For a circular coil of radius R and number of turns N carrying a 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^(2))^(3//2)`
Consider two parallel co-axis circular coils of equal radius R and number of turns N, carrying equal current 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 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 know as Helmholtz coils).

Explanation

Magnetic field in a small region of length 2d about the mid point of the space between the two coils.
`B=(mu_(0)NIR^(2))/(2[((R)/(r)+d)^(2)+R^(2)]^(3//2))+(mu_(0)NIR^(2))/(2[((R)/(r)-d)^(2)+R^(2)]^(3//2))`
`=(mu_(0)NIR^(2))/(2){[(5R^(2))/(4)(1+(4d)/(5R))]^(-(3)/(2))+[(5R^(2))/(4)(1-(4d)/(5R))]^(-(3)/(2))}`
`=0.72(mu_(0)NI)/(R)`