91影视

Consider the figure for the field as:

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

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

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

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

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

$B=B_1+B_2$

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

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

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

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

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

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

Substitute $z-0$in the above expression.

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

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

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

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

Differentiate the equation (2) with respect to z.

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

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

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

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

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

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

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

$d=R$

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.

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

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