Lehner G. Electromagnetic Field Theory for Engineers and Physicists

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274 Basics of Magnetostatics

The field of densely wound toroidal coils, as shown in Fig. 5.14, is also easy

to write. The field for a circular, concentric path inside the coil is

(5.40)

Fig. 5.13

r()

Fig. 5.14

H ds•

∫

2πrH

ϕi

NI==

Fig. 5.15

r()

ϕi

2πr

---------=

5.2 Some Magnetic Fields 275

where N is the total number of turns and I the current in one turn. All fields outside

the coil vanish if we make the idealized assumption that there is no current flows

along the coil in azimuthal direction. Fig. 5.15 illustrates the field as a function of r.

Note that all this does not depend on the shape of the coil’s cross-sectional area.

5.2.4 The Field of a Circular Current and a Magnetic Dipole

Let an azimuthal current I flow in a circular loop as illustrated in Fig. 5.16.

The loop is situated in the x-y plane. Its current density is

(5.41)

Cartesian coordinates allow the use of eq. (5.15) to calculate the vector potential.

(5.42)

Then, using eq. (5.15) gives

Similarly the y-component becomes

Returning to the components in cylindrical coordinates using the relation

one obtains

Fig. 5.16

Iδ rr

–()δz()=

g–

ϕsin=

ϕ .cos=

r()

4π

--------

ϕ'sin δ r' r

–()δz'()r' ϕ'dr'dz'd–

r ϕcos r' ϕ'cos–()

r ϕsin r' ϕ'sin–()

zz'–()

------------------------------------------------------------------------------------------------------------------------------

∫

4π

-------------

ϕ'sin ϕ'd–

2rr

ϕϕ'–()cos–++

-----------------------------------------------------------------------------

2π

∫

r()

4π

-------------

ϕ'cos ϕ'd

2rr

ϕϕ'–()cos–++

-----------------------------------------------------------------------------

2π

∫

ϕcos A

ϕsin+=

A–

ϕsin A

ϕcos+=

276 Basics of Magnetostatics

(5.43)

. (5.44)

One might as well integrate from through . vanishes because of the odd

integrand, does not, however. We will, once again, demonstrate the necessity of

starting from Cartesian coordinates when using eq. (5.15). Had we tried to

calculate and directly from (5.15), the result would not have been (5.44),

but an incorrect one. This would have been the integral without the factor

in the numerator of the integrand. To find an analytic solution for

of (5.44) is not possible. However, the integral is related to the total elliptic

integrals. To see this, one has to resort to some mathematical tricks. First, we

introduce the variable

This makes

Furthermore, define the parameter

(5.45)

Then, obtain

Obviously, this is independent of because integration is always over a whole

period of . Also the symmetry of the problem does not allow a -

dependency of . This allows one to pick a convenient value, for example,

. The integration is then from through :

4π

-------------

ϕϕ'–()sin ϕ'd

2rr

ϕϕ'–()cos–++

-----------------------------------------------------------------------------

2π

∫

0==

4π

-------------

ϕϕ'–()cos ϕ'd

2rr

ϕϕ'–()cos–++

-----------------------------------------------------------------------------

2π

∫

π–+π A

ϕϕ'–()cos A

πϕϕ'–()–

----------------------------=

ϕϕ'– π 2ψ–=

ϕϕ'–()cos 2 ψsin

1–=

4rr

+()

--------------------------------=

4π rr

+()

-------------------------------------------

2 ψsin

1–

1 k

ψsin

–

---------------------------------

2 ψd

π 2⁄ϕ2⁄–

3π 2⁄ϕ2⁄–

∫

ψsin

ϕ 2π= ψπ2⁄–= ψ +π 2⁄=

2π rr

+()

-------------------------------------------

2 ψsin

1–

1 k

ψsin

–

---------------------------------

ψd

π 2⁄–

π 2⁄

∫

π rr

+()

----------------------------------------

2 ψsin

1–

1 k

ψsin

–

---------------------------------

ψd

π 2⁄

∫

5.2 Some Magnetic Fields 277

These two integrals are called the total elliptic integrals of the first and the second

kind, respectively:

(5.46)

. (5.47)

These enable one to write in the following way:

(5.48)

In order to calculate , one needs to know the derivatives of K and E.

(5.49)

We will limit ourselves to the case where the distance from the origin to the field

point is much larger than :

A consequence of (5.45) is that and both, K and E can be replaced by the first

few terms of their respective series expansion

π rr

+()

----------------------------------------

2 ψsin

1–

-----

–





1 k

ψsin

–

-----------------------------------------------------------

ψd

π 2⁄

∫

π rr

+()

----------------------------------------

-----

1 k

– ψsin

1 k

ψsin

–

---------------------------------

ψd

π 2⁄

∫

–

-----1–





ψd

1 k

ψsin

–

---------------------------------

π 2⁄

∫







π rr

+()

----------------------------------------

-----1–





ψd

1 k

ψsin

–

---------------------------------

π 2⁄

∫

-----

1 k

ψsin

– ψd

π 2⁄

∫

–







---





ψd

1 k

ψsin

–

---------------------------------

2⁄

∫

---





1 k

ψsin

– ψd

2⁄

∫

2πr

---------

+()

-----–





---





---





–







---





---





k 1 k

–()

----------------------

---





-------------------–=

---





---





---





–

-------------------------------------------=











+ r

k 1«

278 Basics of Magnetostatics

(5.50)

The correctness of the given terms of the two series can be proven by expanding

the integrands of (5.46) and (5.47) into a series and then integrate them term by

term. Using (5.50) in (5.48) for small values of k gives

i.e.,

Introducing spherical coordinates (Fig. 5.17):

and

This gives

Using

---





---

-----





…++ +=

---





---

-----

–3

-----





– …–=











2πr

---------

+()

+ ⋅=

---

-----

–





---

------

…++ +





---

–

------

– …–





–







⋅

--------

+()

------





--------

+()

32⁄

-----------------------------

≈≈

R θϕ,,

Fig. 5.17

θsin

--------------------=

4π

--------

πθsin

--------------------

≈

BA∇×=

5.2 Some Magnetic Fields 279

gives

Now, we introduce the magnetic dipole moment m

(5.51)

and its vector representation

(5.52)

where the direction of the current and the direction of the area element are

determined by a right handed system. This gives

(5.53)

and

(5.54)

Unfortunately, the definition of m is not standardized and is oftentimes

introduced without the factor . The field components in eq. (5.54) behave as

functions of the location in the same manner as those of the electric dipole field

(2.63). These two equations merge when exchanging with m. The name

magnetic dipole is based on this analogy.

Just like the ideal electric dipole, the magnetic dipole has to be understood as

the limit of an infinite current I while becomes infinitely small, leaving the

R θsin

---------------

θ∂

∂

θsin A

()

4π

--------

π2 θcos

------------------------

---

R∂

∂

()–

4π

--------

πθsin

--------------------

0 .=

m µ

πµ

Ia ==

m µ

Ia =

4π

------

θsin

-----------

4π

-------

θcos

------------

4π

------

θsin

-----------

0 .=

⁄

Fig. 5.19

–

Fig. 5.18

a 0→

m µ

Ia=

I ∞→

280 Basics of Magnetostatics

product finite. The area is not necessarily circular. In the limit of a

vanishing area, its shape is immaterial for the field. What matters is the size of the

area, which the definitions (5.51) and (5.52) already consider (Fig. 5.18).

A scalar potential was suitable to determine the field of an electric dipole

(2.60). This is also the case for the magnetic dipole, for which we have

(5.55)

with the quantities as shown in Fig. 5.19. When comparing (5.54) and (5.55), be

aware that and . The scalar potential is given in a form

which is independent from a coordinate system. The vector potential (5.53) can

also be written in a representation that is independent of a coordinate system:

(5.56)

where we used the vector identity:

Using and in (5.56), we find as of (5.53). The field

produced by the current in a circular loop around a finite area is shown in Fig. 5.20.

At a sufficiently large distance, it is indistinguishable from the field of two

opposite charges of the same magnitude. The magnetic field of an ideal magnetic

dipole corresponds entirely to the electric field of an ideal electric dipole (see

Fig. 2.15).

The magnetic dipole is one of the most important concepts in magnetostatics.

It will be the center piece of several of the following sections. Therefore, we will

summarize the most important formulas, side by side with the electrostatic ones.

Consider a dipole, oriented in the positive z-direction, located at the origin of a

coordinate system. Then for spherical coordinates we have the following relations:

m θcos

4πµ

–

---------------------------------

mrr

–()•

4πµ

–

---------------------------------

H ψ∇–= B µ

H= ψ

mrr

–()×

4π rr

–

------------------------------

4π

------

∇

–

----------------

×–

4π

------

∇

–

----------------

× == =

∇ bϕ()×ϕ∇b×()b ϕ∇()×–=

m me

= r

0= A

Fig. 5.20

5.2 Some Magnetic Fields 281

Independent of any coordinate system, for a dipole (p or m) at location we have

electric magnetic

2p θcos

4πε

-------------------= H

2m θcos

4πµ

--------------------=

p θsin

4πε

-------------------= H

m θsin

4πµ

-------------------=

0= H

D ε

E= B µ

E ϕ∇–= H ψ∇–=

p θcos

4πε

-------------------= ψ

m θcos

4πµ

-------------------=

BA∇×=

m θsin

4πR

---------------=

prr

–()•

4πε

–

--------------------------------= ψ

mrr

–()•

4πµ

–

---------------------------------=

mrr

–()×

4π rr

–

------------------------------=

4π

------

∇

–

----------------

×–=

4π

------

∇

–

----------------

×=

282 Basics of Magnetostatics

Anticipating the outcome of the ensuing discussion, we will give the result for a

uniform dipole layer

5.2.5 The Field of an Arbitrary Current Loop

An arbitrarily shaped current loop as illustrated in Fig. 5.21 can be thought of as

being composed of many dipoles. If the currents in all loops are identical (I), then

they cancel on the inside, even for infinitesimal sections. What remains is the

current I on the boundary. Therefore, the current loop is equivalent to a magnetic

dipole layer, i.e., a layer of magnetic dipoles (in analogy to the electric dipole layer

of Sect. 2.5.3). The area density of the dipole moment is

(5.57)

i.e., a constant. This means that the dipole layer is uniform. In analogy to eq. (2.72)

we obtain therefore

(5.58)

where is the solid angle subtending the current loop as seen from the

observation point. By eq. (2.73), the potential has a discontinuity of when

passing through the electric dipole layer in the positive direction. Similarly, has

a discontinuity of I, when passing through a magnetic dipole layer in the positive

direction. We have seen this result previously in a different form (refer to the

electric magnetic

()

(since )

4πε

------------

Ω=

------=

-------

4πµ

-------------

Ω

4π

------

Ω==

------- Iµ

Fig. 5.21

-------

Ida

--------------- µ

I==

4πµ

-------------

Ω

4π

------

Ω==

Ω

ϕτε

⁄

5.2 Some Magnetic Fields 283

discussion on the scalar magnetic potential of Sect. 5.1. The separating surface

introduced then, now turns out to be a magnetic dipole layer).

Despite the perfect formal analogues, there is a major difference between

electric and magnetic dipole layers. Electric dipole layers are a physical reality. By

contrast, the magnetic dipole layer is a purely formal-fictitious construct and a

substitute for finite current loops. This is obvious from the fact that they may be

placed more or less arbitrarily since only the boundary matters (Fig. 5.22). The

boundary defines the field in a unique way because the solid angle does not

depend on the choice of the separating surface. Positive and negative solid angles

mutually cancel. Notice that has the same sign as .

Cylindrical dipole layers result when the currents only flow in z-direction

(this is the analogue of cylindrical electric dipole layers, see Sect. 2.5.4). Fig. 5.23

allows to determine the angle , which in analogy to eq. (2.85) gives

(5.59)

The same result follows from eq. (5.58) because for the situation of Fig. 5.23 we

have

(5.60)

The fields of cylindric dipole layers are plane, i.e., the fields are independent of z.

Let us calculate the field on the axis of a circular current loop (shown in

Fig. 5.24) as an example on how to apply eq. (5.58). The solid angle needed

here, can be taken from the example of Sect. 2.5.3.

Then

Fig. 5.22

possible separating surfaces

conductor loop

Ω

dΩ dΩθcos

2π

------

α=

Ω 4π

2π

------

2α==

Ω

Ω 2π

2πz

--------------------–=for 0z<

4π

------

Ω

---

2 r

------------------------–==