1

1

(Chapter 5) Magnetostatics5.1 The Lorentz Force Law5.1.1 Magnetic Fields

If charges are at rest, then the problem is one of electrostatics and we need only to consider electric fields.

But, if the charges are in motion, then we have a current and we must also consider magnetic fields.

(Recall “right hand rule” for MagneticFields generated by current-carryingconductors – see the figure.)

Consider a collection of charges.

2

5.1.2 Magnetic Forces

Consider a charge Q, moving with velocity in a magnetic field

The magnetic force is given by the Lorentz force law:

And the total force, if there is also an electric field present, is:

Note the cross product indicates that:

Magnetic Forces do no work.

vr :Br

0,2

1

=⋅=⋅=⇒⊥ ∫∫t

t

b

a

dtvFldFWBvF rrrrrrrr

r

( )BvQFmag

rrr×=

( )BvEQFrrrr

×+=

3

5.1.3 Currents

To measure the motion of charge we have to define a current.

A current is the charge per unit time that passes across a surface.

Current is defined such that a positive current is in the direction of motion of the positive charge.

Current has an associated direction.

is a vector quantity.

Consider a line charge λ, traveling at a velocity

If is a segment of a wire, then

Ir

⇒( )I

vr

vI rrλ=

ldr ( )∫ ×= BldIFmag

rr

∫ ×=∴= BldIFconstI mag

rr

4

λ represents a one dimensional line charge.

We can also have 2-D and 3-D charge distributions in motion.

We define a surface current density

And the Lorentz force law becomes

vdl

IdK rr

rσ=≡

⊥

( ) ( )∫ ∫ ×=×= daBKdaBvFmag

rrrrrσ

vI rrλ=

5

For a volume current, we define “volume current density” as:

Note that can be used to calculate

Then using the divergence theorem:

vda

IdJ rr

rρ==

⊥

∫ ∫ ×=×= ττρ dBJdBvFmag

rrrrr

Jr I

∫ ⋅=s

adJI rr

( ) τdJadJs v∫ ∫ ⋅∇=⋅

rrrr

( ) ∫∫ −=⋅∇vv

ddtddJ τρτ

rr

tJ

∂∂

−=⋅∇⇒ρrr

Outward flow decreases the charge left in the volume. 6

PROBLEM 5.4

Suppose that the magnetic field in some region has the form

xkzB ˆ=r

where k is a constant. Find the force on a square loop (side a), lying on the yz plane and centered at the origin, if it carries a current I, flowing counterclockwise, when you look down the x axis.

x-axis

a•

2

7

( )

( ) ( ) zIkaxydyaIkBldIF

xakByIdylId

a

a

ˆ2

ˆˆ2

ˆ2

;ˆ1

22

2

1 ∫ ∫ =×−=×=

⇒=⋅−=

−

rvr

rr

PROBLEM 5.4 IDENTIFYrelevant

concepts

Lorentz force law for a current-carrying element ( )∫ ×= BldIFmag

rr

x(2)

y

z

I (1)

(3)

(4)B

SET UPDraw a diagram:

Break square loop into four segments,as shown in the diagram.

Vector products (Eq. 1.12)

EXECUTE

8

PROBLEM 5.4 (cont.) EXECUTE (cont.)

( )

( ) ( ) yIkaxzzdzIkBldIF

xkzBzIdzlIda

a

ˆ2

ˆˆ

ˆ;ˆ222

2

2 ∫ ∫ =×=×=

⇒=⋅=

−

rvr

rr

( )

( ) ( ) zIkaxydyaIkBldIF

xakByIdylId

a

a

ˆ2

ˆˆ2

ˆ2

;ˆ3

22

2

3 ∫ ∫ =×−=×=

⇒−=⋅=

−

rvr

rr

( )

( ) ( ) yIkaxzzdzIkBldIF

xkzBzIdzlIda

a

ˆ2

ˆˆ

ˆ;ˆ422

2

4 ∫ ∫ −=×−=×=

⇒=⋅−=

−

rvr

rr

( ) ( ) ( ) ( ) ( )[ ] zIkayzkaIFFFFF ˆˆ)11(ˆ112

22

4321 =−++=+++=rrrrr

9

PROBLEM 5.5 A current I flows down a wire of radius a.

(a) If it is uniformly distributed over the surface, what is the surface current density K ?

(b) If it is distributed in such a way that the volume current is inversely proportional to the distance from the axis, what is J ?

⋅ a

dθ ⊥dl

I

J(a) (b)

10

aI

dldIK

π2==

⊥

θπθ addldIdI == ⊥,

2

rJlet

rJ α

=⇒∝1

∫ ∫ ∫∫ ==== addrrdrdr

JdaIa

παθαθα π

22

00

arIJ

aI

ππα

22=⇒=⇒

PROBLEM 5.5 IDENTIFYrelevant

concepts

⊥

=dldIKSurface current density

Relation between current andvolume current density ∫ ⋅=

s

adJI rr

SET UP & EXECUTE

(a)

Draw diagram (see previous slide)

(b)

11

5.2. The Biot-Savart Law

Where do magnetic fields come from? CurrentsIf the current does not change with time, we say this is a steady current

as opposed to a time-varying current.

Analogy: steady charge – electrostaticssteady currents – magnetostatics

The magnetic field produced by a steady current is given by the Biot-Savart Law.

is the permeability of free space

⇒

( ) ∫ ℜℜ×

= 'ˆ

4 20 dlIrB

rrr

πμ

∫ ℜℜ×

= 20

ˆ'4

ldIr

πμ

0μ 27 /104 AN−×= π 12

The extension to surface and volume integrals is straightforward:

Just like Coulomb’s law, superposition applies to magnetic fields too.

( ) ( )∫ ℜ

ℜ×=

s

darKrB 'ˆ'

4 20

rrrr

πμ

( ) ( ) )39.5('ˆ'

4 20 ∫ ℜ

ℜ×=

v

drJrB τπμ rr

rr

3

13

EXAMPLE 5.5 Find the magnetic field a distance s from a long straight wire carrying a steady current I .

P

s

l′I ldr′

ℜv

α

θ

•

14

EXAMPLE 5.5

IDENTIFYrelevant

principles

Biot-Savart law ( ) ∫ ℜℜ×′

= 20

ˆ

4ldIrBr

rr

πμ

Geometrical and trigonometric relations

SET UP & EXECUTE

P

s

l ′Ildr′

ℜv

α

θ

•θα cossinˆ ldldld ′=′=ℜ×′

r

θπα +=2

( ) θθ

θθ

θθθθθ dsdsldssl 22

22

coscossincos

cossintan =

−−=′⇒==′

2

2

2

cos1coss

s θθ =ℜ

⇒ℜ=

15

EXAMPLE 5.5SET UP & EXECUTE (cont.)

•

Iθ1

θ2

(b)

For the wire segment in Figure:

( )120

22

20

sinsin4

coscos

cos4

,ˆ2

1

θθπμ

θθθ

θπ

μ θ

θ

−=

⎟⎠⎞

⎜⎝⎛⎟⎟⎠

⎞⎜⎜⎝

⎛== ∫

sIB

dss

IBkBBr

( )[ ]sI

sI

Bπμ

πμ

πθπθ

211

4

2,

200

21

=−−=

⇒=−=Infinite wire:

4434421segmentwire

16

EXAMPLE 5.6 Find the magnetic field a distance z above the center of a circular loop of radius R, which carries a steady current I (see the figure).

ℜr

17

EXAMPLE 5.6

SET UP & EXECUTE

horizontalvertical BdBdBdrrr

+=

0=∫ncecircumfere

horizontalBdr

∫∫ ℜ′

== 20 cos

4)( θ

πμ ldIdBzB

ncecircumferevertical

( ) RIldIzBncecircumfere

πθπμθ

πμ

2cos4

cos4 2

02

0

ℜ=′

ℜ= ∫

( ) ( )322

20

22 2cos

zR

IRzB

zRRR

+=⇒

+=

ℜ=

μθ

ℜr

18

PROBLEM 5.9 Find the magnetic field at point P for each of steady current configurations shown in the figure below.

4

19

(a)

For the sides sides do not contribute.

From Example 5.6, the magnetic field a distance z above a circular loop is

Here z=0

Since we have a quarter segment only, So,

Let be +

Direction from the page.

( ) ∫∫×

=×

= 20

20 ˆ'

4'

ˆ4 r

rldIdlr

rIrBrr

rr

πμ

πμ

,0ˆˆ// =×⇒ rldrldrr

( ) ( ) 2/322

20

2 zRRIzB+

=μ

RIB

20μ=⇒

.8

0

RIB μ

=

•

⎟⎠⎞

⎜⎝⎛ −=⇒

baIB 11

80μ

PROBLEM 5.9

SET UP & EXECUTE

ba BBBrrr

+=

bI

aIB

8800 μμ

−=⇒

20

(b)

The two ½ lines are equivalent to a single infinite line

and the ½ circle gives

Direction is into the page,

.4

0

RIB μ

=

( ) ⎟⎠⎞

⎜⎝⎛ +=+=⇒

211

242000

πμμ

πμ

RI

RI

RIPB

.⊗

PROBLEM 5.9 (cont.) SET UP & EXECUTE (cont.)

RIB

πμ2

0=

21

BandBrrrr

×∇⋅∇

5.3 The Divergence and Curl of Magnetic Field Strength

Consider a wire with a steady current

We know that for infinite line

What is the integral for fixed s?

This result is actually general, the loop need not be circular and hence if we have multiple wires passing through the loop

This is the integral form of Ampere’s Law.

.I.

20

sIBπμ

=

ldBrr

⋅

IdlsIldB 0

0

2μ

πμ

==⋅∫ ∫rr

enclosedIldB 0μ∫ =⋅rr

22

Using Stokes Theorem we know

Furthermore,

Differential form of Ampere’s Law

Ampere’s Law is always valid. If a system has a certain symmetry it is useful for calculation.

∫ ∫ ⋅×∇=⋅P s

EqadBldB )57.1.(rrrrr

∫ ⋅=s

enclosed adJI rr

∫ ∫ ⋅=⋅×∇⇒ adJadB rrrrr0μ

JBrrr

0μ=×∇⇒

Closed loop area

23

We can now use the Biot-Savart Law for volume current (Eq. 5.39)

i.e. the magnetic field is a divergenceless field (no sources or sinks).

( ) ( )∫ ℜ

ℜ×′= '

ˆ

4 20 τπμ drJrB

rrrr

( ) ( )∫∫ ⎟⎟

⎠

⎞⎜⎜⎝

⎛

ℜℜ

×⋅∇=⋅∇⇒⎭⎬⎫

⎩⎨⎧

ℜℜ×′

=⋅∇ 'ˆ

4'

ˆ

4 20

20 τ

πμ

τπμ

dJBdrJrBrrrrrr

rrr

Recall vector identity No. 6 (inside the cover page):( ) ( ) ( )BAABBA

rrrrrrrrv×∇⋅−×∇⋅=×⋅∇

( ) ( ) ( )( )⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

=ℜℜ

×∇

≠′=′×∇

62.1.0ˆ

0

2 probsee

rfrJrJr

rrrrrv

to show that 0=⋅∇ Brr

( )

.00ˆ

ˆˆˆ

2

222

=⋅∇⇒=⎟⎟⎠

⎞⎜⎜⎝

⎛

ℜℜ

×⋅∇⇒

⎟⎟⎠

⎞⎜⎜⎝

⎛

ℜℜ

×∇⋅−×∇⋅ℜℜ

=⎟⎟⎠

⎞⎜⎜⎝

⎛

ℜℜ

×⋅∇⇒

BJ

JJJ

rrrr

rrrrrr

24

PROBLEM 5.13

A steady current I flows down a long cylindrical wire of radius a(see the fig.). Find the magnetic field, both inside and outside the wire, if

(a) The current is uniformly distributed over the outside surface of the wire.(b) The current is distributed in such a way that J is proportional to s, the

distance from the axis.

a

s

I

5

25

∫ =⋅ enclosedIldB 0μrr

( ) enclosedIsB 02 μπ =as

sI

BasB

>=

<=

φπμ ˆ2

00

r

r

⇒

PROBLEM 5.13IDENTIFYrelevant

concepts

Ampere’s law ∫ =⋅ enclosedIldB 0μrr

SET UP & EXECUTE

(a)

(b) sJ ∝ ksJ =⇒

( )dssksAdJIa

∫∫ ⋅=⋅=0

2πrr

a

s

ds

( )dssdA π2= φ φ̂

Br

26

PROBLEM 5.13 (cont.)EXECUTE (cont.)

( )

saIJand

aIk

akskdsskdssksIaaa

33

3

00

32

0

23,

23

32

3222

ππ

ππππ

==⇒

===⋅= ∫∫

( )

( ) 3

33

3

3

0 0

0

3232

322

2

asIs

aIsksdsskJdAI

IsBs s

enclosed

enclosed

=====

=

∫ ∫ ππππ

μπ

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

≥

<=⇒

⎪⎭

⎪⎬⎫

=≥

=<

asforsI

asforaIs

BIIas

asIIas

enclosed

enclosed

φπμ

φπ

μ

ˆ2

ˆ2

,

,0

3

20

3

3r

27

5.4 Magnetic Vector Potential5.4.1. The Vector Potential

Recall that if

Similarly, if

Since we can write magnetic field as

is known as the magnetic vector potential.

Just as V is defined to within any function whose curl is zero (That function is gradient of V),

we can always add to any function whose divergence is zero.

Hence we can always define so that

Magnetic vector potential could be calculated:

( )103.1.0 EqnVFF ∇−=⇔=×∇rrrr

( )104.1.0 EqnAFFrrrrr

×∇=⇔=⋅∇

0=⋅∇⇒ BBrrr

ABrrr

×∇=Ar

ArAr

.0=⋅∇ Arr

Scalar potential

Vector potential

( ) ''4

0 τπμ drJA ∫ ℜ

=rrr

28

PROBLEM 5.25

(a) By whatever means you can think of, find the vector potential a distance s (outside the wire) from an infinite straight wire carrying a current I.Check that

.0 BAandArrrrr

=×∇=⋅∇

(b) Find the magnetic potential inside the wire, if it has radius Rand the current is uniformly distributed.

29

PROBLEM 5.25aIDENTIFYrelevant

concepts( ) ( )zrAAzIrI ˆˆ rrrr

=⇒=

Ordinarily, magnetic vector potential is co-directional with current:

Magnetic vector potential is defined by:

ABrrr

×∇=

SET UP

zφs

I

EXECUTE

Draw diagram and set upcylindrical coordinate system (r → s)

( )( ) ( )φφ

ˆˆ1ˆˆ

ssAssA

sA

zsAzAA z

∂∂

−∂∂

=×∇⇒

==rr

r

(Curl in cylindrical coordinates) )81.1.( Eqsee

30

PROBLEM 5.25 a(cont.)EXECUTE (cont.)

( )

( ) azasIsA

sdsIsA

sI

sA

sI

sAAB

ˆln2

22

ˆ2

ˆ

0

00

0

⎟⎠⎞

⎜⎝⎛−=⇒

−=⇒−=∂∂

⇒

=∂∂

−=×∇=

∫

πμ

πμ

πμ

φπμφ

r

rrr

is the integration constant.

EVALUATE

( ) 0ˆˆ =∂

∂=

∂∂

=⋅∇ zzsAz

zAA z

rr

( ) BsI

ssA

sAA z

rrr==

∂∂

−=∂∂

−=×∇ φπμφφ ˆ2

ˆˆ 0

(Divergence in cylindrical coordinates)

However, A is function of s only, A(s):

Magnetic field of an infinite wire

)80.1.( Eqsee

6

31

RI

PROBLEM 5.25bIDENTIFY

relevant concepts

SET UP & EXECUTE

Ampere’s law enclIldB∫ =⋅ 0μrr

Magnetic vector potential has to be

continuous at s = R .

s

constJ =Current distributed uniformly:

( )s

IBIsBldB encl

encl πμ

μπ2

2 00 =⇒==⋅∫

rr

( ) φπ

μπ

ππ ˆ

2 20

2

22

22

RsI

BRsIs

RIsJIencl =⇒===

r

Also (see Problem 5.25a): φ̂dsdAB −=

r

32

PROBLEM 5.25b (cont.)SET UP & EXECUTE (cont.)

20

20

2ˆˆ

2 RsI

sA

sA

RsIB

πμφφ

πμ

−=∂∂

⇒∂∂

−==r

csRI

AsA z +−==⇒ 22

0

4)(

πμ

( ) ( ) cRIR

aRIRARA insideoutside +−=⎟⎠⎞

⎜⎝⎛−⇒= 2

200

4ln

2 πμ

πμ

a is arbitrary, so we adopt: .4

0

πμ I

cRa =⇒=⇒

[ ] ⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

≤−−

≥⎟⎠⎞

⎜⎝⎛−

=⇒RszRs

RI

RszRsI

Aˆ

4

ˆln2

222

0

0

πμπ

μr

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

33

5.4.3. Multipole Expansion of the Vector Potential

Just as we expanded V(r) in powers of 1/r for a given charge distribution, we can expand the vector potential in powers of 1/r for a given current distribution.

Monopole term

no monopole term in the expansion.

Dipole term

m = magnetic dipole moment

monopolesnoA ⇒=⋅∇ 0rr

⇒

( ) 20 ˆ

4 rrmrAdip

×=

rrr

πμ

aIadIm rrr=≡ ∫ is the “vector area” of the loop

(for flat area it is just area enclosed) ar