2 Engineering Example 4

2.1 Magnetic flux

Introduction

The magnetic flux through a surface is given by [maths rendering]  where [maths rendering] is the surface under consideration, [maths rendering] is the magnetic field and [maths rendering] is the vector normal to the surface.

Problem in words

The field generated by an infinitely long vertical wire on the [maths rendering] -axis is given by

[maths rendering]

Find the flux through a rectangular region (with sides parallel to the axes) on the plane [maths rendering] .

Mathematical statement of problem

Find the integral [maths rendering] over the surface, [maths rendering] , [maths rendering] [maths rendering] . (see Figure 11 which shows part of the plane [maths rendering] 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:

{ The surface S defined by x subscript 1 less than or equal to x less than or equal to x subscript 2, z subscript 1 less than or equal to z less than or equal to z subscript 2}

Mathematical analysis

On [maths rendering] ,   [maths rendering]  so   [maths rendering]  and the flux is given by the double integral

[maths rendering]

Interpretation

The magnetic flux increases in direct proportion to the extent of the side parallel to the axis (i.e. along the [maths rendering] -direction) but logarithmically with respect to the extent of the side perpendicular to the axis (i.e. along the [maths rendering] -axis).

Example 25

If [maths rendering] , evaluate [maths rendering] where [maths rendering] is the part of the plane [maths rendering] bounded by [maths rendering] , [maths rendering] .

Solution

Here [maths rendering] and hence [maths rendering] and

[maths rendering]

The integral

[maths rendering]

Similarly [maths rendering] .

Thus [maths rendering]

Key Point 5
  1. An integral of the form [maths rendering] evaluates to a scalar.
  2. An integral of the form [maths rendering] evaluates to a vector.

The vector function involved may be the gradient of a scalar or the curl of a vector.

Example 26

Integrate [maths rendering] where [maths rendering] and [maths rendering] is the area between [maths rendering] and [maths rendering] for [maths rendering] and [maths rendering] . (See Figure 12.)

Figure 12:

{ The area S between y=0 and y=x squared, for 0 less than or equal to x less than or equal to 1 and z=0}

Solution

Here [maths rendering] and [maths rendering] . Thus [maths rendering] and

[maths rendering]

For integrals of the form [maths rendering] , 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 [maths rendering] for [maths rendering] and [maths rendering] being the complete surface (including ends) of the cylinder [maths rendering] , [maths rendering] . (See Figure 13.)

Figure 13:

{ The cylinder rho less than or equal to a, 0 less than or equal to z less than or equal to 1 defining S}

Solution

The integral [maths rendering] must be evaluated separately for the curved surface and the ends.

For the curved surface, [maths rendering] (with the [maths rendering] coming from [maths rendering] the scale factor for [maths rendering] and the fact that [maths rendering] on the curved surface.) Thus, [maths rendering] and

[maths rendering]

On the bottom, [maths rendering] so [maths rendering] and the contribution to the integral is zero.

On the top, [maths rendering] and [maths rendering] and [maths rendering] and

[maths rendering]

So [maths rendering]