the configuration where the combined field from both coils satisfies $\partial\vec{B}/\partial z=0$—that is, where the field profile along the $z$-axis is flat. How large the region in which the field is approximately constant is (both along the $z$-direction and in the $xy$-plane) depends on how far apart the rings are. The specific "Helmholtz coils" configuration, with the separation between the coils equal to the coil radius $R$ maximizes the size of that region, by ensuring that not just $\partial\vec{B}/\partial z$ vanish at the center point, but higher derivatives as well; in fact, the first three derivatives of $\vec{B}$ with respect to the $z$-coordinate vanish there if the coil separation is $R$. So with just two coils, the Helmholtz coil configuration is the best that you can do. You can buy an off-the-shelf undergraduate-level experiment kit for testing the uniformity of the fields as a function of the coil separation.
If the level of uniformity offered by Helmholz coils with separation $R$ is not sufficient, it is generally worth using a full solenoid to generate a more uniform field. In the limit of an infinite, tightly-wound solenoid, the field inside is completely uniform (not merely approximately uniform as for the Helmholtz coils). For solenoids of finite length, the uniformity is obviously again approximate, but near the center of a long circular solenoid, a very, very high level of $\vec{B}$ uniformity can be achieved.