The magnetic field intensity $(H)$ at the centre of a long solenoid having '$n$' turns per unit length and carrying a current $I$, when no material is kept in it is ($\mu_0$ = permeability of free space)

In a toroid,the number of turns per unit length is $1000$ and the current through it is $\frac{1}{4 \pi} \ A$. The magnetic field produced inside (in $Wb/m^2$) will be:

$A$ closely wound solenoid of $1200$ turns and area of cross-section $5 \,cm^2$ carries a current. If the magnetic moment of the solenoid is $1.2 \,J \,T^{-1}$, then the current through the solenoid is (in $\,A$)

$A$ long solenoid carrying a current produces a magnetic field $B$ along its axis. If the current is tripled and the number of turns per cm is halved, then the new value of the magnetic field will be:

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 solenoids $X$ and $Y$ such that the area and length of $Y$ are twice that of $X$ respectively and the magnetic energy stored in both the solenoids is same, then the ratio of magnitude of magnetic fields of the two solenoids $\frac{|B_X|}{|B_Y|}$ is