Chapter 27, Problem 23P

To determine

The proof that $\frac{d{B}_z}{dz}=0,\frac{d^2{B}_z}{d{z}^2}=0\text{and}\frac{d^3{B}_z}{d{z}^3}=0$ .

Answer to Problem 23P

It is proved that $\frac{d{B}_z}{dz}=0,\frac{d^2{B}_z}{d{z}^2}=0\text{and}\frac{d^3{B}_z}{d{z}^3}=0$ .

Explanation of Solution

Formula used:

The expression for total magnetic fieldis given by,

  $B_z=\frac{{\mu }_{0}n{R}^{2}I}{2}\left(z_1^{-1}+z_2^{-1}\right)$

Calculation:

The total magnetic fieldis calculated as,

  $\begin{array}{c}\frac{d{B}_z}{dz}=\frac{d}{dz}\left(\frac{{\mu }_{0}n{R}^{2}I}{2}\left(z_1^{-1}+z_2^{-1}\right)\right)\\ =\left(\frac{{\mu }_{0}n{R}^{2}I}{2}\right)\left(-3z_1^{-4}\frac{d{z}_1}{dz}-3z_2^{-4}\frac{d{z}_2}{dz}\right)\end{array}$ ....... (1)

Differentiate $z_1=\sqrt{{\left(z+\frac{1}{2}R\right)}^2+R^2}$ with respect to $z$ .

  $\begin{array}{c}\frac{d{z}_1}{dz}=\frac{d}{dz}\left(\sqrt{{\left(z+\frac{1}{2}R\right)}^2+R^2}\right)\\ =\frac{1}{2}\left({\left({\left(z+\frac{1}{2}R\right)}^2+R^2\right)}^{-\frac{1}{2}}\right)\left(2\left(z+\frac{1}{2}R\right)\right)\\ =\frac{\left(z+\frac{1}{2}R\right)}{z_1}\end{array}$

Differentiate $z_2=\sqrt{{\left(z-\frac{1}{2}R\right)}^2+R^2}$ with respect to $z$ .

  $\begin{array}{c}\frac{d{z}_2}{dz}=\frac{d}{dz}\left(\sqrt{{\left(z-\frac{1}{2}R\right)}^2+R^2}\right)\\ =\frac{1}{2}\left({\left({\left(z-\frac{1}{2}R\right)}^2+R^2\right)}^{-\frac{1}{2}}\right)\left(2\left(z-\frac{1}{2}R\right)\right)\\ =\frac{\left(z-\frac{1}{2}R\right)}{z_2}\end{array}$

Substitute $\frac{\left(z+\frac{1}{2}R\right)}{z_1}$ for $\frac{d{z}_1}{dz}$ and $\frac{\left(z-\frac{1}{2}R\right)}{z_2}$ for $\frac{d{z}_2}{dz}$ in equation (1).

  $\begin{array}{r}\frac{d{B}_z}{dz}=\left(\frac{{\mu }_{0}n{R}^{2}I}{2}\right)\left(-3z_1^{-4}\left(\frac{\left(z+\frac{1}{2}R\right)}{z_1}\right)-3z_2^{-4}\left(\frac{\left(z-\frac{1}{2}R\right)}{z_2}\right)\right)\\ =\left(\frac{{\mu }_{0}n{R}^{2}I}{2}\right)\left(\frac{-3\left(z+\frac{1}{2}R\right)}{z_1^5}+\frac{-3\left(z-\frac{1}{2}R\right)}{z_2^5}\right)\end{array}$ ....... (2)

Substitute $0$ for $z$ $\sqrt{\frac{5}{4}{R}^2}$ for $z_1$ and $\sqrt{\frac{5}{4}{R}^2}$ for $z_2$ in equation (2).

  $\begin{array}{c}\frac{d{B}_z}{dz}=\left(\frac{{\mu }_{0}n{R}^{2}I}{2}\right)\left(\frac{-3\left(0+\frac{1}{2}R\right)}{\sqrt{\frac{5}{4}{R}^2}}+\frac{-3\left(0-\frac{1}{2}R\right)}{\sqrt{\frac{5}{4}{R}^2}}\right)\\ =0\end{array}$

Differentiate equation (2) with respect to $z$ .

  $\begin{array}{c}\frac{d^2{B}_z}{d{z}^2}=\frac{d}{dz}\left[\left(\frac{{\mu }_{0}n{R}^{2}I}{2}\right)\left(\frac{-3\left(z+\frac{1}{2}R\right)}{z_1^5}+\frac{-3\left(z-\frac{1}{2}R\right)}{z_2^5}\right)\right]\\ =\left(\frac{{\mu }_{0}n{R}^{2}I}{2}\right)\left(-3\right)\left(\begin{array}{l}\left(\frac{1}{z_1^5}-\frac{5\left(z+\frac{1}{2}R\right)}{z_1^6}\left(\frac{d{z}_1}{dz}\right)\right)\\ +\left(\frac{1}{z_2^5}-\frac{5\left(z+\frac{1}{2}R\right)}{z_2^6}\left(\frac{d{z}_2}{dz}\right)\right)\end{array}\right)\end{array}$ ....... (3)

Substitute $\frac{\left(z+\frac{1}{2}R\right)}{z_1}$ for $\frac{d{z}_1}{dz}$ and $\frac{\left(z-\frac{1}{2}R\right)}{z_2}$ for $\frac{d{z}_2}{dz}$ in equation (3).

  $\begin{array}{c}\frac{d^2{B}_z}{d{z}^2}=\left(\frac{{\mu }_{0}n{R}^{2}I}{2}\right)\left(-3\right)\left(\begin{array}{l}\left(\frac{1}{z_1^5}-\frac{5\left(z+\frac{1}{2}R\right)}{z_1^6}\left(\frac{\left(z+\frac{1}{2}R\right)}{z_1}\right)\right)\\ +\left(\frac{1}{z_2^5}-\frac{5\left(z+\frac{1}{2}R\right)}{z_2^6}\left(\frac{\left(z-\frac{1}{2}R\right)}{z_2}\right)\right)\end{array}\right)\\ =\left(\frac{{\mu }_{0}n{R}^{2}I}{2}\right)\left(-3\right)\left(\left(\frac{1}{z_1^5}-\frac{5{\left(z+\frac{1}{2}R\right)}^2}{z_1^7}\right)+\left(\frac{1}{z_2^5}-\frac{5{\left(z-\frac{1}{2}R\right)}^2}{z_2^7}\right)\right)\end{array}$ ....... (4)

Substitute $0$ for $z$ $\sqrt{\frac{5}{4}{R}^2}$ for $z_1$ and $\sqrt{\frac{5}{4}{R}^2}$ for $z_2$ in equation (4).

  $\begin{array}{c}\frac{d^2{B}_z}{d{z}^2}=\left(\frac{{\mu }_{0}n{R}^{2}I}{2}\right)\left(-3\right)\left(\left(\frac{1}{{\left(\sqrt{\frac{5}{4}{R}^2}\right)}^{\frac{5}{2}}}-\frac{5{\left(0+\frac{1}{2}R\right)}^2}{{\left(\sqrt{\frac{5}{4}{R}^2}\right)}^{\frac{7}{2}}}\right)+\left(\frac{1}{{\left(\sqrt{\frac{5}{4}{R}^2}\right)}^{\frac{5}{2}}}-\frac{5{\left(0-\frac{1}{2}R\right)}^2}{{\left(\sqrt{\frac{5}{4}{R}^2}\right)}^{\frac{7}{2}}}\right)\right)\\ =\left(\frac{{\mu }_{0}n{R}^{2}I}{2}\right)\left(-3\right)\left(\frac{1}{{\left(\frac{5}{4}{R}^2\right)}^{\frac{5}{2}}}-\frac{1}{{\left(\frac{5}{4}{R}^2\right)}^{\frac{5}{2}}}+\frac{1}{{\left(\frac{5}{4}{R}^2\right)}^{\frac{5}{2}}}+\frac{1}{{\left(\frac{5}{4}{R}^2\right)}^{\frac{5}{2}}}\right)\\ =0\end{array}$

Differentiate equation (4) with respect to $z$ .

  $\begin{array}{c}\frac{d^3{B}_z}{d{z}^3}=\frac{d}{dz}\left[\left(\frac{{\mu }_{0}n{R}^{2}I}{2}\right)\left(-3\right)\left(\left(\frac{1}{z_1^5}-\frac{5{\left(z+\frac{1}{2}R\right)}^2}{z_1^7}\right)+\left(\frac{1}{z_2^5}-\frac{5{\left(z-\frac{1}{2}R\right)}^2}{z_2^7}\right)\right)\right]\\ =\left(\frac{{\mu }_{0}n{R}^{2}I}{2}\right)\left(-3\right)\left[\begin{array}{c}\left(\begin{array}{l}\frac{-5}{z_1^6}\left(\frac{d{z}_1}{dz}\right)-\frac{10\left(z+\frac{1}{2}R\right)}{z_1^7}-\\ \frac{-35{\left(z+\frac{1}{2}R\right)}^2}{z_1^8}\left(\frac{d{z}_1}{dz}\right)\end{array}\right)+\\ \left(\begin{array}{l}\frac{-5}{z_1^6}\left(\frac{d{z}_2}{dz}\right)-\frac{10\left(z-\frac{1}{2}R\right)}{z_1^7}\\ -\frac{-35{\left(z-\frac{1}{2}R\right)}^2}{z_1^8}\left(\frac{d{z}_2}{dz}\right)\end{array}\right)\end{array}\right]\end{array}$ ....... (5)

Substitute $\frac{\left(z+\frac{1}{2}R\right)}{z_1}$ for $\frac{d{z}_1}{dz}$ and $\frac{\left(z-\frac{1}{2}R\right)}{z_2}$ for $\frac{d{z}_2}{dz}$ in equation (5)

  $\frac{d^3{B}_z}{d{z}^3}=\left(\frac{{\mu }_{0}n{R}^{2}I}{2}\right)\left(-3\right)\left[\begin{array}{c}\left(\begin{array}{l}\frac{-5}{z_1^6}\left(\frac{\left(z+\frac{1}{2}R\right)}{z_1}\right)-\frac{10\left(z+\frac{1}{2}R\right)}{z_1^7}-\\ \frac{-35{\left(z+\frac{1}{2}R\right)}^2}{z_1^8}\left(\frac{\left(z+\frac{1}{2}R\right)}{z_1}\right)\end{array}\right)+\\ \left(\begin{array}{l}\frac{-5}{z_1^6}\left(\frac{\left(z-\frac{1}{2}R\right)}{z_1}\right)-\frac{10\left(z-\frac{1}{2}R\right)}{z_1^7}\\ -\frac{-35{\left(z-\frac{1}{2}R\right)}^2}{z_1^8}\left(\frac{\left(z-\frac{1}{2}R\right)}{z_1}\right)\end{array}\right)\end{array}\right]$

  $=\left(\frac{{\mu }_{0}n{R}^{2}I}{2}\right)\left(-3\right)\left(\begin{array}{l}-\frac{15\left(z+\frac{1}{2}R\right)}{z_1^7}+\frac{35{\left(z+\frac{1}{2}R\right)}^3}{z_1^9}\\ -\frac{15\left(z-\frac{1}{2}R\right)}{z_2^7}+\frac{35{\left(z-\frac{1}{2}R\right)}^2}{z_2^9}\end{array}\right)$........ (6)

Substitute $0$ for $z$ $\sqrt{\frac{5}{4}{R}^2}$ for $z_1$ and $\sqrt{\frac{5}{4}{R}^2}$ for $z_2$ in equation (6).

  $\begin{array}{c}\frac{d^3{B}_z}{d{z}^3}=\left(\frac{{\mu }_{0}n{R}^{2}I}{2}\right)\left(-3\right)\left(\begin{array}{l}-\frac{15\left(0+\frac{1}{2}R\right)}{{\left(\frac{5}{4}{R}^2\right)}^{\frac{7}{2}}}+\frac{35{\left(0+\frac{1}{2}R\right)}^3}{{\left(\frac{5}{4}{R}^2\right)}^{\frac{9}{2}}}\\ -\frac{15\left(z-\frac{1}{2}R\right)}{{\left(\frac{5}{4}{R}^2\right)}^{\frac{7}{2}}}+\frac{35{\left(z-\frac{1}{2}R\right)}^2}{{\left(\frac{5}{4}{R}^2\right)}^{\frac{9}{2}}}\end{array}\right)\\ =0\end{array}$

Conclusion:

Therefore, it is proved that $\frac{d{B}_z}{dz}=0,\frac{d^2{B}_z}{d{z}^2}=0\text{and}\frac{d^3{B}_z}{d{z}^3}=0$ .