4 Engineering Example 2

4.1 Divergence of a magnetic field

Introduction

A magnetic field $\underset{̲}{B}$ must satisfy $\underset{̲}{\nabla }\cdot \underset{̲}{B}=0$ . An associated current is given by:

$\underset{̲}{I}=\frac{1}{{\mu }_0}\left(\underset{̲}{\nabla }\times \underset{̲}{B}\right)$

Problem in words

For the magnetic field (in cylindrical polar coordinates $\rho ,\varphi ,z$ )

$\underset{̲}{B}=B_0\frac{\rho }{1+{\rho }^2}\underset{̲}{\stackrel{̂}{\varphi }}+\alpha \underset{̲}{ẑ}$

show that the divergence of $\underset{̲}{B}$ is zero and find the associated current.

Mathematical statement of problem

We must

1. show that $\underset{̲}{\nabla }\cdot \underset{̲}{B}=0$
2. find the current $\underset{̲}{I}=\frac{1}{{\mu }_0}\left(\underset{̲}{\nabla }\times \underset{̲}{B}\right)$

Mathematical analysis

1. Express $\underset{̲}{B}$ as $\left(B_{\rho },B_{\varphi },B_z\right)$ ; then $$\begin{array}{rcll}\underset{̲}{\nabla }\cdot \underset{̲}{B}& =& \frac{1}{\rho }\left[\frac{\partial }{\partial \rho }\left(\rho B_{\rho }\right)+\frac{\partial B_{\varphi }}{\partial \varphi }+\rho \frac{\partial B_z}{\partial z}\right]& \\ & & & \\ & =& \frac{1}{\rho }\left[\frac{\partial }{\partial \rho }\left(0\right)+\frac{\partial }{\partial \varphi }\left(B_0\frac{\rho }{1+{\rho }^2}\right)+\rho \frac{\partial }{\partial z}\left(\alpha \right)\right]& \\ & & & \\ & =& \frac{1}{\rho }\left[0+0+0\right]=0\text{ as required.}& \end{array}$$
2. To find the current evaluate
   $$\begin{array}{rcll}\underset{̲}{I}=\frac{1}{{\mu }_0}\left(\underset{̲}{\nabla }\times \underset{̲}{B}\right)& =& \frac{1}{{\mu }_0}\frac{1}{\rho }\left|\begin{array}{ccc}\hfill \underset{̲}{\stackrel{̂}{\rho }}\hfill & \hfill \rho \underset{̲}{\stackrel{̂}{\varphi }}\hfill & \hfill \underset{̲}{ẑ}\hfill \\ \hfill \hfill \\ \hfill \frac{\partial }{\partial \rho }\hfill & \hfill \frac{\partial }{\partial \varphi }\hfill & \hfill \frac{\partial }{\partial z}\hfill \\ \hfill \hfill \\ \hfill B_{\rho }\hfill & \hfill \rho B_{\varphi }\hfill & \hfill B_z\hfill \end{array}\right|=\left|\begin{array}{ccc}\hfill \underset{̲}{\stackrel{̂}{\rho }}\hfill & \hfill \rho \underset{̲}{\stackrel{̂}{\varphi }}\hfill & \hfill \underset{̲}{ẑ}\hfill \\ \hfill \hfill \\ \hfill \frac{\partial }{\partial \rho }\hfill & \hfill \frac{\partial }{\partial \varphi }\hfill & \hfill \frac{\partial }{\partial z}\hfill \\ \hfill \hfill \\ \hfill 0\hfill & \hfill B_0\frac{{\rho }^2}{1+{\rho }^2}\hfill & \hfill \alpha \hfill \end{array}\right|& \\ & =& \frac{1}{{\mu }_0\rho }\left[0\underset{̲}{\stackrel{̂}{\rho }}+0\rho \underset{̲}{\stackrel{̂}{\varphi }}+B_0\frac{\partial }{\partial \rho }\left(\frac{{\rho }^2}{1+{\rho }^2}\right)\underset{̲}{ẑ}\right]& \\ & & & \\ & =& \frac{1}{{\mu }_0\rho }{B}_0\frac{1+4{\rho }^3}{{\left(1+{\rho }^2\right)}^2}\underset{̲}{ẑ}& \end{array}$$

Interpretation

The magnetic field is in the form of a helix with the current pointing along its axis (Fig 22). Such an arrangement is often used for the magnetic containment of charged particles in a fusion reactor.

Figure 22:

Example 24

A magnetic field $\underset{̲}{B}$ is given by $\underset{̲}{B}={\rho }^{-2}\underset{̲}{\stackrel{̂}{\varphi }}+k\underset{̲}{ẑ}$ . Find $\underset{̲}{\nabla }\cdot \underset{̲}{B}$ and $\underset{̲}{\nabla }\times \underset{̲}{B}$ .

Solution

$\underset{̲}{\nabla }\cdot \underset{̲}{B}=\frac{1}{\rho }\left[\frac{\partial }{\partial \rho }0+\frac{\partial }{\partial \varphi }{\rho }^{-2}+\frac{\partial }{\partial z}k\rho \right]=\frac{1}{\rho }\left[0+0+0\right]=0$

$\underset{̲}{\nabla }\times \underset{̲}{B}=\frac{1}{\rho }\left|\begin{array}{ccc}\hfill \underset{̲}{\stackrel{̂}{\rho }}\hfill & \hfill \rho \underset{̲}{\stackrel{̂}{\varphi }}\hfill & \hfill \underset{̲}{ẑ}\hfill \\ \hfill \hfill \\ \hfill \frac{\partial }{\partial \rho }\hfill & \hfill \frac{\partial }{\partial \varphi }\hfill & \hfill \frac{\partial }{\partial z}\hfill \\ \hfill \hfill \\ \hfill B_{\rho }\hfill & \hfill B_{\varphi }\hfill & \hfill B_z\hfill \end{array}\right|=\frac{1}{\rho }\left|\begin{array}{ccc}\hfill \underset{̲}{\stackrel{̂}{\rho }}\hfill & \hfill \rho \underset{̲}{\stackrel{̂}{\varphi }}\hfill & \hfill \underset{̲}{ẑ}\hfill \\ \hfill \hfill \\ \hfill \frac{\partial }{\partial \rho }\hfill & \hfill \frac{\partial }{\partial \varphi }\hfill & \hfill \frac{\partial }{\partial z}\hfill \\ \hfill \hfill \\ \hfill {\rho }^{-2}\hfill & \hfill 0\hfill & \hfill k\hfill \end{array}\right|$

$=\underset{̲}{0}$

All magnetic fields satisfy $\underset{̲}{\nabla }\cdot \underset{̲}{B}=0$ i.e. an absence of magnetic monopoles. There is a class of magnetic fields known as potential fields that satisfy $\underset{̲}{\nabla }\times \underset{̲}{B}=\underset{̲}{0}$

Task!

Using cylindrical polar coordinates, find $\underset{̲}{\nabla }f$ for $f={\rho }^{2}zsin\varphi$

Answer

Task!

1. Using cylindrical polar coordinates, find $\underset{̲}{\nabla }f$ for $f=zsin2\varphi$

Answer

Task!

Find $\underset{̲}{\nabla }\cdot \underset{̲}{F}$ for $\underset{̲}{F}=\rho cos\varphi \underset{̲}{\stackrel{̂}{\rho }}-\rho sin\varphi \underset{̲}{\stackrel{̂}{\rho }}+\rho z\underset{̲}{ẑ}$

i.e. $F_{\rho }=\rho cos\varphi ,F_{\varphi }=-\rho sin\varphi ,F_z=\rho z$

1. First find the derivatives $\frac{\partial }{\partial \rho }\left[\rho F_{\rho }\right],\frac{\partial }{\partial \varphi }\left[F_{\varphi }\right],\frac{\partial }{\partial z}\left[\rho F_z\right]$ :

   Answer

   $2\rho cos\varphi ,-\rho cos\varphi ,{\rho }^2$

2. Now combine these to find $\underset{̲}{\nabla }\cdot \underset{̲}{F}$ :

   Answer

   $$\begin{array}{rcll}\underset{̲}{\nabla }\cdot \underset{̲}{F}& =& \frac{1}{\rho }\left[\frac{\partial }{\partial \rho }\left(\rho F_{\rho }\right)+\frac{\partial }{\partial \varphi }\left(F_{\varphi }\right)+\frac{\partial }{\partial z}\left(\rho F_z\right)\right]& \\ & =& \frac{1}{\rho }\left[\frac{\partial }{\partial \rho }\left({\rho }^{2}cos\varphi \right)+\frac{\partial }{\partial \varphi }\left(-\rho sin\varphi \right)+\frac{\partial }{\partial z}\left({\rho }^{2}z\right)\right]& \\ & =& \frac{1}{\rho }\left[2\rho cos\varphi -\rho cos\varphi +{\rho }^2\right]& \\ & =& cos\varphi +\rho & \end{array}$$

Task!

Find $\underset{̲}{\nabla }\times \underset{̲}{F}$ for $\underset{̲}{F}=F_{\rho }\underset{̲}{\stackrel{̂}{\rho }}+F_{\varphi }\underset{̲}{\stackrel{̂}{\varphi }}+F_z\underset{̲}{ẑ}={\rho }^3\underset{̲}{\stackrel{̂}{\rho }}+\rho z\underset{̲}{\stackrel{̂}{\varphi }}+\rho zsin\varphi \underset{̲}{ẑ}$ . Show that the results are consistent with those found using Cartesian coordinates.

1. Find the curl $\underset{̲}{\nabla }\times \underset{̲}{F}$ :

   Answer

   $\frac{1}{\rho }\left|\begin{array}{ccc}\hfill \underset{̲}{\stackrel{̂}{\rho }}\hfill & \hfill \rho \underset{̲}{\stackrel{̂}{\varphi }}\hfill & \hfill \underset{̲}{ẑ}\hfill \\ \hfill \hfill \\ \hfill \frac{\partial }{\partial \rho }\hfill & \hfill \frac{\partial }{\partial \varphi }\hfill & \hfill \frac{\partial }{\partial z}\hfill \\ \hfill \hfill \\ \hfill {\rho }^3\hfill & \hfill {\rho }^{2}z\hfill & \hfill \rho zsin\varphi \hfill \end{array}\right|=\left(zcos\varphi -\rho \right)\underset{̲}{\stackrel{̂}{\rho }}-zsin\varphi \underset{̲}{\stackrel{̂}{\varphi }}+2z\underset{̲}{ẑ}$

2. Find $\underset{̲}{F}$ in Cartesian coordinates:

   Answer

   Use $\underset{̲}{\stackrel{̂}{\rho }}=cos\varphi \underset{̲}{i}+sin\varphi \underset{̲}{j},\underset{̲}{\stackrel{̂}{\varphi }}=-sin\varphi \underset{̲}{i}+cos\varphi \underset{̲}{j}$ to get $\underset{̲}{F}=\left(x^3+x{y}^2-yz\right)\underset{̲}{i}+\left(x^{2}y+y^3+xz\right)\underset{̲}{j}+yz\underset{̲}{k}$

3. Hence find $\underset{̲}{\nabla }\times \underset{̲}{F}$ in Cartesian coordinates:

   Answer

   $\left(z-x\right)\underset{̲}{i}-y\underset{̲}{j}+2z\underset{̲}{k}$

4. Using $\stackrel{̂}{\underset{̲}{\rho }}=cos\varphi \underset{̲}{i}+sin\varphi \underset{̲}{j}$ and $\stackrel{̂}{\underset{̲}{\varphi }}=-sin\varphi \underset{̲}{i}+cos\varphi \underset{̲}{j}$ , show that the solution to part 1. is equal to the solution for part 3.:

Exercises

1. For $\underset{̲}{F}=\rho \underset{̲}{\stackrel{̂}{\rho }}+\left(\rho sin\theta +z\right)\underset{̲}{\stackrel{̂}{\varphi }}+\rho z\underset{̲}{ẑ}$ , find $\underset{̲}{\nabla }\cdot \underset{̲}{F}$ and $\underset{̲}{\nabla }\times \underset{̲}{F}$ .
2. For $f={\rho }^2{z}^{2}cos2\varphi$ , find $\underset{̲}{\nabla }\times \left(\underset{̲}{\nabla }f\right)$ .

Answer