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

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

To manufacture a solenoid of length $\ell$ and inductance $L$,the length of the thin wire required is (Diameter of the solenoid is very small compared to its length,$\mu_0$ is the permeability of free space).

$A$ closely wound solenoid $80\; cm$ long has $5$ layers of windings of $400$ turns each. The diameter of the solenoid is $1.8\; cm$. If the current carried is $8.0\; A$,estimate the magnitude of $B$ inside the solenoid near its centre.

The magnetic field inside a $200$ turns solenoid of radius $10 \ cm$ is $2.9 \times 10^{-4} \ T$. If the solenoid carries a current of $0.29 \ A$,then the length of the solenoid is . . . . . . $\pi \ cm$.

Using Ampere's circuital law,derive the expression for the magnetic field at a perpendicular distance $r$ from an infinitely long current-carrying wire. Explain it.