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_0IR^2N)/(2(x^2+R^2)^(3//2))`
(a) Show that this reduces to the familiar result for field at the centre of the coil.
(b) Consider two parallel coaxial 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_0NI)/(R)` approximately.
[Such as arrangement to produce a nearly uniform magnetic field over a small region is known as Helmholtz coils.]

Explanation

(b) In a small region of length 2d about the mid-point between the coils
`B = (mu_0 IR^2 N)/(2) xx [ { (R/2 + d)^2 + R^2 }^(-3//2) + { (R/2 - d)^2 + R^2 }^(-3//2) ]`
`= (mu_0 IR^2 N)/(2) xx ((5R^2)/(4))^(-3//2) xx [ ( 1 + (4d)/(5R)^(-3//2)) + ( 1 - (4d)/(5R) )^(-3//2) ]`
` = (mu_0IR^2N)/(2R^3) xx (4/5)^(3//2) xx [ 1 - (6d)/(5R) + 1 + (6d)/(5R) ]`
where in the second and third steps above, terms containing `d^2 // R^2` and higher powers of d/R are neglected since `d/R lt lt 1`. The terms linear in d/R cancel giving a uniform field B in a small region:
`B = (4/5)^(3//2) (mu_0 IN)/( R) = 0.72 (mu_0 IN)/(R )`