$A$ thick current-carrying cable of radius $R$ carries current $I$ uniformly distributed across its cross-section. The variation of magnetic field $B(r)$ due to the cable with the distance $r$ from the axis of the cable is represented by:

$A$ long solenoid has a radius $a$ and the number of turns per unit length is $n$. If it carries a current $i$,then the magnetic field on its axis is directly proportional to

$A$ toroid has a non-ferromagnetic 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 ($\mu_{0} =$ permeability of free space).

$A$ 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 $(\mu_0 = 4 \pi \times 10^{-7} \text{ H m}^{-1})$.

The ratio of the magnetic field inside a solenoid at an axial point well inside and at an axial end point is

$A$ long solenoid has $200$ turns per $cm$ and carries a current $i$. The magnetic field at its centre is $6.28 \times 10^{-2} \ Wb/m^2$. Another long solenoid has $100$ turns per $cm$ and it carries a current $i/3$. The value of the magnetic field at its centre is: