The magnetic field intensity $(H)$ at the centre of a long solenoid carrying a current of $2 \ A$ is found to be $1000 \ A/m$. The number of turns per centimeter of the solenoid is: (Use $\mu_0 = 4 \pi \times 10^{-7} \ T \ m \ A^{-1}$)

The number of turns per unit length of a long solenoid is $10 \, \text{turns/cm}$. If its average radius is $5 \, \text{cm}$ and it carries a current of $10 \, \text{A}$, then the ratio of flux densities obtained at the centre and at the end on the axis will be:

$A$ 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 $\mu_0 = 4\pi \times 10^{-7} \text{ SI units}$.)

$A$ current $i$ $A$ flows along an infinitely long straight thin-walled tube. The magnetic induction at any point inside the tube is:

In a toroid,the magnetic field on the axis is to be calculated given the radius $r = 0.5 \, cm$,current $I = 1.5 \, A$,number of turns $N = 250$,and relative permeability $\mu_r = 700$. Find the magnetic field $B$ (in Tesla). (in $.5$)

There are $50$ turns per $cm$ length in a very long solenoid. It carries a current of $2.5 \ A$. The magnetic field at its centre on the axis is . . . . . . $T$.