$A$ current of $2\, A$ flows in a long,straight wire of radius $2\, mm$. The intensity of the magnetic field on the axis of the wire is:

The magnetic flux near the axis and inside the air core solenoid of length $60$ cm carrying current $I$ is $\frac{\pi}{2} \times 10^{-6} \text{ Wb}$. Its magnetic moment will be (cross-sectional area is very small as compared to length of solenoid, $\mu_0 = 4\pi \times 10^{-7} \text{ SI unit}$) (in $\text{ Am}^2$)

The magnetic intensity at the centre of a long current-carrying solenoid is found to be $1.6 \times 10^3 \text{ A m}^{-1}$. If the number of turns is $8 \text{ per cm}$,then the current flowing through the solenoid is $................\, \text{A}$.

$A$ toroid is a long coil of wire wound over a circular core. If $r$ and $R$ are the radii of the coil and toroid respectively,the coefficient of self-induction of the toroid is (The magnetic field in it is uniform and $R >> r$). ($N =$ number of turns of the coil and $\mu_{0} =$ permeability of free space)

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 $.5$)

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: