Question

Derive the expression for the magnetic field due to a current carrying coil of radius r at a distance x from the centre along the X-axis. 

Answer 1

Figure depicts a circular loop carrying a steadycurrent I. The loop is placed in the y-z plane with its centre at the origin O and has a radius R. The x-axis is the axis of the loop. We wish to calculate the magnetic field at the point P on this axis. Let x be the distance of P from the centre O of the loop.
Consider a conducting element dl of the loop. This is shown in the  igure. The magnitude dB of the magnetic field due to dl is given by the Biot-Savart law,

`dB = (mu_0)/(4pi) (I|vec(dl) xx vec(z)|)/(z^3) `  ——— (i)

Now z² = x² + r² .

Further, any element of the loop will be perpendicular to the displacement vector from the element to the  axial point. For example, the element dl in the figure is in the y-z plane whereas the displacement vector r from dl to the axial point P is in the x-y plane.

Hence   `|dvec(l) xx vec(z) |` = z d1 Thus,

`dB = (mu_0)/(4 pi) (Idl)/((x^2 + r^2))`——— (ii)

The direction of dB is shown in the figure. It is perpendicular to the plane formed by`vec(dl) ` and `vec(r)`.has an x - component d`vec(B)_x` and a component perpendicular to x-axis, `vec(dB)_⊥` . When the components perpendicular to the x-axis are summed over, they cancel out and we obtain a null result.

For example, the dB⊥ .component due to dl is cancelled by thecontribution due to the diametrically opposite dl element, shown in figure. Thus, only the x-component survives. The net contributionalong xdirection can be obtained by integrating dB_x = dB cos θ over theloop. For the figure, 

`cos theta = r/((x^2 + r^2 )^(1/2)`——— (iii)

From Eqs. (ii) and (iii),

`dB_x = (mu_0)/(4pi) I.dl r /((x^2 + r^2)3/2`

The summation of elements dl over the loop yields 2πr, the circumference of the loop. Thus, the magnetic field at P due to entire circular loop is  

`vec(B) = B_xhat(i) = (mu_0 Ir^2)/(2(x^2 +r^2)3/2) hat(i)`