If the number of turns per unit length of a solenoid is doubled, what happens to the magnetic field in the solenoid?

The magnetic flux near the axis and inside the air core solenoid of length $60$ cm carrying current $I$ is $\frac{\pi}{2} \times 10^{-6} \text{ Wb}$. Its magnetic moment will be (cross-sectional area is very small as compared to length of solenoid, $\mu_0 = 4\pi \times 10^{-7} \text{ SI unit}$) (in $\text{ Am}^2$)

For a long current-carrying solenoid,the magnetic field inside is $0.6 \ T$. The magnetic energy per unit volume is . . . . . . .

$A$ solenoid of length $50 \ cm$ and radius $10 \ cm$ has two closely wound layers of windings, each with $100$ turns. If a current of $2.5 \ A$ is passing through the windings, the magnetic field (in $10^{-4} \ T$) at a point $5 \ cm$ from the axis is:

Assertion: The magnetic field produced by a current-carrying solenoid is independent of its length and cross-sectional area.
Reason: The magnetic field inside the solenoid is uniform.

Consider two idealized systems: $(i)$ a parallel plate capacitor with large plates and small separation,and $(ii)$ a long solenoid of length $L \gg R$,where $R$ is the radius of the cross-section. In $(i)$,$E$ is ideally treated as a constant between the plates and zero outside. In $(ii)$,the magnetic field is constant inside the solenoid and zero outside. These idealized assumptions,however,contradict fundamental laws as follows: