### 7 Engineering Example 3

#### 7.1 Magnetic field from a line current

Introduction

A charge $Q$ , moving at a steady velocity $\underset{̲}{v}$ produces a magnetic field given by

$\phantom{\rule{2em}{0ex}}\underset{̲}{dB}=\frac{Q{\mu }_{0}}{4\pi {r}^{2}}\phantom{\rule{0.3em}{0ex}}\left(\underset{̲}{v}×\underset{̲}{r}\right)$

where ${\mu }_{0}$ is the permeability of free space ( $4\pi ×1{0}^{-7}\text{H}\phantom{\rule{0.3em}{0ex}}{\text{m}}^{-1}$ ), $\underset{̲}{r}$ is the position vector from the point of interest, $P$ , to the line current.

If, instead of a single charge, a current is used, then it is necessary to integrate over all charges in the current. So, the total magnetic field due to the current is given by

$\phantom{\rule{2em}{0ex}}\underset{̲}{B}={\int }_{C}^{D}\underset{̲}{dB}={\int }_{C}^{D}\frac{{\mu }_{0}I}{4\pi }\phantom{\rule{0.3em}{0ex}}\frac{\underset{̲}{dl}×\underset{̲}{\stackrel{̂}{r}}}{{r}^{2}}$

where $I\underset{̲}{dl}$ is the continuous form of $Q\underset{̲}{v}$ , $\underset{̲}{\stackrel{̂}{r}}$ is a unit vector along $\underset{̲}{r}$ and the current extends from $C$ to $D$ . Note that the field is perpendicular to both the current and the line from the current to $P$ .

Problem in words

Find the magnetic field strength (or magnetic flux density), measured in tesla (T), due to a current $I$ directed vertically downwards, starting at $z=c$ and ending at $z=-d$ . What is the field 1 m from the current when $c=5$ m, $d=10$ m and $I=1$ A?

Mathematical statement of problem

Figure 6:

Here

$\phantom{\rule{2em}{0ex}}I\underset{̲}{dl}=-I\underset{̲}{k}dz$

i.e. $\underset{̲}{dl}=-\underset{̲}{k}\phantom{\rule{0.3em}{0ex}}dz$ (pointing downwards). Imagine (without loss of generality) a point $P$ a distance $h$ from the line current and a distance $z$ below a typical line element of the current. The increment of field is given by

$\phantom{\rule{2em}{0ex}}\underset{̲}{dB}=\frac{{\mu }_{0}I}{4\pi \left({h}^{2}+{z}^{2}\right)}\phantom{\rule{0.3em}{0ex}}\underset{̲}{dl}×\underset{̲}{\stackrel{̂}{r}}$

where $\sqrt{{h}^{2}+{z}^{2}}$ is the distance of $P$ from the typical element. Since $\underset{̲}{dl}=-\underset{̲}{k}\phantom{\rule{0.3em}{0ex}}dz$ and $\underset{̲}{\stackrel{̂}{r}}$ is a unit vector, the magnitude of the vector product is

$\phantom{\rule{2em}{0ex}}\left|\underset{̲}{dl}×\underset{̲}{\stackrel{̂}{r}}\right|=sin\varphi \phantom{\rule{0.3em}{0ex}}dz=\frac{h\phantom{\rule{0.3em}{0ex}}dz}{\sqrt{{h}^{2}+{z}^{2}}}$

and is in a direction which (by the right-hand-rule) points OUT of the page to the right of the line and IN to it on the left. Knowing the direction of the field, now calculate the magnitude: the increment of field is given by

$\phantom{\rule{2em}{0ex}}dB\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}=\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\frac{{\mu }_{0}I}{4\pi \left({h}^{2}+{z}^{2}\right)}\phantom{\rule{0.3em}{0ex}}\frac{h\phantom{\rule{0.3em}{0ex}}dz}{\sqrt{{h}^{2}+{z}^{2}}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}=\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\frac{{\mu }_{0}I}{4\pi }\phantom{\rule{0.3em}{0ex}}h{\left({h}^{2}+{z}^{2}\right)}^{-3∕2}\phantom{\rule{0.3em}{0ex}}dz$

so that the total field at a point is

$\phantom{\rule{2em}{0ex}}B={\int }_{z=-d}^{c}\frac{{\mu }_{0}I}{4\pi }\phantom{\rule{0.3em}{0ex}}h{\left({h}^{2}+{z}^{2}\right)}^{-3∕2}\phantom{\rule{0.3em}{0ex}}dz$

Mathematical analysis

This integral can be evaluated by means of the substitution $z=htanu$ where

$\phantom{\rule{2em}{0ex}}\begin{array}{ccc}z=htanu\hfill & \hfill ⇒\hfill & dz=h{sec}^{2}u\phantom{\rule{0.3em}{0ex}}du\hfill \\ z=c\hfill & \hfill ⇒\hfill & u={tan}^{-1}\left(c∕h\right)={u}_{c}\hfill \\ z=-d\hfill & \hfill ⇒\hfill & u=-{tan}^{-1}\left(d∕h\right)={u}_{d}\hfill \end{array}$

Substituting into the total field integral gives

and $\underset{̲}{B}=B\phantom{\rule{0.3em}{0ex}}\underset{̲}{\stackrel{̂}{\theta }}$ where $\underset{̲}{\stackrel{̂}{\theta }}$ is a unit vector in a circumferential direction around the line current. Now if $I$ = 1 A, $c=5$ m, $d=10$ m and $h=1$ m the magnetic field becomes

$\phantom{\rule{2em}{0ex}}B=1{0}^{-7}\left(\frac{5}{\sqrt{26}}+\frac{10}{\sqrt{101}}\right)=1.98×1{0}^{-7}\text{T}=1.98\phantom{\rule{1em}{0ex}}\text{milli-gauss}.$

Interpretation

Note that if $c$ and $d$ $\to$ $\infty$ then

$\phantom{\rule{2em}{0ex}}B\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}=\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\frac{{\mu }_{0}I}{4\pi h}\left(\frac{c}{\sqrt{\left({h}^{2}+{c}^{2}\right)}}+\frac{d}{\sqrt{\left({h}^{2}+{d}^{2}\right)}}\right)\to \frac{{\mu }_{0}I}{4\pi h}\left[2\right]\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}=\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\frac{{\mu }_{0}I}{2\pi h}$

i.e. the field lines are circles around the line current and the magnetic field strength is inversely proportional to the distance of the point of interest $P$ from the current.

A scalar or vector involved in a vector line integral may itself be a vector derivative as this next Example illustrates.

##### Example 15

Find the vector line integral ${\int }_{C}\left(\underset{̲}{\nabla }\cdot \underset{̲}{F}\right)\phantom{\rule{0.3em}{0ex}}\underset{̲}{dr}$ where $\underset{̲}{F}$ is the vector ${x}^{2}\underset{̲}{i}+2xy\underset{̲}{j}+2xz\underset{̲}{k}$ and $C$ is the curve $y={x}^{2}$ , $z={x}^{3}$ from $x=0$ to $x=1$ i.e. from $\left(0,0,0\right)$ to $\left(1,1,1\right)$ .

##### Solution

As $\underset{̲}{F}={x}^{2}\underset{̲}{i}+2xy\underset{̲}{j}+2xz\underset{̲}{k}$ , $\underset{̲}{\nabla }\cdot \underset{̲}{F}=2x+2x+2x=6x$ .

The integral

$\begin{array}{rcll}{\int }_{C}\left(\underset{̲}{\nabla }\cdot \underset{̲}{F}\right)\phantom{\rule{0.3em}{0ex}}\underset{̲}{dr}& =& {\int }_{C}6x\left(dx\underset{̲}{i}+dy\underset{̲}{j}+dz\underset{̲}{k}\right)& \text{}\\ & =& {\int }_{C}6x\phantom{\rule{0.3em}{0ex}}dx\phantom{\rule{0.3em}{0ex}}\underset{̲}{i}+{\int }_{C}6x\phantom{\rule{0.3em}{0ex}}dy\phantom{\rule{0.3em}{0ex}}\underset{̲}{j}+{\int }_{C}6x\phantom{\rule{0.3em}{0ex}}dz\phantom{\rule{0.3em}{0ex}}\underset{̲}{k}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}& \text{}\end{array}$

The first term is

$\phantom{\rule{2em}{0ex}}{\int }_{C}6x\phantom{\rule{0.3em}{0ex}}dx\phantom{\rule{0.3em}{0ex}}\underset{̲}{i}={\int }_{x=0}^{1}6x\phantom{\rule{0.3em}{0ex}}dx\phantom{\rule{0.3em}{0ex}}\underset{̲}{i}={\left[\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}3{x}^{2}\right]}_{0}^{1}\underset{̲}{i}=3\underset{̲}{i}$

In the second term, as $y={x}^{2}$ on $C$ , $dy$ may be replaced by $2x\phantom{\rule{0.3em}{0ex}}dx$ so

$\phantom{\rule{2em}{0ex}}{\int }_{C}6x\phantom{\rule{0.3em}{0ex}}dy\phantom{\rule{0.3em}{0ex}}\underset{̲}{j}={\int }_{x=0}^{1}6x×2x\phantom{\rule{0.3em}{0ex}}dx\phantom{\rule{0.3em}{0ex}}\underset{̲}{j}={\int }_{0}^{1}12{x}^{2}\phantom{\rule{0.3em}{0ex}}dx\phantom{\rule{0.3em}{0ex}}\underset{̲}{j}={\left[\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}\phantom{\rule{0.3em}{0ex}}4{x}^{3}\right]}_{0}^{1}\underset{̲}{j}=4\underset{̲}{j}$

In the third term, as $z={x}^{3}$ on $C$ , $dz$ may be replaced by $3{x}^{2}\phantom{\rule{0.3em}{0ex}}dx$ so

$\phantom{\rule{2em}{0ex}}{\int }_{C}6x\phantom{\rule{0.3em}{0ex}}dz\phantom{\rule{0.3em}{0ex}}\underset{̲}{k}={\int }_{x=0}^{1}6x×3{x}^{2}\phantom{\rule{0.3em}{0ex}}dx\phantom{\rule{0.3em}{0ex}}\underset{̲}{k}={\int }_{0}^{1}18{x}^{3}\phantom{\rule{0.3em}{0ex}}dx\phantom{\rule{0.3em}{0ex}}\underset{̲}{k}={\left[\frac{9}{2}{x}^{4}\right]}_{0}^{1}\underset{̲}{k}=\frac{9}{2}\underset{̲}{k}$

On summing, ${\int }_{C}\left(\underset{̲}{\nabla }\cdot \underset{̲}{F}\right)\phantom{\rule{0.3em}{0ex}}\underset{̲}{dr}=3\underset{̲}{i}+4\underset{̲}{j}+\frac{9}{2}\underset{̲}{k}$ .

Find the vector line integral ${\int }_{C}f\underset{̲}{dr}$ where $f={x}^{2}$ and $C$ is

1. the curve $y={x}^{1∕2}$ from $\left(0,0\right)$ to $\left(9,3\right)$ .
2. the line $y=x∕3$ from $\left(0,0\right)$ to $\left(9,3\right)$ .
1. ${\int }_{0}^{9}\left({x}^{2}\underset{̲}{i}+\frac{1}{2}{x}^{3∕2}\underset{̲}{j}\right)dx=243\underset{̲}{i}+\frac{243}{5}\underset{̲}{j}$ ,
2. ${\int }_{0}^{9}\left({x}^{2}\underset{̲}{i}+\frac{1}{3}{x}^{2}\underset{̲}{j}\right)dx=243\underset{̲}{i}+\frac{243}{2}\underset{̲}{j}$ .

Evaluate the vector line integral ${\int }_{C}\underset{̲}{F}×\underset{̲}{dr}$ when $C$ represents the contour

$y=4-4x$ , $z=2-2x$ from $\left(0,4,2\right)$ to $\left(1,0,0\right)$ and $\underset{̲}{F}$ is the vector field $\left(x-z\right)\underset{̲}{j}$ .

${\int }_{0}^{1}\left\{\left(4-6x\right)\underset{̲}{i}+\left(2-3x\right)\underset{̲}{k}\right\}=\underset{̲}{i}+\frac{1}{2}\underset{̲}{k}$

##### Exercises
1. Evaluate the vector line integral ${\int }_{C}\left(\underset{̲}{\nabla }\cdot \underset{̲}{F}\right)\phantom{\rule{0.3em}{0ex}}\underset{̲}{dr}$ in the case where $\underset{̲}{F}=x\underset{̲}{i}+xy\underset{̲}{j}+x{y}^{2}\underset{̲}{k}$ and $C$ is the contour described by $x=2t$ , $y={t}^{2}$ , $z=1-t$ for $t$ starting at $t=0$ and going to $t=1$ .
2. When $C$ is the contour $y={x}^{3}$ , $z=0$ , from $\left(0,0,0\right)$ to $\left(1,1,0\right)$ , evaluate the vector line integrals
1. ${\int }_{C}\left\{\underset{̲}{\nabla }\left(xy\right)\right\}×\underset{̲}{dr}$
2. ${\int }_{C}\left\{\underset{̲}{\nabla }×\left({x}^{2}\underset{̲}{i}+{y}^{2}\underset{̲}{k}\right)\right\}×\underset{̲}{dr}$
1. $4\underset{̲}{i}+\frac{7}{3}\underset{̲}{j}-2\underset{̲}{k}$ ,
1. $\underset{̲}{0}$ ,
2. $\underset{̲}{k}$