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

$\underset{̲}{dB}=\frac{Q{\mu }_0}{4\pi r^2}\left(\underset{̲}{v}\times \underset{̲}{r}\right)$

where ${\mu }_0$ is the permeability of free space ( $4\pi \times 10^{-7}\text{H}{\text{m}}^{-1}$ ), $\underset{̲}{r}$ 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

$\underset{̲}{B}={\int \; <!--nolimits\; -->}_C^D\underset{̲}{dB}={\int \; <!--nolimits\; -->}_C^D\frac{{\mu }_{0}I}{4\pi }\frac{\underset{̲}{dl}\times \underset{̲}{\stackrel{̂}{r}}}{r^2}$

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

$I\underset{̲}{dl}=-I\underset{̲}{k}dz$

i.e. $\underset{̲}{dl}=-\underset{̲}{k}dz$ (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

$\underset{̲}{dB}=\frac{{\mu }_{0}I}{4\pi \left(h^2+z^2\right)}\underset{̲}{dl}\times \underset{̲}{\stackrel{̂}{r}}$

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

$\left|\underset{̲}{dl}\times \underset{̲}{\stackrel{̂}{r}}\right|=sin\varphi dz=\frac{hdz}{\sqrt{h^2+z^2}}$

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

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

so that the total field at a point is

$B={\int \; <!--nolimits\; -->}_{z=-d}^c\frac{{\mu }_{0}I}{4\pi }h{\left(h^2+z^2\right)}^{-3∕2}dz$

Mathematical analysis

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

$\begin{array}{ccc}z=htanu\hfill & \hfill \Rightarrow \hfill & dz=h{sec}^{2}udu\hfill \\ z=c\hfill & \hfill \Rightarrow \hfill & u={tan}^{-1}\left(c∕h\right)=u_c\hfill \\ z=-d\hfill & \hfill \Rightarrow \hfill & u=-{tan}^{-1}\left(d∕h\right)=u_d\hfill \end{array}$

Substituting into the total field integral gives

and $\underset{̲}{B}=B\underset{̲}{\stackrel{̂}{\theta }}$ 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

$B=10^{-7}\left(\frac{5}{\sqrt{26}}+\frac{10}{\sqrt{101}}\right)=1.98\times 10^{-7}\text{T}=1.98\text{milli-gauss}.$

Interpretation

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

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

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 ${\int \; <!--nolimits\; -->}_C\left(\underset{̲}{\nabla }\cdot \underset{̲}{F}\right)\underset{̲}{dr}$ 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

The first term is

${\int \; <!--nolimits\; -->}_{C}6xdx\underset{̲}{i}={\int \; <!--nolimits\; -->}_{x=0}^{1}6xdx\underset{̲}{i}={\left[3x^2\right]}_0^1\underset{̲}{i}=3\underset{̲}{i}$

In the second term, as $y=x^2$ on $C$ , $dy$ may be replaced by $2xdx$ so

${\int \; <!--nolimits\; -->}_{C}6xdy\underset{̲}{j}={\int \; <!--nolimits\; -->}_{x=0}^{1}6x\times 2xdx\underset{̲}{j}={\int \; <!--nolimits\; -->}_0^{1}12x^{2}dx\underset{̲}{j}={\left[4x^3\right]}_0^1\underset{̲}{j}=4\underset{̲}{j}$

In the third term, as $z=x^3$ on $C$ , $dz$ may be replaced by $3x^{2}dx$ so

${\int \; <!--nolimits\; -->}_{C}6xdz\underset{̲}{k}={\int \; <!--nolimits\; -->}_{x=0}^{1}6x\times 3x^{2}dx\underset{̲}{k}={\int \; <!--nolimits\; -->}_0^{1}18x^{3}dx\underset{̲}{k}={\left[\frac{9}{2}{x}^4\right]}_0^1\underset{̲}{k}=\frac{9}{2}\underset{̲}{k}$

On summing, ${\int \; <!--nolimits\; -->}_C\left(\underset{̲}{\nabla }\cdot \underset{̲}{F}\right)\underset{̲}{dr}=3\underset{̲}{i}+4\underset{̲}{j}+\frac{9}{2}\underset{̲}{k}$ .

Task!

Find the vector line integral ${\int \; <!--nolimits\; -->}_{C}f\underset{̲}{dr}$ 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 $\left(9,3\right)$ .

Answer

Task!

Evaluate the vector line integral ${\int \; <!--nolimits\; -->}_C\underset{̲}{F}\times \underset{̲}{dr}$ 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 $\left(x-z\right)\underset{̲}{j}$ .

Answer

Exercises

1. Evaluate the vector line integral ${\int \; <!--nolimits\; -->}_C\left(\underset{̲}{\nabla }\cdot \underset{̲}{F}\right)\underset{̲}{dr}$ 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. ${\int \; <!--nolimits\; -->}_C\left\{\underset{̲}{\nabla }\left(xy\right)\right\}\times \underset{̲}{dr}$
   2. ${\int \; <!--nolimits\; -->}_C\left\{\underset{̲}{\nabla }\times \left(x^2\underset{̲}{i}+y^2\underset{̲}{k}\right)\right\}\times \underset{̲}{dr}$

Answer