Derive an expression for the magnetic induction at the centre of a current carrying circular coil using Biot-Savart law.

Explanation

Consider a circular coil of radius r and carry a current i. Consider a small element 'dl'. Let O is the centre of the coil. By using Biot- Savart's law
`dB=(mu_(0))/(4pi).(I dl sin theta)/(r^(2))`
Here angle `vec(dl) and vecr" is "90^(@)(i.e. theta = 90^(@))`
`dB=(mu_(0))/(4pi)(idl)/(r^(2))rarr(1)`
As the field due to all elements of the circular loop have the same direction.
The resultant magnetic field can be obtained by integrating equation (1)
`int dB = int (mu_(0))/(4pi). (i dl)/(r^(2))`
`B=(mu_(0)i)/(4pi r^(2)) int dl" "because ( int dl =2pi r)`
`=(mu_(0)i)/(4pi r^(2))xx2pi r`
`B=(mu_(0)i)/(2r)`
If the circular coil has n turns.
`B=(mu_(0) ni)/(2r)`