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,
$B=\frac{{\mu }_{0}I{R}^{2}N}{2(x^2+R^2{)}^{3/2}}$
(a) Show that this reduces to the familiar result for field at the centre of the coil.

(b) Consider two parallel 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.
$B=0.72\frac{{\mu }_{0}NI}{R}$
[Such an arrangement to produce a nearly uniform magnetic field over a small region is known as Helmholtz coils.]

Answer 1

(a)

At the centre of the coil, $x=0$.

Hence, magnetic field at centre is $B_c=\frac{{\mu }_{o}I{R}^{2}N}{2R^3}$

$B=\frac{{\mu }_{o}NI}{2R}$

(b)

In a small region of length 2d about the mid-point between the coils,
$B=\frac{{\mu }_{0}I{R}^{2}N}{2}\times [{\{{\left(\frac{R}{2}+d\right)}^2+R^2\}}^{-3/2}+{\{{\left(\frac{R}{2}-d\right)}^2+R^2\}}^{-3/2}]$
$=\frac{{\mu }_{0}I{R}^{2}N}{2}\times {\left(\frac{5R^2}{4}\right)}^{-3/2}\times [(1+\frac{4d}{5R})+{(1-\frac{4d}{5R})}^{-3/2}]$
$=\frac{{\mu }_{0}I{R}^{2}N}{2}\times {\left(\frac{4}{5}\right)}^{-3/2}\times [1-\frac{6d}{5R}+1+\frac{6d}{5R}]$
where in the second and third steps above, terms containing $d^2/R^2$ and higher powers of d/R are neglected since $\frac{d}{R}<<1$ The terms linear in d/R cancel giving a uniform field B in a small region:
$B=\frac{{\mu }_{0}I{R}^{2}N}{2}\times {\left(\frac{4}{5}\right)}^{-3/2}\frac{{\mu }_{0}IN}{R}=0.72\frac{{\mu }_{0}IN}{R}$