For a circular coil of radias R and N turns carrying current I, the magnitude of the magnetic field at a point on its at a distance x from its centre is given by
`B=(mu_0 IR^2 N)/(2(x^2+R^2)^(1/2))`
a. Show that this reduces to the familar result for field the centre of the coil . ]

Explanation

` therefore B = (mu_0. IR^2N)/(2(x^2 + R^2)^2)` At the cetre , x=0
`therefore B = (mu_0. IR^3 N)/(2R^3) = (mu_0 IN)/(2R)`. This is the expression for field at the centre of a circular coil.
b. Axial field , `B=(mu_0 NI R^2)/(2(R^2+x^2)^(1/3))`
Here `x=R/2`
`therefore B = (mu_0 NI)/(2) xx (R^2)/((R^2 +R^2/4))=(mu_0NI)/(2) xx (R^2)/(R^3(5/4)^(1/2)) = (mu_0NI)/(2R(5/4)^(1//2))`
`therefore` For two coils , `B=(mu_0 NI)/(R) (4/5)^(1/2) = (0.72 mu_0NI)/(R)`