If a long hollow copper pipe carries a current,then the magnetic field is produced:

$A$ solenoid of length $0.25 \ m$ has a radius of $1 \ cm$ and is made up of $500$ turns. It carries a current of $2.5 \ A$. What is the magnitude of the magnetic field inside the solenoid? (Take $\mu_0 = 4 \pi \times 10^{-7} \ SI$ units).

$A$ solenoid $60 \;cm$ long and of radius $4.0\; cm$ has $3$ layers of windings of $300$ turns each. $A$ $2.0 \;cm$ long wire of mass $2.5\; g$ lies inside the solenoid (near its centre) normal to its axis; both the wire and the axis of the solenoid are in the horizontal plane. The wire is connected through two leads parallel to the axis of the solenoid to an external battery which supplies a current of $6.0\; A$ in the wire. What value of current (in $A$) in the windings of the solenoid can support the weight of the wire? $(g=9.8\; m \,s ^{-2})$

If a copper rod carries a direct current,the magnetic field associated with the current will be

$A$ coil wrapped around a toroid has an inner radius of $20 \,cm$ and an outer radius of $25 \,cm$. If the wire wrapping makes $800$ turns and carries a current of $12 \,A$, the maximum and minimum values of the magnetic field within the toroid are:

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 $\Im(\infty)$?