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

Explain Ampere's circuital law.

$A$ solenoid has a core made of material with relative permeability $400$. The magnetic field produced in the interior of the solenoid is $1.0 \text{ T}$. The magnetic intensity in $SI$ units is $\alpha \times 10^5$. The value of $\alpha$ is . . . . . . . (Free space permeability $\mu_0 = 4\pi \times 10^{-7} \text{ SI units}$.)

$A$ long solenoid carrying a current produces a magnetic field $B$ along its axis. If the number of turns per $cm$ is doubled and the current is made $\left(\frac{1}{3}\right)^{rd}$ of its original value,then the new value of the magnetic field will be:

The correct curve between the magnetic induction $B$ along the axis of a long solenoid due to current flow $i$ in it and distance $x$ from one end is:

$A$ long solenoid has a radius $a$ and the number of turns per unit length is $n$. If it carries a current $i$,then the magnetic field on its axis is directly proportional to