Answer 1

The law which gives the magnetic field at a point due to a current  element, is Biot-Savart Law.
It states that,if we consider an infinitesimal element dl of the
conductor,the magnetic field dB due to this
element can be determined at a point P at a distance r from it, as follows:
Let θ be the angle between dl and the displacement vector r.
According to Biot-Savart’s law,the magnitude of the magnetic field dB is proportional tothe current I, the element length |dl|,and inversely proportional to the square of the distance r. Its direction is  perpendicular to the plane containing dl and r .

Thus, in vector notation, `d vec( B )  α  I (d vec (l)xx vec(r))/(r^3)`

where μ₀ / 4π is a constant of proportionality. The above expression holds when the medium is vacuum.The magnitude of this field is, 

`|d vec(B) | = (mu_0)/(4pi) (Idlsin theta )/(r^2)`

The proportionality constant in SI units has the exact value,

`(mu_0)/(4pi) = 10^(-7) T - m/A`

We call μ₀ the permeability of free space (or vacuum). 

Magnetic field at the centre of a circular loop of radius r carrying current I:

We consider a small element of length dl. The `vec (r)` joining the centre of the circle to this element is  perpendicular to`d vec(l)` . Hence the magnitude of magnetic field at the centre due to  this element (by Biot-Savart Law) is

`dB = (mu_0)/(4 pi) (Idl sin theta)/(r^2) or dB=  (mu_0)/(4 pi) (Idl sin 90^circ)/(r^2)  = (mu_0)/(4pi) (Idl)/(r^2)` 

This field is perpendicular to the plane of the coil (as it is directed along `dvec(l) xx vec (r))`.
Hence, net field due to the entire circular loop, is

`B = int  (mu_0)/(4 pi) (Idl )/(r^2) = (mu_0)/(4 pi) (I)/(r^2) int dl = (mu_0)/(4 pi) xx(I)/(r^2) xx 2 pir`

Or B = `(mu_0I)/(2r)`