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integral along the small contour. If Hthe ow (locally) rotates and if we integrate

along the uid ow, the line integral v dr will be relatively large, so that this line

integral indeed measures the local rotation.

Rotation and shear each contribute to the curl of a vector eld. Let us consider once

again a vector eld of the form (5.6) which is axially symmetric around the z-axis.

In the following we don't require the rotation around the z-axis to be rigid, so that

v(r) in (5.6) is still arbitrary. We know that both the rotation around the z-axis

and the shear are a source of vorticity.

Problem e: Show that for the ow

v(r) = A (5.11)

r

the vorticity vanishes, with A a constant that is not yet determined. Make a

sketch of this ow eld.

The vorticity of this ow vanishes despite the fact that the ow rotates around the

z-axis (but not in rigid rotation) and that the ow has a nonzero shear. The reason

that the vorticity vanishes is that the contribution of the rotation around the z-axis

to the vorticity is equal but of opposite sign from the contribution of the shear, so

that the total vorticity vanishes. Note that this implies that a paddle-wheel does

not change its orientation as it moves with this ow!

5.5. THE MAGNETIC FIELD INDUCED BY A STRAIGHT CURRENT 47

5.5 The magnetic eld induced by a straight current

At this point you may have the impression that the ow eld (5.11) is contrived

in an arti cial way. However, keep in mind that all the arguments of the previous

section apply to any vector eld and that uid ow was used only as an example to

x our mind. As an example we consider the generation of the magnetic eld B by

an electrical current J that is independent of time. The Maxwell equation for the

curl of the magnetic eld in vacuum is for time-independent elds given by:

B = 0J (5.12)

r

see equation (5.22) in ref. 31]. In this expression 0 is the magnetic permeability

of vacuum. It plays the role of a coupling constant since it governs the strength

of the magnetic eld that is generated by a given current. It plays the same role

as 1/permittivity in (4.12) or the gravitational constant G in (4.17). The vector

J denotes the electric current per unit volume (properly called the electric current

density).

For simplicity we will consider an electric current running through an in nite straight

wire along the z-axis. Because of rotational symmetry around the z-axis and because

of translational invariance along the z-axis the magnetic eld depends neither on '

nor on z and must be of the form (5.6). Away from the wire the electrical current J

vanishes.

Problem a: Show that

B = A' :

r^ (5.13)

A comparison with equation (5.11) shows that for this magnetic eld the contribution

B is exactly balanced by the contribution

of the \rotation" around the z-axis to r

B. It should be noted that the magnetic eld derived

of the\magnetic shear" to r

in this section is of great importance because this eld has been used to de ne the

unit of electrical current, the Ampere. However, this can only be done when the

constant A in expression (5.13) is known.

Problem b: Why does the treatment of this section not tell us what the relation

is between the constant A and the current J in the wire?

We will return to this issue in section (7.3).

5.6 Spherical coordinates and cylinder coordinates

In section (4.4), expressions for the divergence in spherical coordinates and cylinder

coordinates were derived. Here we will do the same for the curl because these

expressions are frequently very useful. It is possible to derive the curl in curvilinear

coordinates by systematically carrying out the e ect of the coordinate transformation

from Cartesian coordinates to curvilinear coordinates on all the elements of the

CHAPTER 5. THE CURL OF A VECTOR FIELD

48

involved vectors and on all the di erentiations. As an alternative, we will use the

physical interpretation of the curl given by expression (5.3) to derive the curl in

spherical coordinates. This expression simply states that a certain component of the

H

curl of a vector eld v is the line integral v dr along a contour perpendicular to

the component of the curl that we are considering, normalized by the surface area

bounded by that contour. As an example we will derive for a system of spherical

coordinates the '-component of the curl, see gure (5.5) for the de nition of the

geometric variables.

z

^

r

^

Ï•

rdÎ¸

dr

Î¸ ^

Î¸

dÎ¸

r

y

Ï•

x

Figure 5.5: De nition of the geometric variables for the computation of the curl in spherical

coordinates.

Consider in gure (5.5) the little surface. When we carry out the line integral

along the surface we integrate in the direction shown in the gure. The reason for

^

this is that the azimuth ' increases when we move into the gure, hence ' point

into the gure. Following the rules of a right-handed screw this corresponds with

the indicated sense of integration. The area enclosed by the contour is given by

rd dr. By summing the contributions of the four sides of the contour we nd using

v is given by:

expression (5.3) that the '-component of r

(r v)' = rd1dr (r + dr )(r + dr)d vr (r + d )dr v (r )rd + vr (r )drg :

fv ; ;

(5.14)

In this expression vr and v denote the components of v in the radial direction and

in the direction of ^ respectively.

5.6. SPHERICAL COORDINATES AND CYLINDER COORDINATES 49

Problem a: Verify expression (5.14).

This result can be simpli ed by Taylor expanding the components of v in dr and d

and linearizing the resulting expression in the in nitesimal increments dr and d .

Problem b: Do this and show that the nal result does not depend on dr and d

and is given by:

(r v)' = 1 @r (rv ) 1 @vr :

@ (5.15)

r r@ ;

The same treatment can be applied to the other components of the curl. This leads

to the following expression for the curl in spherical coordinates:

n@ o ^1 n 1 o n o

v = ^ r sin

r1 ^r @

v') ; @v + ' 1 @r (rv ) ; @vr

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