Question

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, ( ) 2 0 2 2 3/2 2 IR N B x R μ = + (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, 0 0.72 NI B R μ = , approximately. [Such an arrangement to produce a nearly uniform magnetic field over a small region is known as Helmholtz coils.]

Answer 1

On the axis of the circular coil of radius R and number of turns N carrying a current I at a distance x from centre the magnetic field is $B=\frac{\mu_0IR^2N}{2(x^2+R^2)^{3/2}}$

where x is the seperation between centre of coil to observation point.

(a) at the centre , x = 0

so, magnetic field , B = $\frac{\mu_0IR^2N}{2R^3}=\frac{\mu_0NI}{2R}$

which is the same result as is obtained by integration at centre of the coil.

(b) The side of the loop where current flows clockwise becomes south pole and where the current appear anticlockwise becomes north pole.

so, direction of magnetic field is from south pole to north pole.

Now, magnetic field of two parallel coils each of radius R and N turns, B = $B_1+B_2$ [ see figure ]

= $\frac{\mu_0IR^2N}{2}\left[\frac{2}{2(x^2+R^2)^{3/2}}\right]$

putting , x = R/2

B = $\frac{\mu_0IR^2N}{2}\left[\frac{2}{2((R/2)^2+R^2)^{3/2}}\right]$

= $\frac{\mu_0IR^2N}{2}\left[\frac{8\times2}{5\sqrt{5}R^3}\right]$

= $\frac{8\mu_0NI}{5\sqrt{5}R}$

= $\bf{0.72\frac{\mu_0NI}{R}}$