$A$ toroid has an iron core with an internal magnetic field of $10 \pi \text{ mT}$,when the current in the winding of $1500 \text{ turns/m}$ is $10 \text{ A}$. Determine the field due to magnetization $(\mu_0 = 4 \pi \times 10^{-7} \text{ H m}^{-1})$.

$A$ toroid has a core of inner radius $r_1$ and outer radius $r_2$,around which $N$ turns of wire are wound. If the current in the wire is $I$,then the magnetic field inside the toroid is $(\mu_0 = \text{permeability of free space})$

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 $................\, \text{A}$.

$A$ solenoid has $N$ turns,length $l$,and cross-sectional radius $r$. If a current $i$ flows through the solenoid,what is the magnetic field at the axial midpoint? (Given $l \simeq r$)

The magnetic field $B$ within a solenoid having $n$ turns per meter length and carrying a current of $i$ amperes is given by:

$A$ coaxial cable consists of an inner wire of radius $a$ surrounded by an outer shell of inner and outer radii $b$ and $c$ respectively. The inner wire carries an electric current $i_o$,which is distributed uniformly across its cross-sectional area. The outer shell carries an equal current in the opposite direction,also distributed uniformly. What will be the ratio of the magnetic field at a distance $x$ from the axis when $(i)$ $x < a$ and $(ii)$ $a < x < b$?