Derive an expression for magnetic field produced at centre of a current carrying coil. How will you determine the direction of this field ?

Explanation

Consider a circular coil of radius R with centre O, lying with its plaen in the plane of paper. Let I be current flowing in circular coil in direction shown in fig. Suppose the circular coil is made of a large no. of current elements each of length dl. Let R= be the radius of circular coil with centre O. According to Biot Savart.s law, the magnetic field at centre of circular coil due to current elemetn `I vec(dl)` is given by, `dB= (mu_(0))/(4pi) (I dl sin theta)/(R^(2))`
since the angle between `vec(dl) and vec(R ) " is " 90^(@)`
`therefore dB= (mu_(0))/(4pi) (I dl sin 90^(@))/(R^(2))`
SInce `sin 90^(@)=1`
`therefore dB= (mu_(0))/(4pi) (I dl)/(R^(2))`

The total magnetic field at point 0 due to current in whole circular coil can be obtained by Integrating above equation,
`B = int dB= int (mu_(0))/(4pi) (I dl)/(R^(2))= (mu_(0)I)/(4pi R^(2)) int dl`
But `int dl`= circumference of coil `=2pi R`
`B= (mu_(0)I)/(4pi R^(2))2pi R= (mu_(0)2pi I)/(4pi R)`
If circular coil consists n turns
`B= (mu_(0))/(4pi) (2pi nI)/(R) = (mu_(0))/(4pi) (I)/(R ) xx 2pi n`
`B= (mu_(0))/(4pi) (2pi n I)/(R )`
Direction of `vec(B)`: In above case, the direction of magnetic field Induction `vec(B)` at point 0, due to current through circular coil is `bot` to the plane of circular coil directed inwards is given by Right Hand rule.