Chapter 4

Magnetostatics

4.1 Introduction

Characteristic difference between electric and magnetic phenomena：

Electric phenomena: Electric charges exist.

Magnetic phenomena : No magenetic charges exist. Typcal example：magnet

Two methods to analyze magnetic phenomena

A Consider a long magnet (or solenoid) and analyze the phenomena as

if there were magnetic charges at the ends of the magnet (

Coulomb ). Sometimes it can be intuitively effective. In practice one

must consider the case where the total “magentic charges” add up to

zero.

B Analyze the phenomena based on the observation that the magnetic

field is produced by electric currents. (Oersted 1819, Ampere ). The

basic empirical law which describes this process is called the Biot-

Savart’s law .

We will basically take the point of view of B. However, historically the theory

was initially developed along the line of A. So first we will briefly touch on

1

this scheme.

4.2 Static magnetic field of Coulomb and the

Lorentz force

Coulomb performed experiments with long magnets as

in the figure, measured the force between the mag-

netic charges, and established the magenetic version

of the Coulomb’s law.

~F = k′qmq′mR2

R

Just as in the case of the Coulomb force between electric

charges, we can think of this as the force on the magnetic

charge q′m by the magnetic field ~B produced by qm

NN

SS

qmq′m

~B(~r) = k′qm

R2R

~F = q′m ~B

More precisely, ~B is called the magenetic flux density. The “density” here

means the density per unit area.

2

2 Lorentz force :

In the 19th century, it was observed that the following force is exerted by a

magnet on an electric charge moving near by

~F = αq~v × ~B , α = a constant

Characteristic propery of this force:

• It depends on the velocity ~v in such a way that it acts only on a moving

charge. Moreover, it is perpendicular to the velocity so that it does

not do any work on the charge. Consequently, there is no concept of

“magnetostatic potential”.

Unit of ~B in MKSA system:

By taking α to be dimensionless, the formula above fixes the dimension of ~B.

Further, by setting α = 1, its unit is specified. The unit so defined is called

Tesla=Weber/m2. So we have the relation

~F = q~v × ~B

[N ] = [C][m

s

][T ]

Tesla = Weber/m2 =Newton · secCoulomb ·m = 104 Gauss

q qq Wb =N ·m · sec

C

In this unit, the strength of the magenetic field of the Earth is 0.3 Gauss

= 3 × 10−5T . Therefore 1 Tesla represents a very strong magnetic flux

density by our daily standards.

Together with the force by the electric field, the total Lorentz force on an

electric charge is given by

~F = q(

~E + ~v × ~B)

3

4.3 Force acting between currents and Biot-

Savart’s law

In 1819, Oersted found that a current produces a mag-

netic field around it, as in the figure. Afterwards,

Ampere constructed a theory of the force acting between

currents in a systematic way.

I

~B

4.3.1 Force between parallel currents

d

d~F

I1 I1

d`

By experiments, it was found that the following force, produced by one of

the currents, is felt by each small segment of the other current.

dF = km2I1I2

dd`

This is an important result, which is the counterpart of the Coulomb’s law

in electrostatics.

Note that the dependence on the distance d between the currents is not like

∼ 1/d2 but like 1/d. The reason for this dependence will become clear

later.

2 The unit and the magnitute of km :

4

Unit of current = Ampere = Coulomb/sec. Therefore from the above

relation we find

Unit of km = Newton/A2

It is now instructive to compare the force law above for the currents and the

Coulomb’s law for the charges.

dF = km2I1I2

dd` ∼ km

Q2

T 2

F = kq1q2

r2∼ k

Q2

L2

We immeidately notice that the ratio k/km has the dimension (L/T )2 =

(velocity)2. This is independent of the choice of the units of the charge and

the current.

Now what is the value of this “velocity” ?

In 1857, a German physicist Weber measured this velocity by indirect exper-

iments. The result he obtained was 300, 000 km/sec !

This was immediately recognized as the speed of light, which had been mea-

sured in a laboratory by

Foucault and Fizeau in 1850 to be 300, 000 km/sec.

Maxwell’s theory ⇒ precise relation k/km = c2

2 Expression of km in the MKSA system of units and its relation

to ε0 :

km is often written as

km =µ0

4π, µ0 = susceptibility of the vacuum

(4.1)

Combining with the aforementioned relations

5

k =1

4πε0

, k/km = c2

we find that ε0 and µ0 are related by

ε0µ0 =1

c2(4.2)

Since k = 10−7c2, the value of km is given by

value of km = 10−7

2 Interpretation of the force between the parallel currents from

the point of view of magnetic field :

We would like to understand the action of the force as a two-step process

current =⇒ magnetic field =⇒ Lorentz force on the other current

Consider a segment of a wire through which a current isflowing

force per unit vol. = ~f = ρ~v × ~B = ~× ~B

force on d` = d~F = ~fSd` = ~× ~BSd`

Now define ~I ≡ ~S = total charge flowing per unit time.Then

d~F = ~I × ~Bd`

S

~B

d`Vol.= Sd`

A better form is

d~F = Id~× ~B I ≡ |~I|

6

Here we have used the relation ~Idl = I ~dl。

4.3.2 Biot-Savart’s law

Let us figure out what the magnetic field ~B produced by a current along

a straight line should be in order to be able to interpret the force between

parallel wires as a Lorentz force.

d~F

I1 I1

d`

d~∝ ~v

~Bd~F

I1 I2

~B

As in the figure, the direction of ~B produced by the total current I1 is easy

to figure out.

To find its magnitude, compare the two expressions of the force namely,

d~F = I2d~× ~B , dF = km2I1I2

dd`

So we should have

| ~B| = km2 | I1 |

d

If we further refine and generalize this expression for the microscopic

magenetic field produced by an infinitesimal current element, we are led

to the Biot-Savart’s law.

7

2 Biot-Savart’s law for current element:

In 1820, Biot and Savart proposed that the in-

finitesimal magnetic field d ~B produced at the po-

sition ~R by an infinitesimal current elemeng Id~ is

given by the following 1/R2 law :

d ~B =µ0

4π

Id~× R

R2

(4.3) θ

d~

~R

d ~B ∝ d~× R

As one can easily check, the direction of the magnetic field is perpendicular to

the direction of the current, which is consistent with the case of the parallel

currents, it is not at all obvious that the result for the parallel currents can

be correctly reproduced form this more microscopic formula. The reason is

that in the case of the parallel currents, one is measuring the force produced

by the entire wire. Thus we must integrate over the current element in

the Biot-Savart’s formula to compute the relevant magnetic field ~B.

2 Integral form :

By integrating the differential form of the Biot-Savart’s law (4.3) along the

current, we obtain the expression

~B =

∮µ0

4π

Id~× R

R2(4.4)

Here∮

denotes the line-integral. This formula is quite useful for actual

computation of the magnetic flux density.

2 Derivation of the force beteen the paralell currents from Biot-

8

Savart’s law :

As a check of the validity of the Biot-Savart’s law,

let us show that the force between the parallel cur-

rents can be derived from this law. As we already

know that the direction is correct, we only need to

focus on the magnitude. Then the magnetic field

produced by d` is given by (see the figure)

dB =µ0

4π

Id` sin θ

R2(4.5)

d`

−dθθ`

θ + dθ

dR

R

Strategy： It is convenient to convert the integral over ` to the

integral over the angle θ. Note that in this conversion as ` increases θ

decreases.

From the figure (neglecting the second order infinitesimal quantities) we have

the relation

d` sin θ = −R dθ (4.6)

As we already remarked, θ decreases as ` increases and hence dθ is negative.

Substituting (4.6) into (4.5) and using R sin θ = d, we get

dB =µ0

4π

I(−Rdθ)

R2= −µ0

4π

Idθ

R= −µ0

4π

I sin θdθ

d

As the integral over ` is from the top to the bottom in the figure, the corre-

sponding integral over θ is from π to 0. Therefore

B =µ0

4π

I

d

∫ 0

π

(− sin θ)dθ =µ0

4π

I

d

∫ π

0

sin θdθ

=µ0

4π

2I

d

(←

∫ π

0

sin θdθ = 2

)

This is precisely the correct magnitude for the magnetic field in order to

explain the force between the parallel currents.

2 General form of the Biot-Savart’s law :

9

To get the Biot-Savart’s formula for a general distribution of electric current

density, all we have to do is to make the following replacement in the integral

form (4.4):

Id~ = ~Id` −→ d`

∫

sectiond2x′~(~x′)

Then we get

~B(~x) =µ0

4π

∫d3x′

~(~x′)× (~x− ~x′)

|~x− ~x′|3

The case of the current flowing in a wire can be considered as a special case

where ~(~x) is non-zero only along the wire.

This formula is the counterpart of the Coulomb’s formula for the electric

field

~E(~x) =1

4πε0

∫d3x′

ρ(~x′)(~x− ~x′)

|~x− ~x′|3but their forms are rather different. We will comment on this further in the

next section.

Exercise 4.1 A circular ring of radius a is placed on the x-y plane cen-

tered at the origin and a steady current of strength I is flowing through it.

Find the magnetic field at the point P = (0, 0, z) right above the origin.

P = (0, 0, z)

radius a

y

x

z

I

10

4.4 From the Biot-Savart’s law to the Maxwell(Ampere)

law

As we saw, the Biot-Savart’s formula computes the magnetic field directly

from the current distribution and corresponds to the Coulomb’s law in elec-

trostatics. Below we will derive the Maxwell equation (or Ampere’s law)

which is more fundamental.

4.4.1 Derivation of ~∇· ~B = 0, expressing the absence ofmagnetic charges

Let us recall the Biot-Savart’s law:

~B(~x) =µ0

4π

∫d3x′

~(~x′)× (~x− ~x′)

|~x− ~x′|3 (4.7)

The magnetic field computed by this formula is produced from the electric

current, not from the magnetic charges. Therefore one expects that ~∇· ~B = 0,

which expresses the absence of magnetic charges. To check this, recall the

important identity which was frequently used in the electrostatics:

~∇ 1

|~x− ~x′| = − ~x− ~x′

|~x− ~x′|3 (4.8)

Using this formula, Biot-Savart’s law can be rewritten as

~B(~x) =µ0

4π

∫d3x′ ~(~x′)×

(−~∇ 1

|~x− ~x′|

)

= −µ0

4π

∫d3x′ ~(~x′)× ~∇ 1

|~x− ~x′| (4.9)

Now take the divergence of both sides:

~∇ · ~B(~x) = −µ0

4π

∫d3x′ ~∇ ·

(~(~x′)× ~∇ 1

|~x− ~x′|

)(4.10)

The integrand is of the form ~∇ · (~C(~x′) × ~D(~x)). Just honestly use the

definitions of the exterior product and the divergence, and noting that ~C(~x′)

11

vanishes upon acted by ~∇, we get the formula

~∇ · (~C × ~D)

=∂

∂x1

(C2D3 − C3D2) +∂

∂x2

(C3D1 − C1D3) +∂

∂x3

(C1D2 − C2D1)

=∂D3

∂x1

C2 − ∂D2

∂x1

C3 +∂D1

∂x2

C3 − ∂D3

∂x2

C1 +∂D2

∂x3

C1 − ∂D1

∂x3

C2

= −(

∂D3

∂x2

− ∂D2

∂x3

)C1 + · · ·

= − ~C · (~∇× ~D) (4.11)

Applying this to (4.10), we get

~∇ ·(

~(~x′)× ~∇ 1

|~x− ~x′|

)= −~(~x′) ·

(~∇× ~∇ 1

|~x− ~x′|

)(4.12)

The RHS vanishes due to a simple identity, to be proven in the exercise 4.1

below.

Exercise 4.2 Prove for an arbitrary function f(~x) the identity

~∇× ~∇f(~x) = 0 (4.13)

In this way we have obtained the Maxwell’s equation expressing the ab-

sence of magnetic charges:

~∇ · ~B = 0 (4.14)

4.4.2 Derivation of the Ampere’s law ~∇× ~B = µ0~

Next we will derive the Ampere’s law for the production of the magnetic

field by a current. For this purpose, take the rotation of the Biot-Savart’s

law (4.9). We obtain

~∇× ~B(~x) = −µ0

4π

∫d3x′~∇×

(~(~x′)× ~∇ 1

|~x− ~x′|

)(4.15)

12

The integrand is of the form ~∇× (~C × ~D).

Exercise 4.3 Prove the following identity:

~∇× (~C × ~D) = (~∇ · ~D)~C − (~∇ · ~C) ~D (4.16)

In the present case, ~C stands for ~(~x′) and vanishes upon differentiating with

respect to xi. Therefore the formula above simplifies to ~∇ × (~C × ~D) =

(~∇ · ~D)~C. Applying this, we get

~∇× ~B(~x) = −µ0

4π

∫d3x′~(~x′)∇2 1

|~x− ~x′|

= −µ0

4π∇2

∫d3x′

~(~x′)

|~x− ~x′| (4.17)

Now recall the Poisson equation which the electrostatic potential φ(~x) satis-

fies:

φ(~x) =1

4πε0

∫d3x′

ρ(~x′)

|~x− ~x′| (4.18)

−∇2φ(~x) = − 1

4πε0

∇2

∫d3x′

ρ(~x′)

|~x− ~x′| =ρ(~x)

ε0

(4.19)

The expression (4.17) is obtained from this by replacing ρ with ~ and 1/ε0

with µ0. Thus ~B must satisfy the following Ampere’s law

~∇× ~B = µ0~ (4.20)

4.5 Integral form of the law of magnetostatics

and its application

When we look back at the Maxwell equation for electrostatic field, we find

that the integral form, namely the Gauss’s law, was easier to understand

13

physically and very useful in practical calculation (with high enough symme-

try) compared with the more fundamental differential form. In view of this

fact, it is natural to ask what corresponds to the Gauss’ law in the case of

magnetostatics. To answer this question, we shall first introduce and prove

the Stokes theorem, which is the counterpart of the Gauss’ theorem in one

less dimensions.

4.5.1 Stokes’ theorem

The Gauss’ theorem relates the area integral of a vector

field to a volume integral of its divergence. Likewise,

the Stokes’ theorem relates the line integral of a vector

field to an area integral of its rotation. For an arbitrary

vector field ~B, it takes the following form:

∫

∂Σ

~B · d~ =

∫

Σ

(~∇× ~B) · ndSΣ

~B

n

d~

Proof：

The proof is very similar to that of the Gauss’ theorem.

First consider the case where the path is around a small rectangle as in the

figure and we will try to rewrite∫

C~B · d~ into an area integral. Decompose

the integral into those along four edges:

∫

C

~B · d~ =

∫

Cx

+

∫

C′x

+

∫

Cy

+

∫

C′y

The ones along the x direction are

∫

Cx

~B · d~ =

∫ x+∆x

x

Bx(x′, y)dx′

∫

C′x

~B · d~ = −∫ x+∆x

x

Bx(x′, y + ∆y)dx′

∆y

∆x(x, y)

y

x

C ′x

Cx

CyC ′y

14

（The minus sign is due to the different directions for d~. ）Adding up, we

get∫

Cx

+

∫

C′x

= −∫ x+∆x

x

[Bx(x′, y + ∆y)−Bx(x

′, y)] dx′

= −∫ x+∆x

x

∆y∂Bx

∂y(x′, y)dx′ +O((∆y)2∆x)

= −∂Bx

∂y(x, y)∆S

Here ∆S = ∆x∆y

Similarly, the conributions from the integrals in the y-direction are∫

Cy

+

∫

C′y

=∂By

∂x(x, y)∆S

Altoghether,∫

C

~B · d~ =

(∂By

∂x− ∂Bx

∂y

)∆S

≡ (~∇× ~B)z∆S = (~∇× ~B) · n∆S

In the last equality we have used the universal expression independent of the

coordinatization. So this proves the theorem for a small rectangle.

For a general finite C, we decompose it into a collection

of small rectangles. In this decomposition, the surface

Σ can be anything as long as its boundary gives the

closed path C. Except for the outer boudary, all the

line integrals cancel inside, whereas the area integrals

add up. In this way we obtain the Stokes’ theorem for

general C.

4.5.2 Integral form of the law of magnetostatics

Just as in the case of the static electric field, we can obtain the integral form

of the Maxwell’s equation for the static magnetic field, using the Gauss’ and

Stokes’ theorems.

15

First recall the theorems of Gauss and Stokes:

Gauss’ theorem

∫

V

d3x ~∇ · ~f =

∫

∂V

dS n · ~f

Stokes’ theorem

∫

Σ

dSn · (~∇× ~f) =

∫

∂Σ

d~ · ~f

(i) Applying Gauss’ theorem to the differntial law ~∇ · ~B = 0, we get

total magnetic flux =

∫

∂V

dSn · ~B = 0

So the total magnetic flux going out of a closed region is zero since there are

no magnetic charges as the source.

(ii) Applying the Stokes’ theorem to the Ampere’s law ~∇× ~B = µ0~, we get

∫

∂Σ

~B · d~ =

∫

Σ

dS n(~∇× ~B) = µ0

∫

Σ

dSn · ~

The rightmost integral expresses the total current flowing out through the

surface Σ. This formula, relating the integral of the magnetic field around a

closed path and the total current flowing through the surface surrounded by

the path, is called the integral form of the Ampere’s law.

As we shall see later, this integral law will be, just like the Gauss’ law for

the electrostatics, a very powerful means to compute the magnetic field itself

when the configuration has suitable symmetry.

16

~j

Σ

∂Σ

• Use of Gauss’ law： Surround the charges by a closed surface• Use of Ampere’s law : Surround the currents by a closed loop

4.5.3 Applications of the integral form of the Ampere’slaw

When a magnetic field is produced by a current flowing through a conducting

wire, the integral form of the Ampere’s law takes a simpler basic form

∮

C

~B · d~ = µ0IC (4.21)

Here, IC = magnitude of the normal component of the current

penetrating the surface surrounded by C

2 Magnetic field produced by an infinitely long straight wire :

As the simplest application of the basic form above, let us compute the mag-

netic field produced by an infinitely long straight wire, which was calculated

previously by the direct use of the Biot-Savart’s law.

17

We already know from symmetry consideration that the

magnetic field ~B(~x) is produced along the circle sur-

rounding the current, with the magnitude B(r) depend-

ing only on r. Applying the basic form of the Ampere’s

law, we immediately get

∮~B · d~ = 2πrB(r) = µ0I

q qq B(r) =µ0

2π

I

r=

µ0

4π

2I

r(4.22)

This coincides with the previous result if we identify r =

d. Note that this computation is astonishingly easier

than the previous calculation.

r

I

2 Magnetic field produced by an infinitely long solenoid :

C1 C2 C3

z

First, we will simplify the calculation by making preliminary observations

mostly based on the symmetry of the configuration:

(i) Since the solenoid is infinitely long, ~B and ~ must be uniform along the

z direction. In other words, they are independent of z. In particular, for

all the components, we have ∂z~B = 0. (Here ∂z ≡ ∂/∂z)

(ii) Consider a circle of radius a in the x-y plane with the center on the

z-axis. Along this circle, we decompose the magnetic field into the tangential

component ~Bt and the normal component ~Bn, namely ~B = ~Bt + ~Bn. From

the obvious axial symmetry, the magnitude of ~Bt and ~Bn are constant on the

18

circle. If we integrate them around the circle, we get

∫~Bn · d~ = 0

and ∫~Bt · d~ = 2πa| ~Bt|

But since there is no current penetrating through this circle, they must vanish

according to the Ampere’s law. Therefore ~Bt = 0 and possibily we can only

have radial component.

Now from (i) we know that Bx, By are independent of z. Therefore they must

be of the form

Bx = cx

rf(r) , By = c

y

rf(r) (4.23)

where c is a constant and f(r) is some function.

(iii) This is obviously a configuration where the magnetic field emanates

from the center. But this must be impossible since there are no magnetic

charges. In fact, computing the magnetic flux going out of the cylinder of

radius a and length l. we get

∫dS ~B · n = l2πacf(a) (4.24)

which should vanish according the Ampere’s law. Therefore we can only have

c = 0 and we conclude Bx = By = 0.

From the above information we learn that the only nonvanishing component

can be Bz = Bz(x, y).

We are now ready to compute Bz using the Ampere’s law.

19

Consider the three closed loops C1, C2, C3 shown in the

figure in the previous page and apply the basic formula

(4.21). Since C1 and C3 do not encircle any currents (

see the figure below) we have

0 =

∮~B · d~

=

∫dzBz(x, y)−

∫dzBz(x

′, y′)

q qq Bz(x, y) = Bz(x′, y′) = const.

This shows that the magnetic field must be constant

inside and outside. But their values can of course be

different.

~B

(x, y) (x′, y′)

To see how they can differ, we now apply the basic law to C2. Then,

if we denote by n = the number of currents per unit length, then

from the figure we get

`Binz − `Bout

z = `nµ0I

q qq Binz −Bout

z = nµ0I

``n currents

Since the magnetic field must vanish at ∞, it vanishes outside throughout.

Combining we get the well-known result

Boutz = 0

Binz = µ0nI

Exercise 4.4 In the consideration above, we have ignored the thick-

ness of the wire forming the solenoid. As a consequence, the magnetic field

jumped discontinuously as we go from outside to inside. Now what hap-

pens if we take into account the thickness of the wire ? Again make use of

20

the Ampere’s law, as hinted in the figure. (The result will be utilized in the

computation of the pressure exerted on the solenoid later. )

d

a

r`

21