State Ampere's circuital law and explain its mathematical expression.

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 $60 \;cm$ 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? $(g=9.8\; m \,s ^{-2})$

$A$ particle of mass $2.2 \times 10^{-30} \,kg$ and charge $1.6 \times 10^{-19} \,C$ is moving at a speed of $10 \,km/s$ in a circular path of radius $2.8 \,cm$ inside a solenoid. The solenoid has $25 \,turns/cm$ and its magnetic field is perpendicular to the plane of the particle's path. The current in the solenoid is (Take $\mu_0 = 4\pi \times 10^{-7} \,H/m$) (in $\,mA$)

The magnetic field inside a long solenoid is:

$A$ solenoid of length $0.5 \ m$ has a radius of $1 \ cm$ and is made up of $250$ turns. It carries a current of $5 \ A$. What is the magnitude of the magnetic field inside the solenoid?