Explain Biot Savart's law for a small current carrying conductor.

Explanation

Let XY= be the conductor carrying current I. Consider a very small element MN of length dl of a conductor carrying current I. Biot-Savart.s law states that the magnetic field induction at a point due to current I in element of length dl, at distance r from the elment is
`dB= (mu_(0))/(4pi) (Idl sin theta)/(r^(2))`
`d vec(B) = (mu_(0))/(4pi) (I (d vec(l) xx vec(r )))/(r^(3))`
The strength of magnetic field dB due to this small current element at a point P, distant r from the element is found to depend upon following quantities.
(i) `dB prop dl`
(ii) `dB prop I`
(iii) `d B prop sin theta`, (where `theta` is angle between `vec(dl) and vec(r )`)
(iv) `d B prop (1)/(r^(2))` on combining, `d B prop (I dl sin theta)/(r^(2))`
`d B = K (I dl sin theta)/(r^(2))`
Where k is constant of proportionality
In S.I units `K= (mu_(0))/(4pi)` where `mu_(0)` is absolute permeabiliyt of free space `dB= (mu_(0))/(4pi) (dl sin theta)/(r^(2))`
Value of `mu_(0) = 4pi xx 10^(-7) T m A^(-1)`
`(mu_(0))/(4pi) = 10^(-7) T m A^(-1)`
`vec(dB)= (mu_(0))/(4pi). (I vec(dl) xx hat(r ))/(r^(2)) (because vec(dl) xx hat(r ) = dl sin theta)`