91Ó°ÊÓ

Consider the figure for the field as:

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

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

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

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

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

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

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

$\begin{array}{l}\frac{∂B}{∂z}=\frac{∂}{∂z}\left[\frac{{\mathrm{μ}}_{0}I{R}^2}{2}(\frac{1}{{[R^2+{(\frac{d}{2}+z)}^2]}^{3/2}}+\frac{1}{{[R^2+{(\frac{d}{2}−z)}^2]}^{3/2}})\right]\\ \frac{∂B}{∂z}=\frac{{\mathrm{μ}}_{0}I{R}^2}{2}[\frac{(−3/2)\left(2\right)(\frac{d}{2}+z)}{{[R^2+{(\frac{d}{2}+z)}^2]}^{5/2}}+\frac{(−3/2)\left(2\right)(\frac{d}{2}−z)(−1)}{{[R^2+{(\frac{d}{2}−z)}^2]}^{5/2}}]\\ \frac{∂B}{∂z}=\frac{3{\mathrm{μ}}_{0}I{R}^2}{2}[\frac{−(\frac{d}{2}+z)}{{[R^2+{(\frac{d}{2}+z)}^2]}^{5/2}}+\frac{(\frac{d}{2}−z)}{{[R^2+{(\frac{d}{2}−z)}^2]}^{5/2}}]\end{array}$…â¶Ä¦(2)

Substitute$z=0$ in the above expression.

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

Therefore, the magnetic field as a function of z is

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

Differentiate the equation (2) with respect to z.

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

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

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

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.

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

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