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:

$A$ 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 intensity $(H)$ at the centre of a long solenoid having '$n$' turns per unit length and carrying a current $I$, when no material is kept in it is ($\mu_0$ = permeability of free space)

Assertion: The magnetic field produced by a current-carrying solenoid is independent of its length and cross-sectional area.
Reason: The magnetic field inside the solenoid is uniform.

Consider a circular current-carrying loop of radius $R$ in the $x-y$ plane with its centre at the origin. Consider the line integral $\Im(L) = \left| \int_{-L}^{L} \vec{B} \cdot d\vec{l} \right|$ taken along the $z$-axis.
$(a)$ Show that $\Im(L)$ monotonically increases with $L$.
$(b)$ Use an appropriate Amperian loop to show that $\Im(\infty) = \mu_0 I$,where $I$ is the current in the wire.
$(c)$ Verify this result directly.
$(d)$ Suppose we replace the circular coil with a square coil of side $R$ carrying the same current $I$. What can you say about $\Im(L)$ and $\Im(\infty)$?

$A$ toroid has a number of turns per unit length $n$ and current $i$. What is the magnetic field inside the toroid?