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

Radii of two parallel co-axial circular coils = R , Number of turns on each coil = N
Current in both coils = l , Distance between both the coils = R
Let us consider point Q at distance d from the centre ,
Then one coil is at a distance of `R/2+d` from point Q
`:.` Magnetic field at point Q is given as `B_(1)=(mu_(0)NIR^(2))/(2[(R/2+d)^(2)+R^(2)]^(3//2))`
Also, the other coil is at a distance `R/2-d` from point Q `:.` Magnetic field due to this coil is given as:
`B_(2)=(mu_(0)NIR^(2))/(2[(R/2-d)^(2)+R^(2)]^(3//2)), `Total magnetic field ,` B=B_(1)+B_(2)`
`=(mu_(0)IR^(2))/2[{(R/2-d)^(2)+R^(2)}^(-3//2)]+{(R/2+d)^(2)+R^(2)}^(-3//2)xxN`
`=(mu_(0)IR^(2))/2[((5R^(2))/4+d^(2)-Rd)^(-3//2)+((5R^(2))/4+d^(2)+Rd)^(-3//2)]xxN`
`=(mu_(0)IR^(2))/2xx((5R^(2))/4)^(-3//2)[(1+4/5 (d^(2))/(R^(2))- 4/5 d/R)^(-3//2)+(1+4/5(d^(2))/(R^(2))+4/5 d/R)^(-3//2)]xxN`
For `d lt lt R` neglecting the factor `(d^(2))/(R^(2))` we get
`=(mu_(0)IR^(2))/2xx((5R^(2))/4)^(-3//2)xx[(1-(4d)/(5R))^(-3/2)+(1+(4d)/(5R)^(-3//2))]xxN=(mu_(0)IR^(2)N)/(2R^(3))xx(4/5)^(3/2)[1-(6d)/(5R)+1+(6d)/(5R)]`
`B=(4/5)^(3//2)(mu_(0)IN)/R=0.72((mu_(0)IN)/R)`
Hence, it is proved that the field on the axis around the mid-point between the coils is uniform.