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 $\frac{\mathbf{\partial }\mathbf{B}}{\mathbf{\partial }\mathbf{z}}$ 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 ${\mathbf{\partial }}^{\mathbf{2}}\mathit{B}\mathbf{/}\mathbf{\partial }{\mathit{z}}^{\mathbf{2}}\mathbf{=}\mathbf{0}$ at the midpoint, and find the resulting magnetic field at the center. [Answer:$\frac{\mathbf{8}{\mathbf{\mu }}_0\mathbf{I}}{\mathbf{5}\sqrt{5}\mathbf{R}}$ ]

Consider the figure for the field as:

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

${\mathbf{B}}_1\mathbf{=}\frac{{\mathbf{\mu }}_0{\mathbf{IR}}^2}{2}\mathbf{\left(}\frac{1}{{\mathbf{[}{\mathbf{R}}^2\mathbf{+}{\mathbf{(}\frac{d}{2}\mathbf{+}\mathbf{z}\mathbf{)}}^2\mathbf{]}}^{3/2}}\mathbf{\right)}$

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

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

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

$\begin{array}{c}B=B_1+B_2\\ B=\frac{{\mu }_{0}I{R}^2}{2}\left(\frac{1}{{[R^2+{(\frac{d}{2}+z)}^2]}^{3/2}}\right)+\frac{{\mu }_{0}I{R}^2}{2}\left(\frac{1}{{[R^2+{(\frac{d}{2}-z)}^2]}^{3/2}}\right)\\ B=\frac{{\mu }_{0}I{R}^2}{2}(\frac{1}{{[R^2+{(\frac{d}{2}+z)}^2]}^{3/2}}+\frac{1}{{[R^2+{(\frac{d}{2}-z)}^2]}^{3/2}})\end{array}$

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

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

Substitute$z=0$ in the above expression.

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

Therefore, the magnetic field as a function of z is

$\frac{{\mu }_{0}I{R}^2}{2}(\frac{1}{{[R^2+{(\frac{d}{2}+z)}^2]}^{3/2}}+\frac{1}{{[R^2+{(\frac{d}{2}-z)}^2]}^{3/2}})$and first derivative of this magnetic field is zero on the z axis.

Differentiate the equation (2) with respect to z.

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

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

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

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.

$\begin{array}{l}B=\frac{{\mu }_{0}I{R}^2}{2}(\frac{1}{{[R^2+{(\frac{R}{2}+0)}^2]}^{3/2}}+\frac{1}{{[R^2+{(\frac{R}{2}-0)}^2]}^{3/2}})\\ B=\frac{8{\mu }_{0}I}{5\sqrt{5}R}\end{array}$

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