For a circular coil of radius R and N turns carrying current I, the magnitude of the magnetic field at a point on its axis at a distance x from its centre is given by,
`B = (mu_(0)IR^(2)N)/(2(x^(2)+R)^(3//2))`
(a) Show that this reduces to the familiar result for field at the centre of the coil.
(b) Consider two parallel co-axial circular coils of equal radius R, and number of turns N, carrying equal currents in the same direction, and separated by a distance R. Show that the field on the axis around the mid-point between the coils is uniform over a distance that is small as compared to R, and is given by,
`B = 0.72 (mu_(0)NI)/(R)`, approximately.
[Such an arrangement to produce a nearly uniform magnetic field over a small region is known as Helmholtz coils.]

Explanation

Magnetic field on the axis of circular coil

magnetic field due to current of element of circular coil of radius r at point P at distance x from its centre is:
`dB=(mu_(0))/(4pi)=(Idl sin 90^(@))/(S^(2))=(mu_(0))/(4pi)(Id1)/((r^(2)+x^(2)))`
Component `dBcos phi` due to current element at point P is cancelled by equal and opposite component `d B cos phi` of another diametrically opposite current element, whereas the sine components `dB sin phi` add up to give net magnetic field along the axis. So, net magnetic field at point P due to entire loop is
`B=int dB sin phi=int_(0)^(2pir)(mu_(0))/(4pi)(Idl)/((r^(2)+x^(2))) . r/((r^(2)+x^(2))^(1//2))`
`B=(mu_(0)Ir)/(4pi(r^(2)+x^(2))^(3/2)) int_(0)^(2pir)`or `B=(mu_(0)Ir)/(4pi(r^(2)+x^(2))^(3//2)) 2pir` or `B=(mu_(0)Ir^(2))/(2(r^(2)+x^(2))^(3//2))`
directed along the axis, towards the coil if current in it is in clockwise direction away from the coil if current in it is in anticlockwise direction. If the magnetic field at the centre of the coil is considered, then x = 0.
`:.B=(mu_(0)IR^(2)N)/(2R^(3))=(mu_(0)NI)/(2R)`
This is the familiar result for magnetic field at the centre of the coil.