The magnetic field on the axis of a circular current loop (Eq. 5.41) is far from uniform (it falls off sharply with increasing z). You can produce a more nearly uniform field by using two such loops a distanced apart (Fig. 5.59).

(a) Find the field (B) as a function of $z$, and show that $\frac{\partial \mathbf{B}}{\partial \mathbf{z}}$is zero at the point midway between them $(z=0)$

(b) If you pick d just right, the second derivative of$B$will also vanish at the midpoint. This arrangement is known as a Helmholtz coil; it's a convenient way of producing relatively uniform fields in the laboratory. Determine $d$such that

${\partial }^{2}B/\partial z^2=0$at the midpoint, and find the resulting magnetic field at the center.

Consider the figure for the field as:

The magnetic field due to the upper loop by using equation $5.41$is given as:

$B_1=\frac{{\mu }_0{\mathrm{ℝ}}^2}{2}\left(\frac{1}{{\left[R^2+{\left(\frac{d}{2}+z\right)}^2\right]}^{3/2}}\right)$

The magnetic field due to the lower loop by using equation $5.41$is given as:

$B_2=\frac{{\mu }_{0}I{R}^2}{2}\left(\frac{1}{{\left[R^2+{\left(\frac{d}{2}-z\right)}^2\right]}^{3/2}}\right)$

The net magnetic field due to both loops is given as:

$B=B_1+B_2$

$B=\frac{{\mu }_{0}I{R}^2}{2}\left(\frac{1}{{\left[R^2+{\left(\frac{d}{2}+z\right)}^2\right]}^{3/2}}\right)+\frac{{\mu }_{0}I{R}^2}{2}\left(\frac{1}{{\left[R^2+{\left(\frac{d}{2}-z\right)}^2\right]}^{3/2}}\right)$

$B=\frac{{\mu }_{0}I{R}^2}{2}\left(\frac{1}{{\left[R^2+{\left(\frac{d}{2}+z\right)}^2\right]}^{3/2}}+\frac{1}{{\left[R^2+{\left(\frac{d}{2}-z\right)}^2\right]}^{3/2}}\right)$

Differentiate the above expression of magnetic field to find the location for zero magnetic field on $z$axis.

$\frac{\partial B}{\partial z}=\frac{\partial }{\partial z}\left[\frac{{\mu }_{0}I{R}^2}{2}\left(\frac{1}{{\left[R^2+{\left(\frac{d}{2}+z\right)}^2\right]}^{3/2}}+\frac{1}{{\left[R^2+{\left(\frac{d}{2}-z\right)}^2\right]}^{3/2}}\right]\right]$

$\frac{\partial B}{\partial z}=\frac{{\mu }_{0}I{R}^2}{2}\left[\frac{(-3/2)\left(2\right)\left(\frac{d}{2}+z\right)}{{\left[R^2+{\left(\frac{d}{2}+z\right)}^2\right]}^{5/2}}+\frac{(-3/2)\left(2\right)\left(\frac{d}{2}-z\right)(-1)}{{\left[R^2+{\left(\frac{d}{2}-z\right)}^2\right]}^{5/2}}\right]$

$\frac{\partial B}{\partial z}=\frac{3{\mu }_{0}I{R}^2}{2}\left[\frac{-\left(\frac{d}{2}+z\right)}{{\left[R^2+{\left(\frac{d}{2}+z\right)}^2\right]}^{5/2}}+\frac{\left(\frac{d}{2}-z\right)}{{\left[R^2+{\left(\frac{d}{2}-z\right)}^2\right]}^{5/2}}\right]$

Substitute $z-0$in the above expression.

$$\begin{gathered} \frac{\partial B}{\partial z}=\frac{3{\mu }_{0}I{R}^2}{2}\left[\frac{-\left(\frac{d}{2}+0\right)}{{\left[R^2+{\left(\frac{d}{2}+0\right)}^2\right]}^{5/2}}+\frac{\left(\frac{d}{2}-0\right)}{{\left[R^2+{\left(\frac{d}{2}-0\right)}^2\right]}^{5/2}}\right] \\ \frac{\partial B}{\partial z}=\frac{3{\mu }_{0}I{R}^2}{2}\left[\frac{-\frac{d}{2}}{{\left[R^2+{\frac{d}{4}}^2\right]}^{5/2}}+\frac{\frac{d}{2}}{{\left[R^2+{\frac{d}{4}}^2\right]}^{5/2}}\right] \\ \frac{\partial B}{\partial z}=0 \end{gathered}$$

Therefore, the magnetic field as a function of $z$is

$\frac{{\mu }_{0}I{R}^2}{2}\left(\frac{1}{{\left[R^2+{\left(\frac{d}{2}+z\right)}^2\right]}^{3/2}}+\frac{1}{{\left[R^2+{\left(\frac{d}{2}-z\right)}^2\right]}^{3/2}}\right)$

and first derivative of this magnetic field is zero on the $\mathrm{z}$axis.

Differentiate the equation (2) with respect to z.

$\frac{{\partial }^{2}B}{\partial z^2}=\frac{\partial }{\partial z}\left(\frac{\partial B}{\partial z}\right)$

$\frac{{\partial }^{2}B}{\partial z^2}=\frac{\partial }{\partial z}\left[\frac{3{\mu }_{0}I{R}^2}{2}\left[\frac{-\left(\frac{d}{2}+z\right)}{{\left[R^2+{\left(\frac{d}{2}+z\right)}^2\right]}^{5/2}}+\frac{\left(\frac{d}{2}-z\right)}{{\left[R^2+{\left(\frac{d}{2}-z\right)}^2\right]}^{5/2}}\right]\right.$

$\left.\frac{{\partial }^{2}B}{\partial z^2}=\frac{3{\mu }_0/R^2}{2}\left[\frac{-1}{{\left[R^2+{\left(\frac{d}{2}+z\right)}^2\right]}^{5/2}}+\frac{-\left(\frac{d}{2}+z\right)(-5/2)\left(2\right)\left(\frac{d}{2}+z\right)}{{\left[R^2+{\left(\frac{d}{2}+z\right)}^2\right]}^{7/2}}\right]+\frac{-1}{{\left[R^2+{\left(\frac{d}{2}-z\right)}^2\right]}^{5/2}}+\frac{\left(\frac{d}{2}-z\right)(-5/2)\left(2\right)\left(\frac{d}{2}-z\right)(-1)}{{\left[R^2+{\left(\frac{d}{2}-z\right)}^2\right]}^{7/2}}\right]$

Substitute $z=0$and $\frac{{\partial }^{2}B}{\partial z^2}=0$in the above expression.

$\left.0=\frac{3{\mu }_0/R^2}{2}\left[\frac{-1}{{\left[R^2+{\left(\frac{d}{2}+0\right)}^2\right]}^{5/2}}+\frac{-\left(\frac{d}{2}+0\right)(-5/2)\left(2\right)\left(\frac{d}{2}+0\right)}{{\left[R^2+{\left(\frac{d}{2}+0\right)}^2\right]}^{7/2}}\right]+\frac{-1}{{\left[R^2+{\left(\frac{d}{2}-0\right)}^2\right]}^{5/2}}+\frac{\left(\frac{d}{2}-0\right)(-5/2)\left(2\right)\left(\frac{d}{2}-0\right)(-1)}{{\left[R^2+{\left(\frac{d}{2}-0\right)}^2\right]}^{7/2}}\right]$

$0=\frac{3{\mu }_0/R^2}{{\left[R^2+{\left(\frac{d}{2}\right)}^2\right]}^{7/2}}\left(d^2-R^2\right)$

$0=\frac{3{\mu }_0/R^2}{{\left[R^2+{\left(\frac{d}{2}\right)}^2\right]}^{7/2}}\left(d^2-R^2\right)$

$d=R$

Therefore, the second derivative is zero at midpoint if both loops are placed at distance equal to the radius of loop.

Substitute $d=R$and $z=0$in equation (1) to find resulting magnetic field.

$$\begin{gathered} 0=\frac{3{\mu }_{0}I{R}^2}{2}\left[\frac{-1}{{\left[R^2+{\left(\frac{d}{2}+0\right)}^2\right]}^{5/2}}+\frac{-\left(\frac{d}{2}+0\right)\left(-5/2\right)\left(2\right)\left(\frac{d}{2}+0\right)}{{\left[R^2+{\left(\frac{d}{2}+0\right)}^2\right]}^{7/2}} \\ +\frac{-1}{{\left[R^2+{\left(\frac{d}{2}-0\right)}^2\right]}^{5/2}}+\frac{\left(\frac{d}{2}-0\right)\left(-5/2\right)\left(2\right)\left(\frac{d}{2}-0\right)\left(-1\right)}{{\left[R^2+{\left(\frac{d}{2}-0\right)}^2\right]}^{7/2}}\right] \\ 0=\frac{3{\mu }_{0}I{R}^2}{{\left[R^2+{\left(\frac{d}{2}\right)}^2\right]}^{7/2}}\left(d^2-R^2\right) \\ 0=\frac{3{\mu }_{0}I{R}^2}{{\left[R^2+{\left(\frac{d}{2}\right)}^2\right]}^{7/2}}\left(d^2-R^2\right) \\ d=R \end{gathered}$$

Therefore, the resulting magnetic field at the midpoint is$\frac{8{\mu }_{0}l}{5\sqrt{5}R}$.