$A$ solenoid of length $50 \ cm$ and radius $10 \ cm$ has two closely wound layers of windings, each with $100$ turns. If a current of $2.5 \ A$ is passing through the windings, the magnetic field (in $10^{-4} \ T$) at a point $5 \ cm$ from the axis is:

Using Ampere's circuital law,derive the expression for the magnetic field at a perpendicular distance $r$ from an infinitely long current-carrying wire. Explain it.

From Ampere's circuital law for a long straight wire of circular cross-section carrying a steady current,the variation of magnetic field in the inside and outside region of the wire is :

$A$ 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? $(g=9.8\; m \,s ^{-2})$

In a co-axial straight cable,the central conductor and the outer conductor carry equal currents in opposite directions. The magnetic field is zero:

$A$ 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: $\mu_0 = 4\pi \times 10^{-7} \text{ SI units}$)