The expression for magnetic induction inside a solenoid of length $L$ carrying a current $I$ and having $N$ number of turns is

$A$ solenoid has a core of a material with relative permeability $400$. The windings of the solenoid are insulated from the core and carry a current of $1 \text{ A}$. If the number of turns is $1000 \text{ per metre}$,the magnetic field $(B)$ is . . . . . . $\text{T}$. (Given: $\mu_0 = 4\pi \times 10^{-7} \text{ SI units}$)

$A$ toroidal solenoid with air core has an average radius $R$,number of turns $N$,and area of cross-section $A$. The self-inductance of the solenoid is (Neglect the field variation across the cross-section of the toroid).

The magnetic field at the centre of a long solenoid having $400$ turns per unit length and carrying a current $i$ is $6.24 \times 10^{-2} \,T$. The magnetic field at the centre of another long solenoid having $200$ turns per unit length and carrying a current $\frac{i}{2}$ is

There are $50$ turns of a wire in every $1$ $cm$ length of a long solenoid. If $4$ $A$ current is flowing in the solenoid,the approximate value of the magnetic field along its axis at an internal point and at one end will be respectively:

The figure shows a long straight wire of circular cross-section (radius $a$) carrying a steady current $I$. The current $I$ is uniformly distributed across this cross-section. Calculate the magnetic field in the region $r < a$ and $r > a$.