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C

S1

x

Figure 7.2: De nition of the geometric variables for problem a.

7.1. STATEMENT OF STOKES' LAW 61

Problem a: Let us verify this property for an example. Consider the vector eld v = r^ .

'

Let the curve C used for the line integral be a circle in the x y-plane with radius

R, see gure (??) for the geometry of the problem. (i) Compute the line integral in

the left hand side of (7.2) by direct integration. Compute the surface integral in the

right hand side of (7.2) by (ii) integrating over a circle of radius R in the x y-plane

(the surface S1 in gure (??)) and by (iii) integrating over the upper half of a sphere

with radius R (the surface S2 in gure (??)). Verify that the three integrals are

identical.

S1

C

S2

Figure 7.3: Two surfaces that are bouded by the same contour C.

It is actually not di cult to prove that the surface integral in Stokes' law is independent

of the speci c choice of the surface S as long as it is bounded by the same contour C.

Consider gure (7.3) where the two surfaces S1 and S2 are bounded by the same contour

v is the same for the two surfaces,

C. We want to show that the surface integral of r

Z Z

i.e. that:

(r v) dS = (r v) dS : (7.3)

S1 S2

We can form a closed surface S by combining the surfaces S1 and S2 .

Problem b: Show that equation (7.3) is equivalent to the condition

I

(r v) dS =0 (7.4)

S

where the integration is over the closed surfaces de ned by the combination of S1

and S2 . Pay in particular attention to the sign of the di erent terms.

Problem c: Use Gauss' law to convert (7.4) to a volume integral and show that the

integral is indeed identical to zero.

The result you obtained in problem c implies that the condition (7.3) is indeed satis ed

and that in the application of Stokes' law you can choose any surface as long as it is

CHAPTER 7. THE THEOREM OF STOKES

62

bounded by the contour over which the line integration is carried out. This is a very

useful result because often the surface integration can be simpli ed by choosing the surface

carefully.

7.2 Stokes' theorem from the theorem of Gauss

Stokes' law is concerned with surface integrations. Since the curl is intrinsically a three-

dimensional vector, Stokes's law is inherently related to three space dimensions. However,

if we consider a vector eld that depends only on the coordinates x and y (v = v(x y))

v points along

and that has a vanishing component in the z-direction (vz = 0), then r

the z-axis. If we consider a contour C that is con ned to the x y-plane, Stokes' law takes

for such a vector eld the form

I Z

(vx dx + vy dy)= (@x vy @y vx ) dxdy : (7.5)

;

C S

Problem a: Verify this.

This result can be derived from the theorem of Gauss in two dimensions.

Problem b: Show that Gauss' law (6.1) for a vector eld u in two dimensions can be

I Z

written as

(u n)ds = (@x ux + @y uy )dxdy

^ (7.6)

C S

where the unit vector n is perpendicular to the curve C (see gure (7.4)) and where

^

ds denotes the integration over the arclength of the curve C.

^

n

^

t

^

v

^

u

Figure 7.4: De nition of the geometric variables for the derivation of Stokes' law from the

theorem of Gauss.

7.3. THE MAGNETIC FIELD OF A CURRENT IN A STRAIGHT WIRE 63

In order to derive the special form of Stokes' law (7.5) from Gauss' law (7.6) we have to

de ne the relation between the vectors u and v. Let the vector u follow from v by a

clockwise rotation over 90 degrees, see gure (7.4).

Problem c: Show that:

vx = and vy = ux : (7.7)

y

;u

We now de ne the unit vector ^ to be directed along the curve C, see gure (7.4). Since

t

a rotation is an orthonormal transformation the inner product of two vectors is invariant

for a rotation over 90 degrees so that (u n) = (v ^).

^ t

Problem d: Verify this by expressing the components of ^ in the components of n and

t ^

by using (7.7).

Problem e: Use these results to show that (7.5) follows from (7.6).

What you have shown here is that Stokes' law for the special case considered in this section

is identical to the theorem of Gauss for two spatial dimensions.

7.3 The magnetic eld of a current in a straight wire

We now return to the problem of the generation of the magnetic eld induced by a current

in an in nite straight wire that was discussed in section (5.5). Because of the cylinder

symmetry of the problem, we know that the magnetic eld is p the direction of the unit

in

^

vector ' and that the eld only depends on the distance r = x2 + y2 to the wire:

B =B(r)^ : (7.8)

'

The eld can be found by integrating the eld equation B = 0 J over a disc with radius

r

r perpendicular to the wire, see gure (7.5). When the disc is larger than the thickness of

R

the wire the surface integral of J gives the electric current I through the wire: I = J dS.

Problem a: Use these results and Stokes' law to show that:

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