2 Engineering Example 4

2.1 Magnetic flux

Introduction

The magnetic flux through a surface is given by $\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\underset{̲}{B}\cdot \underset{̲}{dS}$  where $S$ is the surface under consideration, $\underset{̲}{B}$ is the magnetic field and $\underset{̲}{dS}$ is the vector normal to the surface.

Problem in words

The field generated by an infinitely long vertical wire on the $z$ -axis is given by

$\underset{̲}{B}=\frac{{\mu }_{0}I}{2\pi }\frac{-y\underset{̲}{i}+x\underset{̲}{j}}{x^2+y^2}$

Find the flux through a rectangular region (with sides parallel to the axes) on the plane $y=0$ .

Mathematical statement of problem

Find the integral $\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\underset{̲}{B}\cdot \underset{̲}{dS}$ over the surface, $x_1\le x\le x_2$ , $$ $z_1\le z\le z_2$ . (see Figure 11 which shows part of the plane $y=0$ for which the flux is found and a single magnetic field line. The strength of the field is inversely proportional to the distance from the axis.)

Figure 11:

Mathematical analysis

On $y=0$ ,   $\underset{̲}{B}=\frac{{\mu }_{0}I}{2\pi x}\underset{̲}{j}\text{ and }\underset{̲}{dS}=dxdz\underset{̲}{j}$  so   $\underset{̲}{B}\cdot \underset{̲}{dS}=\frac{{\mu }_{0}I}{2\pi x}dxdz$  and the flux is given by the double integral

Interpretation

The magnetic flux increases in direct proportion to the extent of the side parallel to the axis (i.e. along the $z$ -direction) but logarithmically with respect to the extent of the side perpendicular to the axis (i.e. along the $x$ -axis).

Example 25

If $\underset{̲}{F}=x^2\underset{̲}{i}+y^2\underset{̲}{j}+z^2\underset{̲}{k}$ , evaluate $\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\underset{̲}{F}\times \underset{̲}{dS}$ where $S$ is the part of the plane $z=0$ bounded by $x=\pm 1$ , $y=\pm 1$ .

Solution

Here $\underset{̲}{dS}=dxdy\underset{̲}{k}$ and hence $\underset{̲}{F}\times \underset{̲}{dS}=\left|\begin{array}{ccc}\hfill \underset{̲}{i}\hfill & \hfill \underset{̲}{j}\hfill & \hfill \underset{̲}{k}\hfill \\ \hfill \hfill \\ \hfill x^2\hfill & \hfill y^2\hfill & \hfill z^2\hfill \\ \hfill \hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill dxdy\hfill \end{array}\right|=y^{2}dxdy\underset{̲}{i}-x^{2}dxdy\underset{̲}{j}$ and

$\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\underset{̲}{F}\times \underset{̲}{dS}={\int \; <!--nolimits\; -->}_{y=-1}^1{\int \; <!--nolimits\; -->}_{x=-1}^1{y}^{2}dxdy\underset{̲}{i}-{\int \; <!--nolimits\; -->}_{y=-1}^1{\int \; <!--nolimits\; -->}_{x=-1}^1{x}^{2}dxdy\underset{̲}{j}$

The integral

${\int \; <!--nolimits\; -->}_{y=-1}^1{\int \; <!--nolimits\; -->}_{x=-1}^1{y}^{2}dxdy={\int \; <!--nolimits\; -->}_{y=-1}^1{\left[y^{2}x\right]}_{x=-1}^{1}dy={\int \; <!--nolimits\; -->}_{y=-1}^{1}2y^{2}dy={\left[\frac{2}{3}{y}^3\right]}_{-1}^1=\frac{4}{3}$

Similarly ${\int \; <!--nolimits\; -->}_{y=-1}^1{\int \; <!--nolimits\; -->}_{x=-1}^1{x}^{2}dxdy=\frac{4}{3}$ .

Thus $\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\underset{̲}{F}\times \underset{̲}{dS}=\frac{4}{3}\underset{̲}{i}-\frac{4}{3}\underset{̲}{j}$

Example 26

Integrate $\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\left(\underset{̲}{\nabla }\mathrm{\&varphi;}\right).\underset{̲}{dS}$ where $\mathrm{\&varphi;}=x^2+2yz$ and $S$ is the area between $y=0$ and $y=x^2$ for $0\le x\le 1$ and $z=0$ . (See Figure 12.)

Figure 12:

Solution

Here $\underset{̲}{\nabla }\mathrm{\&varphi;}=2x\underset{̲}{i}+2z\underset{̲}{j}+2y\underset{̲}{k}$ and $\underset{̲}{dS}=\underset{̲}{k}dydx$ . Thus $\left(\underset{̲}{\nabla }\mathrm{\&varphi;}\right).\underset{̲}{dS}=2ydydx$ and

For integrals of the form $\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\underset{̲}{F}\cdot \underset{̲}{dS}$ , non-Cartesian coordinates e.g. cylindrical polar or spherical polar coordinates may be used. Once again, it is necessary to include any scale factors along with the unit normal.

Example 27

Using cylindrical polar coordinates, (see HELM booklet  28.3), find the integral ${\int \; <!--nolimits\; -->}_S\underset{̲}{F}\left(\underset{̲}{r}\right)\cdot \underset{̲}{dS}$ for $\underset{̲}{F}=\rho z\underset{̲}{\stackrel{̂}{\rho }}+z{sin}^2\mathrm{\&varphi;}\underset{̲}{ẑ}$ and $S$ being the complete surface (including ends) of the cylinder $\rho \le a$ , $0\le z\le 1$ . (See Figure 13.)

Figure 13:

Solution

The integral ${\int \; <!--nolimits\; -->}_S\underset{̲}{F}\left(\underset{̲}{r}\right)\cdot \underset{̲}{dS}$ must be evaluated separately for the curved surface and the ends.

For the curved surface, $\underset{̲}{dS}=\underset{̲}{\stackrel{̂}{\rho }}ad\mathrm{\&varphi;}dz$ (with the $a$ coming from $\rho$ the scale factor for $\mathrm{\&varphi;}$ and the fact that $\rho =a$ on the curved surface.) Thus, $\underset{̲}{F}\cdot \underset{̲}{dS}=a^{2}zd\mathrm{\&varphi;}dz$ and

On the bottom, $z=0$ so $\underset{̲}{F}=\underset{̲}{0}$ and the contribution to the integral is zero.

On the top, $z=1$ and $\underset{̲}{dS}=\underset{̲}{ẑ}\rho d\rho d\mathrm{\&varphi;}$ and $\underset{̲}{F}\cdot \underset{̲}{dS}=\rho z{sin}^2\mathrm{\&varphi;}d\mathrm{\&varphi;}d\rho =\rho {sin}^2\mathrm{\&varphi;}d\mathrm{\&varphi;}d\rho$ and

So $\int \; <!--nolimits\; -->{\int \; <!--nolimits\; -->}_S\underset{̲}{F}\left(\underset{̲}{r}\right)\cdot \underset{̲}{dS}=\pi a^4+\frac{1}{2}\pi a^2=\pi a^2\left(a^2+\frac{1}{2}\right)$