Magnetic Field on the Axis of a circular loop ( coil ) carrying Current
Consider a point P on the axis of the current loop at a distance $x$ from the centre O of the loop. Every current element ($d\vec{l}$) of the loop is perpendicular to the direction of $\vec{r}$ (represented by unit vector $\hat{r}$).
Thus, $|d\vec{l} \times \hat{r}| = dl \sin 90^\circ = dl$.
Thus, every element of the current loop is at a distance $r = \sqrt{R^2 + x^2}$ from the point P on the axis of the current loop.
According to Biot-Savart's law, the magnitude of the magnetic field due to current element ($Id\vec{l}$) at a distance $r$ is given by
$ dB = \frac{\mu_0}{4\pi} \frac{I |d\vec{l} \times \hat{r}|}{r^2}$
$dB = \frac{\mu_0}{4\pi} \frac{I dl}{r^2} $.... (1)
The direction of the magnetic field is perpendicular to the plane formed by $d\vec{l}$ and $\hat{r}$ (along PL).
Resolving magnetic field ($d\vec{B}$) into two components:
(i) $dB_x = dB \cos \theta$ along the axis of the current loop (or coil) and away from the centre of the loop.
(ii) $dB_y = dB \sin \theta$, which is perpendicular to the axis of the loop or coil.
As the loop or coil is symmetrical about its axis, so every current element of length $dl$ has another equal and opposite current element on the loop or coil.
Thus, $dB_y = dB \sin \theta$ component of magnetic field due to each current element of the loop is cancelled by the $dB_y = dB \sin \theta$ component of the magnetic field due to the equal and opposite current element on the loop.
Therefore, the sum of $dB_y$ components of all the current elements at point P is zero.
However, $dB_x = dB \cos \theta$ components of magnetic field due to all current elements of the loop are in the same direction and hence added up.
Therefore, magnetic field due to current loop or coil at point P is given by
$ B = \sum dB_x$
$B= \sum dB \cos \theta$
$B= \oint dB \cos \theta $ .....(2)
Now using eqn. (1) and $\cos \theta = \frac{R}{r}$, we get
$B = \oint \left( \frac{\mu_0}{4\pi} \frac{Idl}{r^2} \right) \times \frac{R}{r}$
$B= \left( \frac{\mu_0}{4\pi} \right) \frac{IR}{r^3} \oint dl $
Since $r = \sqrt{R^2 + x^2}$ and $\oint dl = 2 \pi R$ (length of loop)
$ B = \left( \frac{\mu_0}{4\pi} \right) \frac{IR \times 2 \pi R}{(R^2 + x^2)^{3/2}}$
$B= \left( \frac{\mu_0}{4\pi} \right) \frac{2 \pi IR^2}{(R^2 + x^2)^{3/2}}$ ....(3)
The direction of magnetic field is along x-axis (axis of the loop).
Therefore, eqn. (3) can be written as
$ \vec{B} = \left( \frac{\mu_0}{4\pi} \right) \frac{2 \pi IR^2}{(R^2 + x^2)^{3/2}} \hat{i}$ ....(4)
Thus, magnetic field due to a current carrying loop or coil is perpendicular to the plane of the loop or coil and upward.
If coil has N turns, then the magnetic field at a distance x from the centre of the coil on its axis is given by
$ B' = NB $
Using eqn. (3), we get
$ B' = \left( \frac{\mu_0}{4\pi} \right) \frac{2 \pi NIR^2}{(R^2 + x^2)^{3/2}} $....(5)
Special Cases:
(i) If point of observation (point P) is far away from the loop or coil (i.e. x > R), then $R^2$ can be neglected as compared to $x^2$, hence $(R^2 + x^2)^{3/2} = x^3$
Then, equation (3) becomes $B = \frac{\mu_0}{4 \pi} \frac{2 \pi I R^2}{x^3}$
Since area of coil or loop, $A = \pi R^2$,
$B = \frac{\mu_0}{4 \pi} \frac{2 I A}{x^3}$
(ii) Magnetic field at the centre of the current loop or coil. At the centre of current loop, $x= 0$. Hence from eqn. (3), we get
$B = \left( \frac{\mu_0}{4 \pi} \right) \frac{2 \pi I}{R}$
For a loop or coil having N turns, eqn. (8) becomes
$B' = NB = \left( \frac{\mu_0}{4 \pi} \right) \frac{2 \pi NI}{R}$
Variation of magnetic field of a current carrying loop along its axis is shown in figure
