If a long hollow copper pipe carries a current,then the magnetic field is produced:

$A$ closely wound solenoid $120 \ cm$ long has $4$ 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 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)

What current must be passed through a solenoid with $20 \text{ turns/cm}$ to produce a magnetic field of $20 \text{ mT}$ (in $\text{ A}$)? (Given: $\frac{\mu_0}{4\pi} = 10^{-7} \text{ T m/A}$)

$A$ solenoid of length $2 \ m$ carries a current of $20 \ A$. The diameter of the solenoid is $3 \ cm$. If the magnetic field inside the solenoid is $20 \ mT$, then the length of wire forming the solenoid is (assume $\mu_0 = 4 \pi \times 10^{-7} \ H/m$) (in $m$)

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: