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

The figure shows the cross-sectional view of a hollow cylindrical conductor with inner radius $R$ and outer radius $2R$,carrying a uniformly distributed current $i$ along its axis. The magnetic induction at point $P$ at a distance $3R/2$ from the axis of the cylinder will be

$A$ cylindrical conductor of radius $R$ is carrying a constant current. The plot of the magnitude of the magnetic field,$B$ with the distance $d$,from the centre of the conductor,is correctly represented by the figure:

$A$ thick current-carrying cable of radius $R$ carries current $I$ uniformly distributed across its cross-section. The variation of magnetic field $B(r)$ due to the cable with the distance $r$ from the axis of the cable is represented by:

$A$ $25\,cm$ long solenoid has radius $2\,cm$ and $500$ total number of turns. It carries a current of $15\,A.$ If it is equivalent to a magnet of the same size and magnetization $\vec M$ (magnetic moment per unit volume),then $\left| {\vec M} \right|$ is

$A$ long solenoid carrying current $I_1$ produces a magnetic field $B_1$ along its axis. If the current is reduced to $20 \%$ and the number of turns per $cm$ is increased five times,then the new magnetic field $B_2$ is equal to: