Using Biot - Savart's law, derive an expression for magnetic field at any point on axial line of a current carrying circular loop. Hence, find magnitude of magnetic field intensity at the centre of circular coil.

Explanation

Consider a circular loop of radius R, carrying a current I. Plane of the coil is Y-Z plane (i.e., perpendicular to the plane of paper), while the axis of the loop OX lies in the plane of paper. Consider the circular loop to be divided into large number of current elements `Ivec(dt)`. Two such elements `N_(1)M_(1) and N_(2)M_(2)`, diametrically opposite to each other, produce the magnetic fields `vec(dB_(1)) and vec(dB_(2))` perpendicular to both `vecr and vec(dl)` and given by right hand rule as,
`|vec(dB_(1))|=|vec(dB_(2))|=(mu_(0))/(4pi).(Idl sin 90^(@))/(r^(2))=(mu_(0))/(4pi)(Idl)/((R^(2)+x^(2)))`
These fields may be resolved into components along x -- axis and y - axis. Obviously components along y - axis balance each other but components along x - axis are all summed up. Hence, magnetic field due to whole current loop will be :
`B=oint dB sin phi=oint(mu_(0))/((R^(2)+x^(2))).(R)/(sqrt((R^(2)+x^(2))))=(mu_(0)IR)/(4pi(P^(2)+x^(2))^(3//2))oint dl`
`=(mu_(0)IR)/(4pi(R^(2)+x^(2))^(3//2)).2piR=(mu_(0)IR^(2))/(2(R^(2)+x^(2))^(3//2))`

For a point at the centre of circular coil x = 0, so value of field B will be
`B=(mu_(0)IR^(2))/(2R^(3))=(mu_(0)I)/(2R)`