Biot-Savart Law

It states that if a conductor carry current I having length l produces magnetic field that is directly proportional to I sinϴ dl and inversly proportional to square of r.
Let us consider a wire of length lcarries conductor I through it. Let us consider a point P at a distance r from an elementary length dl And ϴ be the angle made by r with dl. Then there will be magnetic field produced at P.
Now Biot and Savart kaw states that:
Magnetic field produced at p due to elementry length dl is
1) directly proportional to I
dB α I
2)directly proportional to sine of ϴ
dB α sinϴ
3)directly proportional to dl
dB α dl
4)inversly proportional to the square of r
dB α 1/r^2
Therefor
dB α Idlsinϴ/r^2
dB = KIdlsinϴ/r^2
dB = Uo/4π * Idlsinϴ/r^2

Application of Biot-Savart's Law
1) Magnetic field at centre of circular coil
Let us consider a circular coil of radius r carrying current in the direction as shown in figure. To calculate the magnetic field at centre of this coil, consider an elementary length dl which makes an angle of ϴ with radius vector r.
Now according to Biot-savart law the magnetic field at the centre of coil can be given as,
dB = Uo/4π * Idlsinϴ/r^2
Now, total magnetic field at centre due to whole coil can be given as,
B = 0[2πr U0/4π Idlsinϴ/r^2
B = U0Isinϴ/4πr^2 0[2πr dl
B = UoI/4πr^2 [l] 0-2πr
B = UoI/2r
if there are n coils
B = U0nI/2r

2)Magnetic field on axis of coil
Let us consider a circular coil of radius a which carry current I through it. Let its plane be perpendicular to the plane of paper such that its axis lied on plane of paper. Let us consider a point p at a distance x from centre of coil on its axis. Let us consider an elementary length dl on coil and r be the distance between p and dl. Let β be the angle between dl and r.
Now, magnetic field at P due to this elementary length dl is
dB = Uo/4π * Idlsinϴ/r^2
Here we can take
β =π/2
dB =Uo/4π * Idl/r^2--------------1
Now direction of dB is perpendicular to both dl and r. Let it be along PR which can be resloved into tow componentd dBcosϴ along perpendicular to axis and dBsinϴ along axix. Since coil is symmetric about axis. These perpendicular components, i.e dBcosϴ will cancel each other and therefore their contribution is zero. SO the magnetic field is due to the component along x-axis.
The magnetic field due to whole coil at point P can be given as:
B = [dBsinϴ
B = [Uo/4π * Idlsinϴ/r^2
B = Uo/4π * Isinϴ/r^2 0[2πr dl
B = Uo/4πr^2 Isinϴ*2πa
B = UoIa^2/2r^3
B = UoIa^2/2(a^2+x^2)^3/2

Case I
when P lies at the centre of coil x =0
B = UoIa^2/2a^3
B = UoI/2a
Case II
when P lies at very far disrance a<<<<B = UoIa^2/2x^3

3) Magnetic field due to long cunductor.
Let us consider a long conductor carrying current I. Consider a point P at a distance a from centre of conductor where magnetic field is to be found.
Let us consider on elementary length dl,AB. Let P be at a distance r from A and AB subtend an angle dα at P. Again let r makes an angle α with conductor. Now draw a perpendicular AC from A to BP. The we can have dB = Uo/4π * Idlsinϴ/r^2-----------1
Now in /_\ ACP
sindα = AC/AP
or dα = AC/r
AC = rdα-----------2

Again in /_\ ABC,
sinα = AC/AB = Ac/d1
AC = dlsinα

From 2 and 3
dlsinα = rdα
r = a/sinα
dB = Uo/4π * I/a * sinαdα

Now magnetic field at point P due to whole conductor can be given as
B = α1{α2 Uo/4π * I/a * sinαdα

If we consider the conductor as infinitely long than we can have
B = UoI/4a 0{π sinαdα-----------7
B = UoI/4πa *2
B = UoI/2πa