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M&I
Chapter 16
Electric potential
Energy of a single particle
q
Particle energy 2mc K
2 212
mc mv for v c
final initial externalK K K W
The kinetic energy of a single particle can be changed
if positive or negative work is done on the particle by
an external force.
external externalW F r
M&I
16.1
Then
Potential energy associated with pairs of interacting particles
sys ext intK W W Q
For a system of interacting particles:
orsys int extK W W Q
if 2 0mc
Write intW U
change in potential energy U
Thensys extK U W Q
Systems of charged particles:
sys 0K U
…in most cases considered here, ext 0 and 0W Q
Uniform electric field between plates
proton electric 0K U
+
+
+
+
+
+
E
electricF
electric 0final initialK K U
increases decreases
int 0final initialK K W
0final initialK K F r
0final initial xK K eE x
Proton moving to the
right in a uniform
electric field
r
M&I
16.2
Uniform electric field between plates
electron electric 0K U
decreases increases
int 0final initialK K W
0final initialK K F r
( ) 0final initial xK K e E x
Electron moving to
the right in a uniform
electric field
0final initial xK K eE x
+
+
+
+
+
E
r
electricF
Electric Potential DifferenceM&I
16.3
electric x xU eE x e E x
In the previous examples:
For the proton:
For the electron: electric x xU eE x e E x
Let potential difference: xV E x
Then electricU q V Units of : volts (V)V
If is not parallel to Er
then ( )x y zV E x E y E z
or V E r
+
r
E
The electron volt
If an electron moves through a potential difference of one volt
then there is a change in the electric potential energy whose
magnitude is
-19 -19(1 volt) (1.6 10 C)(1 J/C) = 1.6 10 jouleU e
-191.6 10 joule = 1 electron volt = 1 eV
1 keV = 1000 eV
1 MeV = 106 eV
Sign of potential difference and direction of electric fieldM&I
16.4
A B
E
r1.
A B
E
r2.
3. A B
E
r
0V
V E r
0E r
0V
0E r
0V
0E r
Example: Field and potential
B A
cos
V V V
E r
E r
30
2 m
B
A
E
r
100 N C-1= (100)(2)cos(30 )
= 173 volts
Work done by an external
agent in moving a proton
from A to B:
ext electricW U q V
=(1.6 10-19)( 173) = 2.76 ×10-17 J or 173 eV
Potential difference in a non-uniform fieldM&I
16.5
If we move through two (or more) regions where the
electric field is different, then final initialV V V E r
if each is small enough that is uniform in the region
through which it passes.
r E
Remember that x y zE x E y E zE r
For example:
B A
C A B C
1 1 2 2
1 2( ) ( )x C A x B C
V V V
V V V V
E x x E x x
E r E r
If the electric field in a
region varies continuously,
then we need to integrate:
final initial
f
iV V V dE r
Note that … potential difference is independent of path
or x y zV E dx E dy E dz
Change of electric potential in a non-uniform field …2
dr
E
+
i
f
M&I uses the notation2
1V dE l
M&I
16.6
… and for a round trip 0V
C
C
AA
V V V dE r
Path independent
f f f
i i i
f x y z
x y zi x y z
V d E dx E dy E dzE r
Hence write:
M&I
16.7
Electric field inside and outside a current carrying wire
The charges are
moving, hence the
electric field E inside
the wire is non-zero.
0V though the wire
Hence along
path 2 (in the air) …
… hence E is non-zero
in the air.
0V
The electric field is uniform in this region.
S is at (2, 2, 0) m and T is at (2, 0, 0) m
What is along a path from S to T?
1. +150V
2. 150V
3. +300V
4. 300V
5. +600 V
6. 600 V
-1ˆ300 N CE j
x x
x
R S
T
E
V
1 2 3 4 5
The electric field is uniform in this region.
S is at (0, 0, 0) m and T is at (0, 2, 0) m
What is along a path from S to T?
1. +200V
2. 200V
3. +400V
4. 400V
5. +800 V
6. 800 V
-1ˆ400 N CE j
x x
x
R S
T
E
V
1 2 3 4 5
The electric field is uniform in this region.
S is at (0, 0, 0) m and T is at (0, 2, 0) m
What is along a path from S to T?
1. 0 V
2. 300 V
3. 500 V
4. 600V
5. 1000 V
-1ˆ ˆ200 300 N CE i j
x x
x
R S
T
E
V
1 2 3 4 5
The electric field is uniform in this region.
S is at (0, 0, 0) m and T is at (0, 2, 0) m
What is the magnitude of the electric
field in this region?
1. 250 V m-1
2. 500 V m-1
3. 750 V m-1
4. 1000 V m-1
x x
x
R S
T
EV along a path from S to T is 500 V.
1 2 3 4 5
The electric field is uniform in this region.
R is at (3, 2, 0) m and T is at (8, 0, 0) m
What is along a path from R to T?
-1ˆ ˆ200 400 N CE i j
x x
x
R S
T
E
V
1. +200V
2. 200V
3. +800V
4. 800V
4. +1000 V
6. 1000 V
1 2 3 4 5
The electric field is uniform in this region.
R is at (3, 2, 0) m and S is at (5, 2, 0) m
What is along a path from R to S?
1. 0 V
2. 400V
3. +400V
4. 800V
5. +800 V
-1ˆ400 N CE j
x x
x
R S
T
E
V
1 2 3 4 5
Without doing any calculations, what is the sign of VB VA ?
1. positive
2. negative
3. 0
1 2 3 4 5
What is VB VA ?
1. –20 V
2. –10.05 V
3. –8.06 V
4. –0.1 V
5. none of the above
1 2 3 4 5
The potential at one locationM&I
16.8
final initialV V V
set Vinitial = 0 at “infinity”
For the potential at a distance r from a single point charge q
2
0 0 0
1 1 1
4 4 4
rr r
r r x
q q qV V V E dx dx
x x r
If we know the value of the potential at location A, then if
we place a charge q at A, then the potential energy of the
system isA AU qV
For two point charges separated by a distance r: 1 2
0
1
4
q qU
r
V(x,y)
x
y
The electric potential in 2D.
V(x,y) around a positive
point charge.
Lines of equipotential
0
1
4r
qV
r
Lines of equipotential around an electric dipole
Lines of equipotential
How much work would
you need to do to move a
charge … … from here
… to here?
+Q
+q
r4r
a b
c
AB
C
Insert either > or < or =
below for each.
1. EA EB
2. VA VB Va Vb
3. VC VB Vc Vb
4. Q q
5. EB EA
6. EB EC
7. Eb Ea
8. VA Va
9. EA Eb
Electrical potential energy
For each situation below, decide qualitatively whether the initial
or final situation has higher electrical potential energy.
All charges are either +q or q.
initial final
(a)
(b)
(c)
Uinitial = Ufinal
<
>
Uinitial = Ufinal
<
>
Uinitial = Ufinal
<
>
+
-
+
- -
-
-
+
- -
+
- +- - +-
-
Electrical potential energy …2
initial final
(d)
(e)
(f)
Uinitial = Ufinal
<
>
Uinitial = Ufinal
<
>
Uinitial = Ufinal
<
>
+
-
+
-
-
-
+ -
-+
-
+-
+ +
-
+-
Electric potential and electric field
Shown below are examples of the variation of the electrical
potential along the x-direction. Draw arrows representing the
direction and relative magnitude of the electric field at positions
A and B on the x-axis.
V(x)
0x
(a)
A B
A B x
V(x)
0x
(b)
A B
A B x
Electric potential and electric field …2
V(x)
0x
(c)
A B
A B x
V(x)
0x
(f)
A B
A B x
V(x)
0x
(d)
A B
A B x
V(x)
0x
(e)
A B
A B x
Electric potential and electric field …3
V(x)
0x
(g)
A B
A B x
V(x)
0x
(j)
A B
A B x
V(x)
0x
(h)
A B
A B x
V(x)
0x
(i)
A B
A B x
Important worked example: A disk and a spherical shell
A thin spherical (plastic) shell
carries a uniformly distributed
negative charge –Q1.
A thin circular (glass) disk
carries a uniformly distributed
positive charge +Q2.
Find the potential difference
V2 – V1.
… choose a path (straight line) from 1 to 2
… neglect the polarization of the plastic and
glass since both object are made of thin
material.
2 2 2
2 1 shell disk net shell disk1 1 1
V V V V V d d dE r E r E r
due to shell:V
surface of shell 1 0V V since inside shellshell 0E
Outside shell: 1shell 2
0
1ˆ
4
Q
rE r
2
2 surface of shell shell3
21
230
1 1
0 1 1
1( )
4
1
4
V V d
Qdx
x
Q Q
R d R
E r
We move opposite to the direction of the field, therefore 0V
Check:
3
Why is there a sign here?
due to disk:V
Since and : 2d R1 2R R
2
2 2disk
02
Q RE
2 22 2
2 2 2 22 1 disk 1
1 10 0
( ) ( )2 2
Q R Q RV V d dx d RE r
We move opposite to the direction of the field, therefore 0V
Check:
due to both shell and disk:V
2
1 1 2 22 1 1
0 1 1 0
1 ( )4 2
Q Q Q RV V d R
R d R
A metal in static equilibrium
For a metal in static equilibrium:
final initial 0V V V
… for any two locations inside the metal
Therefore final initialV V
… the potential inside the metal is constant
but not necessarily zero!
(and E = 0)
i
f
A metal slab inside a capacitor
A s
s = 3 mm, ΔV = 6 voltsV
Es
= 2000 V m-1
+
+
+
+
+
A
s1Q 1Q
+
+
+
+
+
A
1Q 1Q
1 mm
+
+
+
+
+
2Q2Q
Insert 1 mm metal slab without
touching sides of capacitor …
… which then polarizes …
Start with a charged capacitor
Since E inside slab is zero, 2 1Q Q
Why?
But in the air gaps, E is unchanged
left rightV V = (2000 V m-1)(0.001 m)
= 2 volts
capacitorV = 2 V + 0 V + 2 V = 4 volts
Potential difference in an insulator
Again start with a charged capacitor …
+
+
+
+
+
A
s QQ
Eplates
+
+
+
+
+
A
Q QEplates
+ +
+ +
+ +
+ +
+ +
… now insert an insulator …
What is the effect on the
electric field inside the
capacitor?
M&I
16.9
Inside the plastic, is complex
Outside the plastic … consider around the closed path shown
E r… will be positive outside the plastic
r
r
… therefore the average field inside the plastic must point to the left
E r
Edipoles
+
+
+
+
+A
Q Q
+ +
+ +
+ +
+ +
+ +
EplatesEdipoles
Enet
… result is that the electric field
inside the capacitor is reduced.
applied
netK
EE
where K is the dielectric constant
Electric field inside a capacitor with a dielectric constant =0
Q
AK
K always > 1
andvacuum
insulator
VV
K
Effect of dielectric: … decreases the electric field
… decreases the potential difference
Dielectric constants for various insulators
Vacuum 1 (by definition)
Air 1.0006
Typical plastic 5
Sodium chloride 6.1
Water 80
Strontium titanate 310
Energy density associated with electric field = J m-32102E
… general result … do it yourself …
M&I
16.10
Originally ΔV was –1000 volts.
A metal slab is inserted into the
capacitor.
Now ΔV = VB – VA =
1) + 1000 volts
2) +500 volts
3) 0 volts
4) –500 volts
5) –1000 volts
1 2 3 4 5
With a plastic slab in the capacitor:
Now ΔV = VB – VA =
1. between –500 and –1000 volts
2. between +500 and+ 1000 volts
3. –500 volts
4. +500 volts
5. not enough information to tell
1 2 3 4 5
Potential along the axis of a ring
Potential of distributed charges …
... with radius R and total charge Q
2 2R z
z
R Each point charge q on the
ring contributes:
2 20
1
4
qV
R z
Adding up the potential contributed by all the point charges:
ring2 2 2 2 2 2
0 0 0
1 1 1 1
4 4 4
q QV q
R z R z R z
M&I
16.8
Potential along the axis of a uniformly charged disk
E
j
i
R
r
... with radius R and total charge Q
k
r
z
ring2 2 2 2
0 0
1 1 2
4 4
q Q r rV
Az r z r
2 200
1
2
RQ rdrV
A z r
2 2
00
1
2
RQz r
A
2 2
0
1
2
QV z R z
A
Check:2 2
0
11
2z
V Q zE
z A z R
M&I
16.12
What is VB VA?
1. 270 V
2. 18 V
3. 6 V
4. –6 V
5. –18 V
6. –270 V
1 2 3 4 5
What is VB VA?
1. +1350 V
2. –1350 V
3. +3375 V
4. –3375 V
5. none of the above
1 2 3 4 5
VP VQ is:
1. positive
2. negative
3. zero
4. not enough information to tell
1 2 3 4 5
Along the semicircular path, VB VA is:
1. positive
2. negative
3. zero
4. not enough information to tell
1 2 3 4 5
Which of the following quantities are zero?
1. VC – VA
2. VD – VC
3. VB – VD
4. VC – VA and VB – VD
5. VC – VA and VB – VD and VD – VC
6. none of the above
1 2 3 4 5
Along the straight path through the metal sphere, VB VA is:
1. positive
2. negative
3. zero
4. not enough information to tell
1 2 3 4 5
M&I
Chapter 17
Magnetic Field
… Electric fields E … generated by the presence of charge
(stationary or moving)
… Magnetic fields B … generated by moving charge…
Electron current i
… the number of electrons per second that enter a
section of a conductor
M&I
17.1
Simple circuits
… refer to laboratory exercise on circuits ...
Detecting magnetic fields
Do it yourself … and record results
… What is the effect of the magnetic field of the Earth?
M&I
17.2
Oersted, 1820
• magnitude of B magnitude of current
• zero current zero B
• direction of B is perpendicular to direction of current
• direction of B above the wire is opposite to direction of B
below the wire
B
Detecting magnetic fields …2
The Biot-Savart law for a single moving charge
“Careful experimentation” …
Units of B : tesla (T)
0
2
ˆ
4
q
r
v rB
0
-7 -1
permeability of free space
= 4 10 T m A
0
4= 10-7 T m A-1 exactly
+
r
v
B
q
r
M&I
17.3
where and G G A G B
ˆ ˆ ˆ ( ) + ( ) + ( ) y z z y z x x z x y y xA B A B A B A B A B A BA B i j k
easy to remember:alwaysˆ ˆ ˆ
x y z
x y z
A A A
B B B
i j k
In polar form in 2D:
where is the angle between tails of and .
kBA ˆ sinAB
A
The cross product
B
A
B
What is the direction of ...
A. < 0, 0, 3> < 0, 4, 0> ?
B. < 0, 4, 0> < 0, 0, 3> ?
C. < 0, 0, 6> < 0, 0, 3> ?
1.
2.
3.
4.
5. zero magnitude
ii
k
k
1 2 3 4 5
What is the direction of magnetic field at the observation location?
A. B. C.
1.
2.
3.
4.
5. zero magnitude
ii
k
k
1 2 3 4 5
At the observation location the magnetic field due to the proton
is in the z direction.
What is a possible direction for the velocity of the proton?
1.
2.
3.
4.
5. zero magnitude
j
j
k
k
1 2 3 4 5
At the observation location the magnetic field due to the electron
is in the x direction.
What is a possible direction for the velocity of the electron?
1.
2.
3.
4.
5. zero magnitude
j
j
k
k
1 2 3 4 5
Relativistic effectsM&I
17.4
0
2
ˆ
4
q
r
v rB
v : velocity of source or observer?
+
vor
+v
Retardation
+ v
B
+ 0v
0?B
+ 0v
0B
t1 t2 t3
(no t in here)
Electron current i
Distance traveled by electron sea in time = v tt
Drift speed of mobile electrons = v
Number of mobile electrons in shaded cylinder =
A
v t
v
E
nAv t
where n is the number of mobile electrons per unit volume
Electron current i is the rate at which electrons pass a
section of a wire (number of electrons per second) = nAv
M&I
17.5Metal wire of cross sectional area A.
Free electrons move under influence
of E.
Conventional current I
... runs in the opposite direction to electron current
... defined as the amount of charge (in coulombs) passing a
point per second
... given by the number of holes per second multiplied by the
(positive) charge associated with one hole
I q nAv
In metals, q e
Units of I: ampere (A)
M&I
17.5
I enAv
The Biot-Savart law for currents
0
2
ˆ
4
I
r
l rB
0
4= 10-7 T m A-1 exactly
Consider a small thin wire of length and cross sectional area A.
If there are n moving charges per unit volume, then there are
moving charges in this volume.nA l
l
Then the total contribution to =qv nA l q v q nAv l I l
Now can write
where is a vector with
magnitude pointing in the
direction of the conventional
current I
l
l
r
l
B
I
r
M&I
17.6
For each situation below, determine the direction of the
magnetic field at point P caused by the current in the short
section of wire in the dashed box.
P
I
I
P
1
2
3
IP
A B
C
1. into the page
2. out of the page
3. zero
1 2 3 4 5
P1 2
I
IP
For each situation below, determine the direction of the
magnetic field at point P caused by the current in the short
section of wire in the dashed box.
1 2 3 4 5
1. into the page
2. out of the page
3. zero
D E
1
2
3
Magnetic field of a straight wire
Magnitude of : r
Then22
ˆ ˆˆ
x y
r x y
r i jr
22r x y
ˆ ˆx yr i j
B
y r
j
i
y
x
k
r
I
ˆyl j
0
2 2
ˆ ˆ
4
I y
x y
j rThen magnetic field
due to small piece only =
B
0
2 2 22
ˆ ˆ ˆ
4
I y x y
x y x y
j i j
... of length L, carrying current I
M&I
17.7
Magnetic field of a straight wire …2
0
2 2 22
ˆ ˆ ˆ
4
I y x y
x y x y
j i jB
ˆ ˆ ˆ ˆx y xj i j k
32
0
2 2
ˆ4
I x y
x y
B k
Let and integrate over entire length L of wire
(only Bz is non-zero):
32
2
0
2 2
2
4
L
z
L
dyB Ix
x y
Tables of integrals
2
2
0 0
2 2 2 22 ...
4 4 2
L
L
z
y LIB Ix
x x y x x L
0y
Magnetic field of a straight wire …3
0
22
4 2wire
LIB
r r L
Check the result ... units? ... direction?
Special case :L r
Then
2 22 2 2 2r r L r L r L
can write r = x
0 2
4wire
IB
r
Another special case :r L0
2
4wire
I lB
r
Direction of ? ... use right hand rulewireB
I
wireB
Magnetic field of a straight wire …4
wireB curls around the wire
0
22
4 2wire
LIB
r r L
I
B
Go to worksheets ...
Electric currents produce magnetic fields (1 & 2)
VPython script
Bwire_with_r.py
Magnetic field along the axis of a circular loop of wire
... with radius R and current I
ˆ ˆR zr j k
Magnitude of : r 2 2r R z
Then2 2
ˆ ˆˆ
R z
r R z
r j kr
Then magnetic field
due to small piece only =0
2
ˆ
4
I
r
l r32
0
2 2
ˆ ˆ ˆ
4
R R zI
R z
i j k
ˆRl i
M&I
17.8
B
j
i
k
r
z
l
R
r
Magnetic field along the axis of a circular loop of wire …2
32
0
2 2
ˆ ˆ ˆ
4
R R zI
R z
i j kB
2
ˆ ˆ ˆ
ˆ ˆ
R R z
zR R
i j k
j k
32
2
0
2 2
ˆ ˆ +
4
zR RI
R z
j kB
32
2
0
2 2
4z
IRB
R z
B
R
zzB
r
Only Bz will be non-zero:
B_loop_with_r_dB.py
See:
32
2
0
2 2
2 =
4loop
I RB
R z
Check the result ... units? ... direction?
Special case: centre of the loop, z = 0
B
I
0 2 =
4loop
IB
R
Another special case :z R3 32 22 2 2 3R z z z
2
0
3
2 =
4loop
R IB
z
3 32 2
22 2
0 0
2 2 2 2
0
= 2
4 4z
I R I RB d
R z R z
Now let and integrate around loop0loop of wire …3
Magnetic field at other locations outside the loop
B_loop_xy_xz.py
… use a computer program …
What is the direction of magnetic field at the observation location?
A. B.
1.
2.
3.
4.
5. zero magnitude
ii
k
k
1 2 3 4 5
Which components of at the observation location are nonzero?
1. z
2. y
3. x
4. y & z
5. x & y
6. x & z
7. all components
1B
1 2 3 4 5
Which components of at the observation location are nonzero?
1. z
2. y
3. x
4. y & z
5. x & y
6. x & z
7. all components
2B
1 2 3 4 5
What is the direction of magnetic field at location A? ... and B?
1) +x
2) –x
3) +y
4) –y
5) +z
6) –z
7) zero magnitude
1 2 3 4 5
For each situation below, determine the
direction of the magnetic field at point P
caused by the current in the entire wire.
1 2 3 4 5
P
I I into page
P
P
I
P
I
I
A B CP
D E1. into the page
2. out of the page
3.
4.
5. zero
Magnetic dipole moment,
I
B Bμ
2
0
3
2 =
4axis
R IB
r
for r R
Write 0
3
2 =
4axisB
r
Where the magnetic dipole moment = IA
In an applied magnetic field, a current-carrying loop rotates
so as to align the magnetic dipole moment with the field. μ
M&I
17.9
A
The magnetic field of a bar magnet
… what about magnetic monopoles?
S N
M&I
17.10
Be careful of
pictures like
this ...
0
3
2 =
4axisB
r… for both the bar magnet and ring of current
The magnetic field of the Earth
BEarth at Cape Town52.6 10 T
Magnetic dipole caused by a current loop
Determine the direction and magnitude of the magnetic dipole
moment produced by each current loop shown below:
I = 2.0 A
15 cm
20 cm
I = 2.0 A
r = 20 cm
I = 2.0 A
r = 20 cm
A B C
Worked example: A circuit in the Antarctic
AI
Say that a circuit containing a ¾ loop of wire
of radius 5 cm lies on a table in a lab in the
Antarctic. There is a 5 ampere current in the
wire. Say that you have a bar magnet with
magnetic moment 1.2 A m2. How far above
location A (at the centre of the loop), and in
what orientation, should you hold the bar
magnet such that the net magnetic field at A
is zero. Take the Earth’s magnetic field at the
Antarctic to be 6 10-5 T.
points out of the page (out of ground at Antarctic)
points out of the page (out of table)
Therefore bar magnet needs to be orientated with its north pole
downward (into the page).
EarthB
circuitB
d l
R
r
A
k axis out5
Earthˆ6 10 T B k
straight wires 0B
03/4 loop 2
ˆ
4
I dd
r
l rB B
Put the origin at the centre of the loop.
Rr (constant)
At all locations thereforeˆd l r ˆ sind dl dl Rdl r2 2
20 0 03/4 loop 2 2 2 2
2 2
= =
4 4 4
I Rd IR IRd
R R RB
503/4 loop
3 ˆ ˆ = ... = 4.7 10 T 4 2
I
RB k k
Antarctic loop …2
5
Earthˆ6 10 T B k
5
3/4 loopˆ= 4.7 10 T B k
Therefore want
4
Earth 3/4 loopˆ1.06 10 T B B k
4
magnetˆ1.06 10 T B k
40magnet 3
21.06 10 T
4 zB
1
130 7 -1 2 3
4
magnet
2 1 10 T m A (2)(1.2 A m )4 0.13 m 1.06 10 T
zB
Antarctic loop …3
The atomic
structure
of magnets
M&I
17.11
Each atomic current loop
contributes an amount
of magnetic field:2
0 0
3 3
2 2 =
4 4
R I
r r
B B
2 2
e e evI
RT R
v
and 2A R
2 1
2 2
evIA R eRv
R
Bohr atomic model ...
Estimating the magnetic dipole moment:
a simple model of the atom
+R
2
net
d vm m F
dt R
v
2 2
2
0
1
4
v em
R R
2
2
0
1
4
ev
R m… get v 1.6 106 m s-1 for R 10-10 m
19 10 6 -112
1(1.6 10 C)(10 m)(1.6 10 m s )
2eRvThen
23 21.3 10 A m per atom
Estimating the magnetic dipole moment:
quantized angular momentum
Orbital angular momentum: L Rmv
1 1 1
2 2 2
e eeRv Rmv L
m mThen
L is quantized in units of = 1.05 10-34 J s
23 20.9 10 A m per atom
191342
31
1 (1.6 10 C)(1.05 10 Js)
2 (9 10 kg)
e
m
The modern theory of magnets
... Bohr model too simplistic, really
Situation closer to ...
... information about location of electron is probabilistic
... spherically symmetric probability distributions average to zero
... non spherically symmetric probability distributions (p, d, f)
orbitals can contribute a non-zero magnetic dipole moment
... most atoms also have more than one electron!
Spin
The electrons themselves also have spin ... which contributes a
significant magnetic dipole moment.
... but it is problematic to think of the electron as a spinning ball
of charge ...
... protons and neutrons in nuclei also have spin, but magnetic
dipole moment is much smaller, and can be ignored for this
purpose ...
where m = mp or mn
... but not for nuclear magnetic resonance (NMR)
... and the technology of magnetic resonance imaging (MR)
1
2
e
m
Alignment of the atomic magnetic dipole moments
Most materials have no net orbital or spin magnetism.
In some materials (e.g. iron, nickel, cobalt, ...) the orbital and
spin motions of neighbouring atoms line up with each other
and can produce a sizable magnetic field
... “ferromagnetic” materials.
... explained by quantum mechanics
... alignment due to electric interactions between atoms, not
magnetic interactions.
Magnetic domains
... many of the individual atomic magnetic dipole moments are then
aligned wit the external field ... causing a significant field associated
with the material ...
In an ordinary piece of iron that is not a
magnet, the material can be thought of being
made up of a “patchwork” of small regions
called magnetic domains within which the
alignment of the atomic magnetic dipole
moments is nearly perfect ...
If the external field is removed, this induced magnetism may
remain... can be destroyed by external force or heating.
... but normally these domains are randomly
orientated ... net magnetic effect is not
significant ... if the iron is placed within an
external magnetic field, the domains nearly
aligned with the field tend to grow, and
others might rotate to align with the field
The magnetic field inside a solenoid
(by the application of the Biot-Savart law)
Tougher mathematics ... try it yourself ... ...
otherwise see later (Ampere’s Law)
If : L R0
z
NIB
LI
L
N loops
RI
solenoid_drag.py
M&I
17.13
M&I
Chapter 18
A Microscopic View of Electric Circuits
electron current i = no. of electrons per second passing a point
electron current flows in direction opposite to
conventional current I = no. of coulombs per second =
conventional current flows in the direction of
E
E
q i
“Static equilibrium” : no charges are moving
“Steady state” : charges are moving, but their velocities do not
change (significantly) over time (and there is no change in the
deposits of excess charge anywhere)
Current in different parts of a circuitM&I
18.2
Consider a simple circuit:
+-+ - A
B
1. iA = iB 2. iA > iB 3. iA < iB
What is being “used up” in the light bulb?
What is a light bulb?
1 2 3 4 5
The current node rule
In a steady state, the electron current entering a node in a circuit is
equal to the electron current leaving that node.
... consequence of the principle of conservation of charge
Also known as the “Kirchhoff node rule”
i1i2
i4
i3i1 = i2 = (i3+i4)
i1= 5 A
i4 = 1 A
i3= 8 A
i2 = ?
i1= 5 A
i4 = 6 A
i3= 8 A
i2 = ?
But i3 need not be equal to i4
M&I
18.3The start-stop motion of electrons in a wire
In order for electrons to move in a wire (i.e. for there to be a
current), there must be an electric field present to drive the sea of
mobile electrons.
Why is a (constant) electric field necessary …?
… and what is the source of the electric field in the wire?
... the mobile electrons are constantly colliding with the lattice of
atomic cores, increasing the thermal motion of the atoms.
... electrons cannot “push” each other through the wire!
Why is a field necessary?
The Drude model
A mobile electron in a metal, under the influence of an electric
field inside the metal, accelerates, gains energy, but then
collides with the lattice of atomic cores, which is vibrating
because of its own thermal energy.
The electron then gets accelerated again, collides, …
The metal heats up as a result of this process.
time
Speed of
a single
electron
v
v = “drift” speed
The Drude model …2
Momentum principle: net eEt
pF
If an electron loses all its momentum in a collision, 0p p eE t
If speed of electron << c, writee e
p eE tv
m m
Averaging over all collisions:e
eE tv uE
m
where is the electron “mobility” e
e tu
m
Different metals have different electron mobilities.
Then electron current: i nAv nAuE
Electric field and drift speed in different elements of a circuit
Consider a part of a circuit where a wire leads into a thinner
section made of the same material ...
thinvthickv
Since thin thicki i
thin thin thick thicknA v nA v or thick
thin thick
thin
Av v
A
The electrons move faster in the thinner section of wire.
... hence the electric field is larger in the thinner section.
Direction of electric field in a wire
The current is the same in all parts
of a series circuit, hence the
electric field E must be the same in
every part of the wire in a circuit
in a direction parallel to the wire at
every location, even if the wire
twists and turns …
E
… E must also be uniform
across a cross section of the
wire … E
Convince yourself by thinking
about
A B
D C
A
ABCDA
A
V dE l
What charges make the electric field in the wires?
In a steady state circuit ...
... there must be an electric field in the wires
... the magnitude of the electric field must be
the same throughout a wire of the same
geometry and material
... the direction of the electric field at every
location must be along the wire, since
the current follows the wire.
M&I
18.4
Consider a very simple circuit consisting of a bulb connected
by long wires to a battery…
Does the bulb shine any differently depending on
where the bulb is in relation to the battery?
… No … !
… so where is the excess charge that creates the electric field
that drives the current in the circuit ?
A mechanical battery
v
v
A “conveyor belt”
replenishes
electrons that have
left the negative
plate and travelled
around the circuit
to the positive
plate.
Connected a bent Nichrome wire
across the terminals of a
mechanical battery …
Think about E due to plates of battery
and at points 1, 2, 3, 4, 5v
vE
Huh !?
A mechanical battery ...2
Excess electrons
build up here on
the surface of the
wire
Excess positive
charge builds up
here on the
surface of the
wireEbattery
Ebends +
+
+
+
A mechanical battery ...3
Ebattery
Ebends +
+
+
++
+
This is an example of “feedback”
… until Ebends > Ebattery and net field electric field points to the left
A mechanical battery ...4
E
+ ++
+
+
+
+ +
+
+ +
+
v
Charge build-up will occur at many points in the wire (not only at
bends) until in the steady state every point in the circuit will have
the same magnitude of E
A mechanical battery ...5
The distribution of
excess surface
charge in a circuit
can be quite
complicated …
Remember that the
real situation is in
3D.
Think about this
simple case: What is the direction of the electric field here?
Typical electric fields: 5 V m-1
… and only about 106 electrons per cm of wire near the negative
end of a 6 volt battery
Connecting a circuit: the initial transient
gap
At t = 0, there is a gap in the circuit
... and E = 0 everywhere in the wire
M&I
18.5
Connecting a circuit: the initial transient …2
Look more closely at the
gap region and consider
electric field inside the
wire due to surface
charges in gap region:
Net electric field inside the
wire must be zero, hence
other charges must
contribute Eother, as shown:
Egap faces
Eother
Connecting a circuit: the initial transient …3
Now close the gap.
Charges on facing ends of wire
neutralize each other, and net
field is given by Eother only…
But there is a large unstable
discontinuity in surface charge
distribution:
Electrons will move under the
influence of Eother …
After a fraction of a nanosecond,
the new distribution might look
like this (a more gradual change
in the charge distribution):
Connecting a circuit: the initial transient …4
All this happens at the speed of light.
The electrons do not have to move
very far in order to effect a significant
surface charge distribution.
The electric field is still zero at other
locations in the circuit (information
hasn’t yet reached these regions!)
After a few nanoseconds the
rearrangement of charges will have
extended to all parts of the circuit …
… leads to the “steady state” situation
where E has uniform magnitude
everywhere ..
E = 0
0E
If a typical electron drift speed is around 5 10-5 m s-1, why does
the light come on “immediately” when you throw the switch?
M&I
18.6Feedback
Feedback during the initial transient produces the right amounts of
surface charge to create the appropriate steady state field.
… it also maintains these steady-state conditions …
… feedback leads to current equalisation …
Two cases:
- - - -
i1 i2
i1 > i2i1
i1 < i2
i2
i1 i2
Negative surface charge
buildup until i1 = i2
- - - -
+ + +i1 i2
Positive surface charge
buildup until i1 = i2
+ + +
What happens if we bend a wire which is carrying a current?
i- - - - -
- - - - -
-
-
-
-
-
-
---
--
Extra charge builds up on the
bend until there are enough
there to repel on-coming
elections just enough too make
them turn the corner, without
running into the side of the
wire.
In summary …
Feedback in a circuit leads to surface charges and steady
state current: inside a metal.
Feedback in static electricity situations leads to static
equilibrium: inside a metal.
0E
0E
Surface charge and resisitorsM&I
18.7
Consider a circuit container a “resistor”
comprising a thin section of Nichrome wire …
Charge will build up at various places on the
wire, as discussed before, but in particular, a
significant amount of charge will build up on
either side of the thin section. Why?“resistor”
i
Ewire
The electric field in the resistor
needs to be high enough for
there to be the same current in
the resistor as elsewhere in the
circuit.Eresistor
thick thick thin thinnA uE nA uE
or thickthin thick
thin
AE E
A
What about a wide resistor? It would need to be made of a
different material (say carbon) to the (Nichrome) wire, and
hence will have a different mobility u.
The steady state electric field in the carbon needs to be much
larger than the wire, hence electrons will tend to build up not
only on the outer surfaces of the wire and resistor, but also on
the interfaces between the wire and resistor in order to make
an electric field of large enough magnitude.
i
+ + + + + +
+ + + + + +
+
+
+
Ewire Ewire
Eresistor
A wide resistor: charges on the interface
Energy in a circuitM&I
18.8
Consider the path of a single electron as it moves around a
circuit: energy gained as it moves across the mechanical battery,
then lost in collisions with atomic cores …
Or we can think about the energy per unit charge gained or lost in
a trip around the circuit.
* We know that over any path the round-trip potential
difference must be zero.
The loop rule (energy conservation)
1 2 3 ... 0V V V along any closed path in a circuit
This is essentially the energy principle,
but on a per unit charge basis.
Potential difference across a battery NCF
Turn on the belt (with no external
circuit) and transport electrons from
the left, to the right hand plate.
The belt exerts a “non-Coulomb”
force on each electron.
Charge build up on the plates. These
charges exert a “Coulomb” force
on each electron being transported.
Eventually and the
motor cannot pump any more charge
and the plates are charged up as
much as they can be.
NCF C CeF E
of platesCE
NCF
C NCF FC C NCF eE F
of platesCE
Potential difference across a battery …2
If the distance between the plates of the mechanical battery
is s and the electric field EC of the charged plates is uniform
between the plates, then the potential difference across the
battery isNC
battery C
F sV E s
e
The quantity is the energy input per unit charge (a
property of the battery and is called the emf of the battery.)
The emf of a battery is measured in volts, although it is not
a potential difference.
NCF s e
Role of a battery:
A battery maintains potential difference across the terminals
of the battery, and this potential difference is numerically
equal to the battery’s emf.
Internal resistanceConnect a wire across the
terminals of the battery
... for a steady state, the transport
of electrons in the battery must
equal the current in the wire.
If there is no resistance to the
movement of charge in the battery,
then
C CeF ENCF
v
C NCF F
However, in any real battery there is
“internal resistance.”
The drift velocity in the battery: NCC
Fv u E
e
Since FNC is fixed, the maximum drift speed is when EC =0, which
means there is no charge on the ends of the battery and
We will assume (“ideal battery”) that u is high inside the battery, so
is reasonable even if FC is nearly as large as FNC, and
hence [... see later how to deal with real batteries ...]
v
0batteryV
emfbatteryV
Field and current in a simple circuit
In the situation alongside, the
electric field inside the
mechanical battery points in
the opposite direction to the
electric field in the
neighbouring wires ...
Starting at the negative plate and
going anti-clockwise ...
... potential increase of +emf across the battery
... then a potential drop of EL along the wire of length L.
For the round trip: 0battery wireV V
or emf
emf ( ) 0 EL EL
... gives a way of determining E and hence I enAuE
E
FNC
Einside battery
Two different paths
Following the dashed path
through wires L2 and L3:
Potential rise: +E2L2 along L2
Potential rise: +E3L3 along L3
Potential drop: emf through the battery
And along path through L1 and L3:
For the round trip: +E2L2 +E3L3 emf = 0
For the round trip: +E1L1 +E3L3 emf = 0
This implies that E1L1 = E2L2
... which makes sense since with the same starting and ending
points the two wires have the same potential difference
Also i3 = i1 + i2 due to the current node rule.
L3
L1
L2
i1
i2
i3 = i1 + i2
i3i3
E3
E2
E1
General use of the loop rule
Consider one loop of
a multi-loop circuit:
loop
1
loop
1
loop
2
B C D
A F E
B C
A F
V2 = VC VB
V1 =
VB VA
V3 =
VF VC
V4 = VA VF 0
V1 + V2 + V3 + V4 = 0
Any round trip potential
difference must be zero:
Hence:
(VB VA)
+ (VC VB)
+ (VF VC)
+ (VA VF) = 0
Energy conservation circuits
M&I
18.9L1 L3
L2
A D
B C
E1
E1E2
E3
E4
E3
E1
V VB VA
VC VB
VA VD
E
E1
E2
E3
E4
Consider the circuit shown
which contains a (thin) resistor:
The electric field is the (negative)
gradient of the potential.
V1 + V2 + V3 + Vbattery = 0
( E1L1) + ( E2L2) + ( E3L3) + emf = 0
VD VC
Going around the circuit:
Applications of the theoryM&I
18.10
• The current node rule (conservation of charge):
In the steady state, for many electrons flowing
into and out of a node:
Electron current: net iin = net iout where
Conventional current: net Iin = net Iout where
i nAuE
I q nAuE
• The loop rule (conservation of energy):
In the steady state, for any round-trip path:
1 2 3 ... 0V V V
1. iA > iB
2. iA = iB
3. iA < iB
1 2 3 4 5
What comprises a current in a circuit?
1. Electrons push each other through the wire
2. Since there is no friction, no force is needed to keep
electrons moving
3. A nonzero electric field inside the wire keeps the
electrons moving
1 2 3 4 5
1. i1 > i2
2. i1 = i2
3. i1 < i2
4. Not enough information
1 2 3 4 5
1. v1 > v2
2. v1 = v2
3. v1 < v2
4. Not enough information
1 2 3 4 5
1. E1 = 4*E2
2. E1 = (1/4)*E2
3. E1 = (1/16)*E2
4. E1 = 16*E2
5. Not enough information
A1 = 4*A2
1 2 3 4 5
n1 = (1/3)*n2
1. E1 = 3*E2
2. E1 = (1/3)*E2
3. E1 = (1/9)*E2
4. E1 = 9*E2
5. Not enough information
1 2 3 4 5
1. Nothing will change
2. The right bend will become negative
3. The right bend will become positive
In the next tiny fraction of a
second, what will happen at
the RIGHT bend in the
wire?
1 2 3 4 5
1. Nothing will change
2. The left bend will become negative
3. The left bend will become positive
In the same tiny fraction of
a second, what will happen
at the LEFT bend in the
wire?
1 2 3 4 5
1. Inside the wire
2. On the surface of the wire
3. Both inside the wire and on the surface of the wire
Where will the excess
positive charge of the
right bend be located?
1 2 3 4 5
1. Esurface = 0
2. Esurface to the right
3. Esurface to the left
At location 4, what is the
direction of E due only to
the charges on the
surface of the wire?
1 2 3 4 5
The wires have the same
length L and cross-sectional
area A, but are from different
materials.
1. E2 = emf/(1.5*L)
2. E2 = emf/L
3. E2 = emf/(2*L)
4. E2 = 1.5*emf/L
Same u’s, but n1 = 2*n2
1 2 3 4 5
What is the pattern of electric field in this steady-state circuit?
1. 2.
3. 4.
1 2 3 4 5
What charges make the
electric field inside the
wire in this circuit?
1. The moving electrons inside the wire
2. Charges on the battery and the surface of the wire
3. Only charges on the battery
4. Only charges on the surface of the wire
1 2 3 4 5
Circuit 1: 1 battery, NiCr wire
length L
cross-sectional area A
electric field E1 inside wire
Circuit 2: 1 battery, NiCr wire
length (3L)
cross-sectional area A
electric field E2 inside wire.
Which statement is correct?
1. E1 = E2
2. E1 = 3*E2
3. E1 = E2/3
1 2 3 4 5
Circuit 1: 1 battery, NiCr wire
length L
cross-sectional area A
electric field E1 inside wire
Circuit 2: 1 battery, NiCr wire
length L
cross-sectional area (4A)
electric field E2 inside wire.
Which statement is correct?
1. E1 = E2
2. E1 = 4*E2
3. E1 = E2/4
1 2 3 4 5
Which statement is correct?
1. i1 = i2
2. i1 = 4*i2
3. i1 = i2/4
Circuit 1: 1 battery, NiCr wire
length L
cross-sectional area A
Circuit 2: 1 battery, NiCr wire
length L
cross-sectional area (4A)
1 2 3 4 5
Energy conservation (loop) equation:
1. +emf – E*(2L1 + L2) = 0
2. +emf + E*(2L1 + L2) = 0
3. +emf – 2E1L1 – E2L2 = 0
4. +emf + 2E1L1 – E2L2 = 0
5. None of the above
1 2 3 4 5
Current conservation (node) equation:
1. i1 = 2*i2
2. 2i1 = i2
3. i1 = i2
4. i1 = (A2/A1)*i2
5. None of the above
1 2 3 4 5
What is E2?
1. 50.4 V/m
2. 12.86 V/m
3. 3.15 V/m
4. 0.788 V/m
5. None of the above
emf = 1.5 V
n = 9 1028 electrons/m3,
u = 7 10-5 (m/s)/(V/m)
L1 = 0.2 m, L2 = 0.05 m
A1 = 9 10-8 m2, A2 = 1.5 10-8 m2
1 2 3 4 5
Applications of the theory
Bulbs in parallel
...refer to laboratory on circuits ...
Connect two identical light
bulbs in parallel with a
battery ...
Both shine with same
brightness ...i3 = i1 + i2
i1
i2
Remember “brightness” equates to “resistance”
For a path through one bulb:
And the other:
2 emf 0EL
(L = filament length)2 emf 0EL
Electric field is thus the same in each light bulb: 2 emf
EL
Bulbs in parallel ...2
Why does the current divide through parallel resistors?
Consider the circuit alongside ... in
steady state ...
...containing two wide resistors in
parallel ...
Electrons move into the dead-end
until the surface charge there
becomes so negatively charged
that no more electrons can enter.
(Effect is that the wire seems
slightly wider at the junction, but
no more electrons move into the
dead-end branch.)
Bulbs in parallel ...3
Now complete the parallel
connection ... leads to a
rearrangement of the surface
and interface charges.
Some electrons now take the
upper branch and some the
lower branch.
There is also a larger current
through the battery and a larger
gradient of surface charge along
the wires to drive the larger
current.
Current in each branch depends on the mobility in each branch ...
... surface charge might build up differently in each branch
... and hence a different current in the branches.
Very important !
Work through ....
M&I Example problem: A circuit and a wide wire
No shortcuts!
A circuit and a wide wire …2
