State Biot-Savart law Using Biot-Savart law, derive an expression for the magnetic field at the centre of a circular coil of number of turns ‘N’, radius ‘r’ carrying a current.

Biot-savart law states that, the magnetic field induction dB at a point due to current element is 

                $dB = \frac{{\mu }_o}{4\pi } \frac{I dl sin\theta }{r^2}$ 
where,
I.dl is the current element,
r is the distance of point from the current elemnt and,
$\mathrm{\theta } \mathrm{is} \mathrm{the} \mathrm{angle} \mathrm{between} \stackrel{\rightharpoonup }{\mathrm{dl} }\mathrm{and} \stackrel{\rightharpoonup }{\mathrm{r}}.$

Expression for magnetic field at the centre of a circular coil :

Consider a circular coil of radius 'r' with center O. Let, I be the current flowing in the circular coil. Assume, that the current coil is made of a large number of current elements each of length 'dl' .
current element = I.dl 
Now, as per Biot Savart law we have,  

               $dB = \frac{{\mu }_o}{4\pi } \frac{I dl sin\theta }{r^2}$
Since, the angle between $\stackrel{\rightharpoonup }{\mathrm{dl}} and \stackrel{\rightharpoonup }{r}$ is 90^0.
Therefore,
                     $dB = \frac{{\mu }_o}{4\pi } \frac{I dl}{r^2}$ 
Thus, 

B= $\int dB = \int \frac{{\mu }_o}{4\pi } \frac{I dl }{r^2} = \frac{{\mu }_o}{4\pi } \frac{I}{r^2}\int dl$
But, 
$$\begin{gathered} \int dl = total length of circular coil \\ = circumference of current loop = 2\pi r \end{gathered}$$ 

Therefore,
                  $\mathrm{B} = \frac{{\mathrm{\mu }}_{\mathrm{O}}}{4\mathrm{\pi }} \frac{2\mathrm{\pi nI}}{\mathrm{r} }$ 
where,
n is the number of turns in the coil.