Let R= be the radius of circular loop. I= be the current flowing anticlockwise, OX= be the axis of loop
P = be the point on the axis of loop, where we want to find magnetic field. OP = x
`CP= DP = r= sqrt(R^(2) + x^(2))`
`I vec(dl)`= be the current element. According to Biot Savart.s law, the magnitude of magnetic field due to current element `I vec(dl)` at a distance r is given by
`dB= (mu_(0))/(4pi) (I|vec(dl) xx hat(r )|)/(r^(2))= (mu_(0))/(4pi) (Id l)/(r^(2))` ...(1)
Resolving Magnetic field `(vec(dB))` into two rectangular components:
(i) `dB_(x)= d B cos theta`, along the axis of current loop and away from centre of loop.
(ii) `dB_(y)= d B sin theta`, which is `bot` to axis of loop or coil. `d B_(y)= d B sin theta` component of magnetic field due to each current element of loop is cancelled by `d B._(y)= d B. sin theta` component of magnet field due to equal and opposite current element on loop.
`therefore` Sum of d `B_(y)` component of all current element is zero. `dB_(x)= d B cos theta` components of magnetic field due to all current elements of loop are in same direction and hence added up. Therefore magnetic field at P is given by
`B= Sigma d B_(x)= Sigma dB cos theta= oint d B cos theta` ...(2)
Now, `cos theta= (R )/( r)`
and `dB= (mu_(0))/(4pi) (Idl)/(r^(2))`
from (2), `B= oint ((mu_(0))/(4pi)) (Idl)/(r^(2)) xx (R )/(r)`
`=((mu_(0))/(4pi)) (IR)/(r^(3)) oint dl`
Since `r= sqrt(R^(2) + x^(2)) and oint dl = 2pi R`
`therefore B= ((mu_(0))/(4pi)) (IR xx 2pi R)/((R^(2) + x^(2))^(3//2))`
`=(mu_(0))/(4pi) (2pi IR^(2))/((R^(2) + x^(2))^(3//2))` ...(3)
The direction of magnetic field is along x axis
`therefore` equation (3) can be written as `vec(B)= ((mu_(0))/(4pi)) (2pi IR^(2))/((R^(2) + x^(2))^(3//2))`
If coil has N turns, then the magnetic field at a distance x from centre of coil on its axis is given by `B.= NB`
using equation (3) `B.= ((mu_(0))/(4pi)) (2pi NIR^(2))/((R^(2)+ x^(2))^(3//2))`
Special cases: (i) If point of observation is far away from the loop or coil i.e., `(x gt gt gt R)` then `R^(2)` can be neglected as compared to `x^(2)`.
hence `(R^(2) + x^(2)) ~~ x^(2)`, then equation (3) becomes
`B= (mu_(0))/(4pi).(2pi IR^(2))/(x^(3))`
SInce area of coil or loop is given by, `A= pi R^(2)`
`B= (mu_(0))/(4pi) (2IA)/(x^(3))`
(ii) At the centre of the coil, x= 0
`therefore B= (mu_(0))/(4pi) (2pi IR^(2))/((R^(2))^(3//2))`
`rArr B= (mu_(0))/(4pi) (2pi I cancel(R^(2)))/(R^(cancel3))`
`rArr B= (mu_(0))/(4pi) (2pi I)/(R )`
