Download Lecture Notes 13: Steady Electric Currents, Magnetic Field, B

UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 13
Prof. Steven Errede
LECTURE NOTES 13
ELECTRIC CURRENTS
Before launching into a full-blown discussion of magnetic phenomena, we first want to discuss
electric currents…
The free electric current I free at a given point in space (call this point r , defined relative to a
local origin of coordinates) is defined as the time-rate of change of free charge Qfree at that point
in space.
dQ free
I free =
or more explicitly:
dt
dQ free ( r , t )
I free ( r , t ) =
= instantaneous electric current at point r at time t
dt
The S.I. unit of current I is the Ampere, in honor of André Marie Ampere, for his 1820 work on
understanding the nature of electric currents.
1 Ampere ≡ 1 Coulomb of charge per second (crossing/passing through an imaginary surface)
An electric current implies that electric charge is in motion, relative to an observer whose rest
frame is in the fixed coordinate system with ϑ at its origin:
ẑ
Q
r
Moving charge Q relative to origin, ϑ
ŷ
ϑ
x̂
n.b. ∃ no absolute reference frame anywhere in the universe – we can only speak of relative
motions between objects.
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
1
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 13
Prof. Steven Errede
Line Currents:
Consider an infinitely long straight filamentary (i.e. infinitesimally thin) wire. Imagine that
this wire is electrically charged with λ free Coulombs/meter of free charge per unit length, as
shown below in the figure:
Free charge
λ free Coul/m
dx
x̂
If the charge is static, i.e. not moving, then there is no electric current I flowing in the wire.
However, if a potential difference ΔV is imposed across the ends of the wire, then a line current
will begin to flow by an amount:
I free = λ free v
where v = (relative) speed of charge moving down the wire, i.e. relative to the wire itself.
But speed is just the magnitude of the velocity, so actually, this relation is a vector relation:
I free ( r , t ) = λ free v ( r , t )
(assuming λ free = constant, i.e. λ free ≠ λ free ( r , t ) )
Now, an infinitesimal segment dx = vdt instantaneously carries charge dq = λ free dx = λ free vdt
past an observation point P in the time interval dt:
I free =
dq free
dt
=
λ free dx
dt
= λ free v where v =
Vectorizing this (assuming λ free = constant): I =
λ dx
dt
dx
dt
= λ v or more generally:
dr ( t )
= λ free v ( r , t ) for line currents
dt
n.b. in principle λ free = λ free ( r , t ) also!
I free ( r , t ) = λ free
In a (circular) particle accelerator, e.g. a cyclotron, charged particles (e.g. protons) circulate in
an evacuated ring of radius R at speeds nearly that of the speed of light i.e. v p < c .
~
Suppose R = 1 meter, and also suppose that we have I = 1 Ampere of (proton) current circulating
in the cyclotron, and assume further that the protons are uniformly distributed around the
circumference of the machine. If we additionally assume (for simplicity’s sake) that
v p = c = 3 × 108 m / sec then:
I 1 Ampere
=
but: 1 Ampere = 1 Coulomb / sec
c 3 ×108 m / s
10−8 Coulombs sec
=
= 0.333 ×10−8 Coulombs
i
meter
meters
3
sec
I p = λ free c or: λ free =
∴ λ free
2
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 13
Prof. Steven Errede
The circumference of the proton accelerator is C = 2π R = total length of line charge
Coulombs
2π
× 2 × π × 1 meter =
× 10−8 Coulombs
3
meter
If there are N p = total # protons circulating in the cyclotron accelerator, then:
∴ QTOT
= λ freeC = 0.333 ×10−8
p
Np =
QTOT
p
qp
2π
×10−8 Coulombs
3
=
1.602 ×10−19 Coulombs
1.3 × 1011 protons circulating in the cyclotron
Thus, Ip = 1 Ampere of proton current circulating in this cyclotron corresponds to N p
1.3 × 1011
protons circulating in the cyclotron (if they are all traveling at the speed of light, c).
Surface Currents:
Imagine an infinitely long, straight, hollow conducting cylindrical tube of radius R with
infinitesimally thin walls (of thickness δ R ) carrying σ free Coulombs/meter2 of (initially)
stationary electric charge. A potential difference ΔV is placed across the length of this hollow
tube, causing an electric surface current to flow, of magnitude I Amperes.
An infinitesimal section of this hollow tube dz = vdt instantaneously carries charge
dq free = σ free 2π Rdz = σ free dA past an observation point P (on the tube somewhere) in the time
interval dt. Then:
I≡
dq free
=
σ free 2π Rdz
= 2π Rσ
dt
dt
∴ I = ( 2π R ) σ free v = Cσ free v
dz
dt
v=
but
dz
dt
In principle, σ free = σ free ( r , t )
Vectorizing this, assuming σ free = constant (i.e. σ free ≠ σ free ( r , t ) ) then:
I ( r , t ) = Cσ free v ( r , t ) = ( 2π R ) σ free v ( r , t ) for a hollow conducting tube of radius R.
However, note that we can also define a surface current density K ( r , t ) as:
K (r ,t ) ≡
(
)
I (r,t ) I (r,t )
Amperes
=
= σ free v ( r , t )
meter
C
2π R
In principle, σ free = σ free ( r , t )
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
3
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 13
Prof. Steven Errede
Instead of a surface current flowing on a long, hollow conducting tube of radius R, suppose
we had a surface current flowing on a flat (i.e. planar) conductor of width W. This is simply
equivalent to e.g. cutting the long hollow conducting tube (with infinitesimally thin walls of
thickness δ R ) and unrolling it out into a flat plane. Then the width W of the flat sheet =
circumference C of the original tube, i.e. W = C = 2πR and thus:
dz
dq free = σ free dA = σ freeWdz
σ free
δR
dq free = σ free dA
v
= σ freeWdz
W
dA = Wdz = Wvdt
(dz = vdt)
ẑ
v=
I≡
dq free
dt
=
σ freeWdz
dt
= σ freeW
dz
dt
dz
= σ freeWv
dt
Vectorizing this (assuming σ = constant, i.e. σ free ≠ σ free ( r , t ) ), the current flowing through the
flat sheet is:
I ( r , t ) = σ freeW v ( r , t ) for a planar conducting sheet of width W
n.b. in principle σ free ≠ σ free ( r , t )
Here again, we can also define a lineal surface current density K ( r , t ) as:
K (r ,t ) ≡
I (r,t )
= σ free v ( r , t ) Amperes / meter ⇐ n.b. not Amperes / m2!!
W
This can also be written in differential form as:
dI ( r , t )
K (r ,t ) ≡
= σ free v ( r , t ) Amperes / meter
dW
n.b. in principle σ free ≠ σ free ( r , t )
To explicitly tie in with Griffiths book: dW = d
Then:
4
W
I ( r , t ) = ∫ K ( r , t ) dW = ∫
0
0
⊥
K (r ,t ) d
⊥
⊥
or W =
⊥
(see p. 211-212).
and K ( r , t ) ≡
dI ( r , t ) dI ( r , t )
=
dW
d ⊥
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 13
Prof. Steven Errede
Volume Currents:
Consider an infinitely long conducting circular rod of radius R of cross-sectional area
A⊥ = π R 2 . For simplicity, let us assume that the volume free charge density ρ free ( r , t ) =
constant , i.e. ρ free ( r , t ) = ρ ofree ( Coulombs m3 ) = uniform volume free charge density initially
charging this conductivity rod. (n.b. this is impossible for static electric charge, but is not
impossible when the electric charge is moving, en-mass!)
We again place a potential difference ΔV across the ends of the rod, and a volume current I
flows in the rod.
dq
ρ dV ρ free A⊥ dz
dz
dz
I ≡ free = free
=
= ρ free A⊥
but: v =
dt
dt
dt
dt
dt
∴ I = ρ free AV = ρ 0 AV (here)
free
Vectorizing this: I ( r , t ) = ρ free A⊥ v ( r , t ) for a conducting rod of cross-sectional area A⊥
If ρ free = ρ free ( r , t ) then more generally: I ( r , t ) = ρ free ( r , t ) A⊥ v ( r , t )
Here again, we can define an areal current density J ( r , t ) as:
J (r ,t ) ≡
I (r,t )
= ρ free ( r , t ) v ( r , t ) Amperes/m2
A⊥
We can also define this in differential form as:
dI ( r , t )
J (r ,t ) ≡
dA⊥
Then:
I ( r , t ) = ∫ J ( r , t )idA⊥
where:
S
A⊥ = cross-sectional
area of conductor
dA⊥ is an infinitesimal crosssectional area element of A⊥
ˆ ⊥
dA⊥ = ndA
ˆ ⊥ → dA = ndA
ˆ
Since we’re taking a dot product, we can drop the “ ⊥ ” subscript on dA⊥ = ndA
(Always remember / keep in mind that A (here) is the cross-sectional area of the conductor!!)
Thus:
I ( r , t ) = ∫ J ( r , t )idA
S
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
5
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 13
Prof. Steven Errede
From electric charge conservation (i.e. the empirical fact that electric charge can neither be
created, nor destroyed), the total charge per unit time leaving a volume V is:
∫ J ( r , t )idA = ∫ ( ∇i J ( r , t ) ) dτ
S
V
(by the divergence theorem)
n.b. Closed surface integral!
Because electric charge is conserved, whatever electric charge flows out of / flows into the
surface S must come from / go into the volume V respectively, i.e.:
⎛ ∂ρ free ( r , t ) ⎞
d
∫V ∇i J ( r , t ) dτ = − dt ∫V ρ free ( r , t ) dτ = − ∫V ⎜⎝ ∂t ⎟⎠ dτ
e.g. a current flowing out through surface S → decrease in the charge density in volume V
(
)
Since this holds for any volume (arbitrary), then integrands on LHS = RHS:
Electric Charge Conservation:
∇i J ( r , t ) = −
∂ρ free ( r , t )
∂t
⇐ Continuity Equation
“Dictionary” for point, line, surface and volume currents:
n
∑(∼) q v
∫ ( ∼ ) Idl
i i
i =1
∫
line
Correspondence to:
surface
dq ~ λ dl
( ∼ ) KdA
∫
volume
dq ~ σ dA
( ∼ ) Jdτ
dq ~ ρ dτ
Griffiths Example 5.4
a.) A current I is uniformly distributed over a wire of radius R and circular cross-section
A = π R 2 . Find the volume current density J .
Since I is uniformly distributed over cross-sectional area A of wire → J must also be uniform:
I
I
J= =
A π R2
b.) If J ( r ) = krzˆ where r = radial distance from cylindrical symmetry axis and k = constant
Thus J is not uniform/constant here. Compute I from: I = ∫ J ( r )idA where dA = rdϕ dr zˆ
S
End-View of Conductor:
2π
I = ∫ J ( r )idA = ∫ dϕ ∫
0
S
= 2π k ∫
R
0
R
0
2π
R
( krzˆ )i( rdrzˆ ) = ∫0 dϕ ∫0
kr 2 dr
( zˆi zˆ = 1)
rdϕ
r=R
2π 3
2π 3
r dr =
kr
=
kR
3
3
r =0
2
dr
r
ẑ out R
of page
6
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 13
Prof. Steven Errede
MAGNETISM AND THE MAGNETIC FIELD B
The Lorentz Force on a Charged Particle, E and B -Fields:
When a charged particle is moving in an external magnetic field B , ∃ (there exists) a magnetic
force acting on the particle Fm ( r ) which (in MKSA (SI) units) is:
Fm ( r ) = q v ( r ) × B ( r ) where:
q = electric charge of the particle
v ( r ) = laboratory velocity of charged particle at r
B ( r ) = magnetic field intensity at r
If the charged particle is also in an external E -field, then ∃ an electric force acting on the particle
Fe ( r ) which is:
Fe ( r ) = qE ( r )
Thus the net force acting on a moving charged particle simultaneously in both an E and B field
is (by the principle of linear superposition):
FTOT ( r ) = Fe ( r ) + Fm ( r ) = qE ( r ) + qv ( r ) × B ( r ) ⇐ Lorentz Force
Note that Fe ( r ) is along E ( r ) (i.e. Fe ( r ) E ( r ) ) while Fm ( r ) is ⊥ to v ( r ) and also B ( r ) !
→ Magnetic forces do no work, because Wm = ∫ Fm ( r )id = 0 , since Fm ( r ) is ⊥ to d !
C
B into paper:
+Q
F+
B
Fm+
v
v
Fm−
F−
−Q
Cross-product v × B of Fm = q v × B is defined by the right-hand rule:
Curl fingers of your right hand for the cross product,
right hand thumb points in the direction of F / q = v × B
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
7
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 13
Prof. Steven Errede
Note that the Lorentz Force FTOT = Fe + Fm = qE + qv × B is valid for all velocities 0 ≤ v ≤ c
- even for fully relativistic particles!
Now FTOT = qE + qv × B is the total force acting on a charged particle in the Lab frame, which
contains both an E -field and a B -field. Note that if v → 0, then Fm → 0 also.
What does this look like to a charged particle in its own rest frame?
If we make a (Galilean) transformation into the rest frame of a moving charged particle, then
there is only an electric field E ′ seen by the particle!
E′ = E + v × B
The force on the charged particle in the charged particle’s own rest frame is:
′ = qE ′ = qE + qv × B = FTOT
FTOT
⇒ Net force acting on the particle in the charged particle’s own rest frame
≡ Net force acting on the particle in the lab frame!
Note that this expression for E ′ is true only for non-relativistic velocities, i.e. v c ≡ β
i.e. v << c = 3 × 108 m/sec
1
Thus, suppose a charged particle is moving in a uniform B -field (No E -field is present in lab)
′ = qE ′ = qv × B thus E ′ = v × B here!
Then: FTOT = qv × B and FTOT
⇒ A charged particle moving in magnetic field B (only) in the lab frame “sees” an electric field
E ′ = v × B in its own rest frame!!
F ′, E ′
F ′, E ′
v
or:
v
q
q
B
B into paper
Conversely, an electrically charged particle moving in the lab with velocity v generates a
(solenoidal) magnetic field B in the lab frame! (n.b. lines of B defined by right-hand rule!):
B = B ( r ) ϕˆ
B out of paper
v , zˆ out of paper
v zˆ
q
or:
B into paper
8
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 13
Prof. Steven Errede
The macroscopic magnetic field B generated by a moving charged particle is a solenoidal
magnetic field, i.e. B = B ( r ) ϕˆ !
For v << c (i.e. β = v/c << 1) the B -field at a point r away from a moving charged particle is:
1
B (r ) = 2 v × E (r )
c
where E (here) is the electrostatic field of charged particle as observed in its own rest frame, i.e.
1 q
E (r ) =
rˆ
4πε o r 2
(
Then: B ( r ) =
Now: c =
⎛ rˆ ⎞
v ×⎜ 2 ⎟
4πε o c
⎝r ⎠
q
2
1
ε o μo
c2 =
∴
but:
)
rˆ
r
r
= 3 = 3 since r = r rˆ = rrˆ
2
r
r
r
1
ε o μo
where:
ε o = electric permittivity of free space = 8.85 ×10−12 Farads/m,
μo = magnetic permeability of free space ≡ 4π ×10−7 N/Ampere2 (= Henrys/m)
∴ B (r ) =
q ε o μo ⎛
r ⎞ ⎛ μo ⎞ ⎛
rˆ ⎞
r ⎞
⎛ μo ⎞ ⎛
⎜ v × 3 ⎟ or: B ( r ) = ⎜ 4π ⎟ ⎜ qv × r 3 ⎟ = ⎜ 4π ⎟ ⎜ qv × r 2 ⎟
r ⎠
4π ε o ⎝
⎠ ⎝
⎠
⎝
⎠⎝
⎠⎝
This is the (macroscopic) magnetic field observed in the lab frame due to a charged particle
moving with velocity v with v << c.
Note one similarity between the macroscopic E and B -fields of a point electrically charged
particle:
⎛ 1 ⎞ ⎛ qrˆ ⎞
E (r ) = ⎜
⎟⎜ 2 ⎟
⎝ 4πε o ⎠ ⎝ r ⎠
Both fields decrease as 1/r2 from point charge!
⎛ μ ⎞⎛ qv × rˆ ⎞
B (r ) = ⎜ o ⎟⎜ 2 ⎟
⎝ 4π ⎠ ⎝ r ⎠
For a point charged particle moving with velocity v (v << c) in the lab:
⎛ μ ⎞⎛ qv × rˆ ⎞
B (r ) = ⎜ o ⎟⎜ 2 ⎟
⎝ 4π ⎠ ⎝ r ⎠
q
B
×
r
v
P ( r ) = observation point
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
9
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 13
Prof. Steven Errede
We emphasize that the macroscopic B -field seen in the lab frame is generated by the motion of
the electrically charged particle through space-time. Note that the B -field appears only in the
lab frame. In the rest frame of the charged particle, the only field it sees is its own electric field!
The phenomenon of magnetism and magnetic fields is not solely a property of moving
electrically charged particles/is not solely a property of electromagnetism!
Moving strong, weak and/or gravitational charges will (respectively) also produce strong, weak
and/or gravitational magnetic fields too!!!
Furthermore, c, the speed of “light” c = 1
ε o μo is not “just” the maximum allowable/maximum
possible speed for the electromagnetic interaction, but it is also the maximum possible speed for
any/all of the four known fundamental forces of nature, i.e. the E & M, strong, weak and
gravitational forces. Thus “c” = speed of “light” is actually a misnomer, because it applies to
any/all of the fundamental forces of nature! c is the maximum speed that anything may travel in
this universe – thus the true physics origin(s) of c have nothing to do with electromagnetism per
se, but everything to do with the very structure of space-time, i.e. the vacuum (“empty” spacetime) itself !!! (n.b. However, the microscopic vacuum is not “empty”!!)
Since c, the speed of “light” is related to the macroscopic parameters ε o and μo associated with
the E&M aspects of the vacuum c = 1
ε o μo and c is the same for any/all of the fundamental
forces of nature, i.e. ce = cs = cw = cg = c = 3 × 108 m / sec , there must also be analogous
macroscopic quantities to that of ε o and μo for the macroscopic “electric” and “magnetic”
properties of the vacuum associated with each of the other fundamental forces of nature, i.e.
⎧
⎫ ⎧
⎫
⎫
⎧
⎫ ⎪
⎪ ⎪
⎪ ⎧⎪
⎪
⎪⎪
⎪
1 ⎪ ⎪
1 ⎪ ⎪
1 ⎪ ⎪
1 ⎪
c ≡ ⎨ce =
⎬ ≡ ⎨cs =
⎬ ≡ ⎨cw =
⎬ ≡ ⎨cg =
⎬
ε o μo ⎪ ⎪
ε s μs ⎪ ⎪
ε w μw ⎪ ⎪
ε g μg ⎪
⎪
⎪⎩
⎪
⎪ ⎪⎩
E &M ⎪
strong ⎪
weak
⎭ ⎪
gravity ⎪
⎭
force ⎭
force
⎩
⎩
⎭
Note that: ε o ≠ ε s ≠ ε w ≠ ε g ← “electric” permittivities not necessarily equal/identical
μo ≠ μ s ≠ μ w ≠ μ g ← “magnetic” permeabilities not necessarily equal/identical
Thus, we from this perspective, we can see that e.g. for the E&M force, the macroscopic B -field
associated with an electrically charged particle moving through space-time is associated with the
response of the vacuum (i.e. space-time and its structure - at the microscopic level) to the
passage of the electrically charged particle through space-time! This says something very deep
about the fundamental nature of our universe!
Just as electromagnetism (E&M) has electric and magnetic fields, so do the strong, weak and
gravitational interactions also have “electric” and “magnetic” fields too!
10
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 13
Prof. Steven Errede
For the weak interactions (responsible for radioactivity and β-decay), there exist “weak” charges,
and there is “weak” electricity and “weak” magnetism – i.e. static “weak” electric field(s)
associated with the “weak” charge(s) and “weak” magnetic field(s) associated with moving
‘weak” charge(s)!
For QCD (Quantum Chromo-Dynamics) – (i.e. the strong interactions / nuclear forces) there
exist strong charges with associated so-called “chromo-electric” fields and “chromo-magnetic”
fields!
For gravity, gravitational charge = mass! There exist gravito-electric and gravito-magnetic
fields. The “every-day” gravitational force we experience living on the surface of our own planet
is due to the gravito-electric field of the Earth! The tides on the earth are due (primarily) to the
gravito-electric field of the moon!
Thus it can be seen that “electric charge”, “electric” and “magnetic” fields are not solely the
“property” of electromagnetism; indeed these are fundamental aspects/properties of any/all/each
of the four known forces of nature!!!
Macroscopic “Electric” fields are associated with the “charges” of each fundamental force.
Macroscopic “Magnetic” fields are produced when a “charge” (of any kind) moves through
space-time!
This undergraduate physics course is devoted to studying the myriad phenomena associated
with just one of the four fundamental forces known to exist (today) in nature (n.b. are there
more??) What we teach you in this class is relevant (in many ways) to all forces of nature!!!
The (static) electric field associated with a point electric charge q, in its own rest frame, at the
microscopic level, is comprised of (large numbers of) virtual photons (= quanta of the
electromagnetic field).
Virtual photons carry momentum p = h / λ where h = Plank’s constant = 6.626×10−34 Joulesec and λ = (DeBroglie) wavelength of the virtual photon.
While virtual photons also carry kinetic energy (like p 2 2mc {non-relativistically}), they
have zero total energy, since the (total energy)2 is E 2 = p 2 c 2 + m 2 c 4 = 0 ; thus a complex relation
exists between the momentum p and mass m of virtual photons: pc = imc 2 (where i = −1 ), and
E = 0 also implies that virtual photons have no vibrational/oscillatory frequency f associated with
them, since E = hf = 0 for virtual photons – i.e. for virtual photons, the relation c = f λ does not
exist (whereas this relation does exist/holds for real photons). Note that f = 0 for virtual photons
does in face make physical sense, since zero-frequency virtual photons are associated with the
static macroscopic electric field E ( r ) of a point electric charge, q.
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
11
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 13
Prof. Steven Errede
Both real and virtual photons have associated with them electric and magnetic field vectors,
which are orthogonal (i.e. perpendicular) to each other:
Real Photons:
E ⊥ B ⊥ ( p, v = c )
Virtual Photons:
Static E -field:
Static B -field:
(
B ⊥ E p, v = c
E
p, v = c
B
Static Radial Macroscopic E -Field of
a “classical” point electric charge, q:
)
(
E ⊥ B p, v = c
)
E
E , p, v < c
B, p, v < c
B
“Static” “Radial” Microscopic E -Field of
a “classical” point electric charge, q:
E , p, v < c
+q
+q
Bnet = 0
Microscopically, averaging over statistically significant numbers of virtual photons, (with
manifest quantum mechanical/wave-like behavior – {n.b. individual photons do not follow/obey
classical particle trajectories!}) a “classical” point electric charge q in its own rest frame has:
A radial electric field: E ( r ) = E ( r ) rˆ
A radial (inward/outward) momentum field: p ( r ) = p ( r ) rˆ (n.b. pnet = 0 )
Depends on sign of electric charge, q
A null magnetic field ( Bnet = 0 , virtual photon B -field vectors statistically cancel each other out).
{n.b. A real electron has a (point) intrinsic magnetic dipole moment μ = e / me c (relativistic
effect!) and corresponding non-zero macroscopic (and thus microscopic) magnetic dipole field!}
Note that, as with any statistical average, that on short time scales there are moment-to-moment
fluctuations on all of these quantities!
12
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 13
Prof. Steven Errede
For a moving “classical” point electric charge (i.e. neglect/ignore the intrinsic point magnetic
dipole moment of real electron), a net macroscopic magnetic field B (= statistical average over
microscopic virtual photons) is observed in the lab frame (but not in the rest frame of the point
electric charge). The relative motion of the electric charge and observer breaks the rotational
invariance associated with the electric charge at rest!
End view – point charge coming toward reader:
E , p, v < c
Bnet = B ( r ) ϕˆ ≠ 0
+q
The macroscopic B -field associated with a moving point electric charge q arises from Einstein’s
relativity & the fundamental nature/fundamental aspects of space-time itself!
The Macroscopic Magnetic “Induction” / Magnetic Intensity / Magnetic Field, B
N − sec
N
=
≡ Teslas = Weber / m 2
Units of B in (SI) MKSA system:
Coulomb − meter Ampere meter
1 Tesla = 104 Gauss (“old” cgs units of B)
The macroscopic magnetic induction B is defined in terms of the force acting on a test charge
QT moving with velocity v from Fm ( r ) = QT v ( r ) × B ( r ) .
A Weber is the (SI) MKSA unit of magnetic flux, Φ m :
Φ m ≡ ∫ B idA - counts B -field lines passing through a surface S.
S
(SI units: Weber = Volt-sec = Tesla-m2)
Magnetic flux is defined analogous to that for electric flux:
Φ e ≡ ∫ E idA - counts E -field lines passing through a surface S.
S
(SI units: (Volts/m)*m2 = Volt-meters)
If the surface S is a planar area A, then the magnetic flux Φ m = B i A ← A = area through which
lines of B pass through.
Note that:
1Weber = 1 Volt-sec.
∴
1 Tesla = 1 Newton/Ampere-meter = 1 Weber/m2 = 1 Volt-sec / m2
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
13
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 13
Prof. Steven Errede
Macroscopic Magnetic Force Acting on a Wire Carrying a Steady Current I:
Consider a filamentary wire carrying steady current I immersed in a magnetic field, B :
B points into paper
d
A
I
vd
A
d is (infinitesimal) line segment
of wire defined to be ║ to direction
of (conventional) current flow, I .
Section of conducting wire
By convention (blame goes to none other than Benjamin Franklin!):
The drift velocity vd of +ve electric charge carriers in a conductor gives the direction of flow of
current in the wire:
vd d I
{In a real wire, electrons (q = e−) comprise current. The e− flow in the direction opposite to that
of conventional current. The electron was discovered in 1897 by J.J. Thompson, long after
Benjamin Franklin’s time!)
The elemental magnetic force dFm acting on a infinitesimal line segment d
of current-carrying
wire in the presence of an external, macroscopic B -field is:
dFm = dQTOT vd × B
Where dQTOT = total free charge contained within infinitesimal volume dV = Ad associated
with infinitesimal line segment d of wire of cross sectional area A containing electric charge
carriers moving with a drift velocity vd in/along the wire.
dQTOT = qndV = qn( Ad )
What is dQTOT??
q = charge of individual carriers
n = # charge carriers / unit volume (= # density, #/m3)
dV = elemental volume containing QTOT charge
A = cross-sectional area of wire
dl = elemental length of infinitesimal line segment of wire
Line segment:
db
dV
A
dV = Adb
(
)
(
)
dFm = dQTOT vd × B = qndV vd × B = qn ( Ad
Now d
) ( vd × B ) = nq ( Ad ) ( vd × B )
vd , so d × B points in the same direction as vd × B , i.e.:
(
(d
) (v × B)
× B) = d (v × B)
vd d × B = d
= vd
(
d
d
because vd = vd
and d = d
)
∴
dFm = nqAvd d × B
14
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
UIUC Physics 435 EM Fields & Sources I
ITOT =
What is nqAvd ??
d
i.e.
dt
Fall Semester, 2007
(
) = nqA d
nqA d
dQTOT
=
dt
dt
= vd = vd
Lecture Notes 13
dt
Prof. Steven Errede
= nqA vd = nqAvd !!!
∴ I = nqA vd = nqAvd
∴ dFm = Id × B = Elemental magnetic force on infinitesimal line segment d
of wire
carrying a (steady) electric current I in an external magnetic field B .
The net magnetic force acting on a wire carrying a steady (i.e. constant) electric current I in an
external magnetic field, B is obtained by summing up all the infinitesimal force contributions
dFm along the length of the wire, i.e. integrating:
Fm = ∫ dFm = ∫ Id
C
( r ) × B ( r ) = I ∫C d ( r ) × B ( r )
d
B
(r )
C
r
A
ϑ (origin)
The net force acting on a straight wire carrying steady current I in external field B :
Fm = ∫ dFm = I ∫ d
C
(r ) = d
Let: d
Fm , ẑ
=L
( r ) × B ( r ) = I ∫ =0 d ( r ) × B ( r )
= d xˆ
B ( r ) = Bo yˆ (uniform)
and
B into paper, ŷ
b=0
Fm
I
Fm (up, ẑ )
B into paper ( ŷ )
b=L
d = d xˆ
d = d xˆ
d × B = ( d xˆ ) × ( Bo yˆ ) = d Bo xˆ × yˆ = d Bo zˆ (i.e. up)
= zˆ
Thus the net magnetic force acting on a straight wire of length L carrying steady current I in the
+ x̂ direction, immersed in a uniform field B = Bo yˆ is:
L
Fm = IBo zˆ ∫ dx = IBo Lzˆ
0
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
15
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 13
Prof. Steven Errede
Suppose the current-carrying wire is not straight, but instead e.g. is a closed current loop:
Then (here): Fm =
∫
C
(r )× B (r )
Id
In general, B ( r ) may not be uniform,
I
B into paper
C
but is a function of position, r
d , vd
i.e. in general: B = B ( r )
Note that in general, d = d
d ′, vd′
I
(r )
too!
I
r
Fm =
∫
C
(r )× B (r )
Id
In general, both d
for a closed loop
r′
ϑ
( r ) and B ( r ) can/will be functions of position
However, suppose B is uniform, e.g. B = Bo zˆ
Then: Fm =
I
constant
But:
∫
C
∴ Fm =
16
d
∫
C
{∫
C
d
( r )} ×
Bo zˆ
constant
( r ) ≡ 0 !!!
Id
(r )× B (r ) ≡ 0
for uniform/constant B (Not true for non-uniform B = B ( r ) !)
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 13
Prof. Steven Errede
The Macroscopic Magnetic Torque Acting on a Circuit
The infinitesimal (or elemental) magnetic torque dτ m acting on an infinitesimal line segment d
carrying a steady current I:
(
dτ m = r × dFm = r × Id × B
)
τ m = ∫ dτ m
Total torque:
C
Total torque for a current-carrying wire:
τm = I ∫ r ×(d × B)
C
(
For a closed steady current loop/closed steady current-carrying circuit: τ m = I ∫ r × d × B
C
)
Define A = Anˆ = vectorial area enclosed by loop (defined by contour C),
and:
Define n̂ = unit normal vector, n.b. direction is defined by right-hand rule of taking contour C.
n̂
A = Anˆ
B into paper
A
C
d
I
r
ϑ
(
) (
)
Thus: τ m = I ∫ r × d × B = I A × B ⇐ {if B ≠ B ( r ) }
C
n.b. Units of area
Components of torque, τ m :
If B is uniform
τ m = I ∫ ⎡⎣ r × ( d × B ) ⎤⎦ = I ( Ay Bz − Az By )
x
C
x
C
y
τ m = I ∫ ⎡⎣ r × ( d × B ) ⎤⎦ = I ( Az Bx − Ax Bz )
y
τ m = I ∫ ⎡⎣ r × ( d × B )⎤⎦ = I ( Ax By − Ay Bx )
z
C
z
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
17
UIUC Physics 435 EM Fields & Sources I
Since:
(
(
⎡r × ( d
⎣
)⎤⎦
)⎤⎦
× B )⎤
⎦
⎡r × d × B
⎣
⎡r × d × B
⎣
x
y
z
Fall Semester, 2007
Lecture Notes 13
Prof. Steven Errede
= y ( dxBy − dyBx ) − z ( dzBx − dxBz ) = ydxBy − ydyBx − zdzBx + zdxBz
= z ( dyBz − dzBy ) − x ( dxBy − dyBx ) = zdyBz − zdzBy − xdxBy + xdyBx
= x ( dzBx − dxBz ) − y ( dyBz − dzBy ) = xdzBx − xdxBz − ydyBz + ydzBy
Ax = Ai xˆ , Ay = Ai yˆ , Az = Ai zˆ
A is an areal vector whose x, y, z-components are the areas Ax, Ay, Az enclosed by C and
projected onto the y-z, z-x, and x-y planes, respectively.
Thus for a uniform B field:
τ m = I ( A× B)
Define: m ≡ IA = IAnˆ = magnetic dipole moment associated with a current loop of area A.
(SI Units: Ampere-meters2)
It can be shown geometrically that:
m ≡ IA =
Thus:
2A =
∫
C
r ×d
or:
A=
1
2
∫
C
r ×d
1
I r ×d
2 ∫C
If an electric current I is not confined to a zero-diameter/filamentary wire, but instead is
associated with an extended medium, then define current density J (Amps/m2) such that:
I = J iA
(
)
Id = J i A d
= J ( Ad
A, nˆ
I, vd
)
J
= JdV
I , J , vd , d , A all parallel
Hence, we see that:
1
1
m ≡ IA = I ∫ r × d =
2 C
2
∫
C
( ) = 12 ∫ r × ( JdV )
r × Id
V
Thus, an infinitesimal volume element dV of current-carrying conductor has associated with it an
infinitesimal magnetic dipole moment of:
dm =
1
r × JdV
2
and thus:
m = ∫ dm =
1
r × JdV
2 ∫V
{This result will be very useful for discussing magnetic phenomena in the future…}
18
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 13
Prof. Steven Errede
Griffiths Example 5.1: Cyclotron Motion
Consider an electrically charged particle with charge Q moving in a uniform magnetic field
B = − Bo zˆ (down) with velocity v ≠ vz zˆ (i.e. has velocity components only in x-y plane).
Magnetic (i.e. Lorentz) force acting on the moving charged particle is Fm = Qv × B
Side View:
Fm (into page) (by the Right-Hand Rule)
v (in x-y plane)
v × B (RHR)
B = − Bo zˆ (down)
Fm = Fm ( − rˆ ) {i.e. Fm is radially inward}
Fm = QvB = ma = m
v ( in x − y plane ) ⊥ B = − Bo zˆ ⇒ v × B = vB ( − rˆ )
v2
R
Magnetic force provides centripetal (i.e. radial inward) acceleration – bends particle around in a
circle of radius R:
Top View:
( B into paper)
R
v ( t = 0 ) = −vxˆ 1
Fm ( t = 0 ) = −QvBo yˆ
Q
ŷ
x̂
Initially, suppose v ( t = 0 ) = −vxˆ .
⎛
⎞
Then: Fm ( t = 0 ) = Qv ( t = 0 ) × B = Q ( −vxˆ ) × ( − Bo zˆ ) = +QvBo ⎜ xˆ × zˆ ⎟ = −QvBo yˆ
⎝ =− yˆ ⎠
As time increases from t = 0, can explicitly show that the orbit of charged particle in this B-field
lies on circle of radius R; motion is CCW around circle.
For Non-Relativisitic Motion:
Momentum p = mv and p = p = mv, v = v . Since Fm = QvB = m v 2 R then: QB = mv R
or: p = mv = QBR
If measure the radius of curvature of a charged particle (and +/− sign of curvature!) then know:
a.) momentum p of the charged particle and b.) charge-sign of the charged particle
⇒ Very important “tool” for use in particle / high energy / nuclear physics experiments!!
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
19
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 13
If the charged particle’s velocity vector v = vx xˆ + v y yˆ + vz zˆ
B = − Bo zˆ,
has a component parallel to B ( = Bo zˆ ) (i.e. vz ≠ 0 ) then
Prof. Steven Errede
zˆ
v
this z-component of the motion (here) is unaffected by B
because Fm = Qv × B = 0 for v B - then charged particle
moves in a helix / spiral in ẑ as shown in the figure on the right:
Griffiths Example 5.2: Cycloid Motion
Suppose we now include an electric field E so that ∃ a force Fe = QE on the charged particle
(charge Q) in addition to the magnetic/Lorentz force, Fm = Qv × B
E ⊥ B as shown in the figure below:
We orient uniform E in the ẑ direction, i.e. E = Eo zˆ
We orient uniform B in the x̂ direction, i.e. B = Bo xˆ
ẑ E
ϑ
a
b
c
d
ŷ
x̂
B
The charged particle is released from rest (i.e. v ( t = 0 ) = 0 ) at the origin ϑ . Initially the E -field
accelerates it due to Fe = QE = ma . However as soon as it acquires a finite velocity the B -field
bends it, due to Fm = Qv × B . Initially, for time t just after t = 0, the electric field gives the
(
)
charged particle a vinit = vz and then vz × B ⇒ yˆ direction. As the charged particle curves over
to the ŷ direction, the charged particle begins to lose speed (kinetic energy) as it curves over
more. Its speed / kinetic energy actually goes to zero when it touches the ŷ axis at point y = a.
Then the process starts all over again . . . the process repeats over and over….
Note that there is no force acting on the charged particle in the x̂ direction – only in the ẑ
direction (due to E & B ) and the ŷ direction (due to B {only}).
∴ v ( t ) = v y ( t ) yˆ + vz ( t ) zˆ = ( 0, v y ( t ) , vz ( t ) ) = ( 0, y ( t ) , z ( t ) )
{n.b. y ( t ) =
xˆ
yˆ
zˆ
v×B = 0
y
z = Bo zyˆ − Bo yzˆ
Bo
20
∂y
∂z
( t ) and z ( t ) = ( t ) }
∂t
∂t
0 0
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
UIUC Physics 435 EM Fields & Sources I
Fall Semester, 2007
Lecture Notes 13
Prof. Steven Errede
Newton’s Second Law: F = ma = m ( yyˆ + zzˆ )
(
)
F = QE + Qv × B = Q E + v × B = ma with E = Eo zˆ and
B = Bo xˆ
Then:
F = Q E + v × B = Q ( Eo zˆ + Bo zyˆ − Bo yzˆ ) = ma = m ( yyˆ + zzˆ )
(
)
Treating the ŷ and ẑ components separately (since they are independent):
QBo z = my and QEo − QBo y = mz
We define the cyclotron angular frequency as:
ωc ≡
fc =
Define the cyclotron frequency as:
QBo
( radians / sec )
m
ωc
2π (
Hz )
ωc = 2π f c
⎛E
⎞
and z = ωc ⎜ o − y ⎟ ⇐ coupled differential equations!
⎝ Bo
⎠
The general solutions to these coupled diff eq’s are:
⎛E ⎞
y ( t ) = C1 cos (ωc t ) + C2 sin (ωct ) + ⎜ o ⎟ t + C3
⎝ Bo ⎠
Then: y = ωc z
z ( t ) = C2 cos (ωc t ) − C1 sin (ωc t ) + C4
The charged particle started from rest ( y ( 0 ) = z ( 0 ) = 0 ) at t = 0 at the origin ϑ (y (0) = z (0) = 0)
These are the four boundary conditions which determine / define constants C1, C2, C3, and C4:
E
E
y ( t ) = o (ωc t − sin ωc t ) and z ( t ) = o (1 − cos ωc t )
ωc Bo
ωc Bo
E
Define: R ≡ o
ωc Bo
Then:
y ( t ) = R (ωct − sin ωc t )
and
z ( t ) = R (1 − cos ωc t )
Then using the trigonometric identity sin 2 ωc t + cos 2 ωc t = 1 , we obtain:
( y − Rωct ) + ( z − R )
2
2
= R 2 ⇐ equation of a circle whose center is ( x, y, z ) = ( 0, Rωc t , R )
Eo
Bo
The motion of the charged particle is such that it is analogous to a point on the rim of a bicycle
E
wheel rolling down the ŷ -axis at constant speed v y = ωc R = o
Bo
The circle travels in the ŷ -direction at constant speed v y = ωc R =
This curve is called a cycloid of motion – the overall motion is not in the direction of E but
actually perpendicular to it (because B ⊥ E here)!!!
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2008. All Rights Reserved.
21