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TRANSCRIPT
M&IChapter 16
Electric potential
Energy of a single particle
q
Particle energy 2mc K= +2 21
2mc mv= + for v c
final initial externalK K K WΔ = − =
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 = ΔF ri
M&I16.1
Then
Potential energy associated with pairs of interacting particles
sys ext intK W W QΔ = + +
For a system of interacting particles:
or sys int extK W W QΔ − = +
if ( )2 0mcΔ =
Write intW U− = Δchange in potential energy U
Then sys extK U W QΔ + Δ = +
Systems of charged particles:
sys 0K U∴Δ +Δ =
…in most cases considered here, ext 0 and 0W Q= =
Uniform electric field between plates
proton electric 0K UΔ + Δ =
−
+
+++
+
+
−−−
−E
electricF
( ) electric 0final initialK K U− + Δ =
increases decreases
( ) int 0final initialK K W∴ − − =
( ) 0final initialK K∴ − − Δ =F ri
( ) 0final initial xK K eE x∴ − − Δ =
Proton
moving to the right
in a uniform
electric field
ΔrM&I16.2
Uniform electric field between plates
electron electric 0K UΔ + Δ =
decreases increases
( ) int 0final initialK K W− − =
( ) 0final initialK K∴ − − Δ =F ri
( ) ( ) 0final initial xK K e E x∴ − − − Δ =
Electron
moving to the right
in a uniform
electric field
( ) 0final initial xK K eE x∴ − + Δ =
−
−
+++
+
+
−−−
−E
Δr
electricF
Electric Potential DifferenceM&I16.3
( )electric x xU eE x e E xΔ = − Δ = − Δ
In the previous examples:
For the proton:For the electron: ( )electric x xU eE x e E xΔ = + Δ = − − Δ
Let potential difference: xV E xΔ = − Δ
Then electricU q VΔ = Δ Units of : volts (V)VΔ
If is not parallel to EΔr
then ( )x y zV E x E y E zΔ = − Δ + Δ + Δ
or VΔ = − ΔE ri
+
Δr
E
θ
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 eΔ = = × ×
-191.6 10 joule = 1 electron volt = 1 eV×
1 keV
= 1000 eV1 MeV
= 106
eV
Sign of potential difference and direction of electric fieldM&I16.4
A B
E
Δr1.
A B
E
Δr2.
3. A BE
Δr
0V∴Δ >
VΔ = − ΔE ri
0Δ >E ri0V∴Δ <
0Δ <E ri
0V∴Δ =0Δ =E ri
Example: Field and potential
B A
cos
V V V
E r θ
Δ = −
= − Δ= − Δ
E ri
30°
2 m
B
A
EΔr
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= Δ = Δ
=(1.6 ×
10-19)(−173) = −
2.76 ×10-17 J or −173 eV
Potential difference in a non-uniform fieldM&I16.5
If we move through two (or more) regions where the electric field is different, then
final initialV V VΔ = − = − Δ∑E ri
if each is small enough that is uniform in the region through which it passes.
Δr E
Remember that x y zE x E y E zΔ = Δ + Δ + ΔE ri
For example:
( ) ( )B A
C A B C
1 1 2 2
1 2( ) ( )x C A x B C
V V VV V V V
E x x E x x
Δ = −
= − + −
= − Δ − Δ= − − − −
E r E ri i
If the electric field in a region varies continuously, then we need to integrate:
final initial
f
iV V V dΔ = − = −∫ E ri
Note that …
potential difference is independent of path
or x y zV E dx E dy E dzΔ = − − −∫ ∫ ∫
Change of electric potential in a non-uniform field …2
drE
+
i
f
M&I uses the notation2
1V dΔ = −∫ E li
M&I16.6
… and for a round trip 0VΔ =
C
C
A AV V V dΔ = − = −∫ E ri
Path independent
f f f
i i i
f x y z
x y zi x y zV d E dx E dy E dzΔ = − = − − −∫ ∫ ∫ ∫E ri
Hence write:
M&I16.7
Electric field inside and outside a current carrying wire
The charges are moving, hence the electric field E inside the wire is non-zero.
0VΔ > though the wire
Hence along path 2 (in the air) …… hence E is non-zero in the air.
0VΔ >
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. +150V2. −150V3. +300V4. −300V5. +600 V6. −600 V
-1ˆ300 N C= −E j
x x
x
R S
T
E
VΔ
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. +200V2. −200V3. +400V4. −400V5. +800 V6. −800 V
-1ˆ400 N C=E j
x x
x
R S
T
E
VΔ
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 V2. −300 V3. −500 V4. −600V5. −1000 V
-1ˆ ˆ200 300 N C= − −E i j
x x
x
R S
T
E
VΔ
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
EVΔ 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 C= − −E i j
x x
x
R S
T
E
VΔ
1. +200V2. −200V3. +800V4. −800V4. +1000 V6. −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 V2. −400V3. +400V4. −800V5. +800 V
-1ˆ400 N C= −E j
x x
x
R S
T
E
VΔ
1 2 3 4 5
Without doing any calculations, what is the sign of VB
−
VA
?
1. positive2. negative3. 0
1 2 3 4 5
What is VB
−
VA
?
1. –20 V2. –10.05 V3. –8.06 V4. –0.1 V5. none of the above
1 2 3 4 5
The potential at one locationM&I16.8
final initialV V VΔ = −set Vinitial
= 0 at “infinity”
For the potential at a distance r from a single point charge q
20 0 0
1 1 14 4 4
rr r
r r xq q qV V V E dx dxx x rπε πε πε∞ ∞ ∞
∞
= − = − = − = =∫ ∫
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 is
A AU qV=
For two point charges separated by a distance r: 1 2
0
14
q qUrπε
=
V(x,y)
x
y
The electric potential in 2D. V(x,y) around a positive point charge.
Lines of equipotential
0
14r
qVrπε
=
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
r 4r
a b
c
A B
CInsert 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 energyFor 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 …2V(x)
0x
(c)
A B
A B xV(x)
0x
(f)
A B
A B x
V(x)
0x
(d)
A B
A B xV(x)
0x
(e)
A B
A B x
Electric potential and electric field …3V(x)
0x
(g)
A B
A B xV(x)
0x
(j)
A B
A B x
V(x)
0x
(h)
A B
A B xV(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 1V V V V V d d dΔ = − = Δ + Δ = − = − −∫ ∫ ∫E r E r E ri i i
due to shell:VΔ
surface of shell 1 0V V− = since inside shellshell 0=E
Outside shell: 1shell 2
0
1 ˆ4
Qrπε−
=E r
2
2 surface of shell shell3
21
230
1 1
0 1 1
1 ( )4
14
V V d
Q dxx
Q QR d R
πε
πε
∴ − = −
−= − −
⎛ ⎞− −= −⎜ ⎟+⎝ ⎠
∫
∫
E ri
We move opposite to the direction of the field, therefore 0VΔ >Check:
3
Why is there a sign here?−
due to disk:VΔ
Since and : 2d R 1 2R R2
2 2disk
02Q Rπ
ε≈E
2 22 2 2 2 2 22 1 disk 11 1
0 0
( ) ( )2 2
Q R Q RV V d dx d Rπ πε ε
∴ − = − = − − = +∫ ∫E ri
We move opposite to the direction of the field, therefore 0VΔ >
Check:
due to both shell and disk:VΔ
21 1 2 2
2 1 10 1 1 0
1 ( )4 2
Q Q Q RV V d RR d R
ππε ε
⎛ ⎞− −∴ − = − + +⎜ ⎟+⎝ ⎠
A metal in static equilibrium
For a metal in static equilibrium:
final initial 0V V VΔ = − =
… 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 voltsVEsΔ
= = 2000 V m-1
−
−
−−
−
+
++
++
A
s1Q− 1Q+
−
−
−−
−
+
++
++
A
1Q− 1Q+
1 mm
+
++
++
−
−
−−
−
2Q−2Q+
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 VΔ = Δ = (2000 V m-1)(0.001 m)= 2 volts
capacitorV∴Δ = 2 V + 0 V + 2 V = 4 volts
Potential difference in an insulator
Again start with a charged capacitor …
−
−
−−
−
+
++
++
A
s Q−Q+
Eplates
−
−
−−
−
+
++
++
A
Q+ Q−Eplates
−
+ −
+
−
+ −
+
−
+ −
+
−
+ −
+
−
+ −
+
… now insert an insulator …
What is the effect on the electric field inside the capacitor?
M&I16.9
Inside the plastic, is complex
Outside the plastic …
consider around the closed path shown
ΔE ri… will be positive outside the plastic
Δr
Δr
… therefore the average field inside the plastic must point to the left
ΔE ri
Edipoles
−
−
−−
−
+
++
++A
Q+ Q−
−
+ −
+
−
+ −
+
−
+ −
+
−
+ −
+
−
+ −
+
EplatesEdipoles
Enet
… result is that the electric field inside the capacitor is reduced.
appliednet K=
EE
where K is the dielectric constant
Electric field inside a capacitor with a dielectric constant =0
QAKε
K always > 1
and vacuuminsulator
VVK
ΔΔ =
Effect of dielectric: …
decreases the electric field… decreases the potential difference
Dielectric constants for various insulators
Vacuum 1 (by definition)Air 1.0006Typical plastic 5Sodium chloride 6.1Water 80Strontium titanate
310
Energy density associated with electric field
= J m-32102 Eε
… general result …
do it yourself …M&I16.10
Originally ΔV was –1000 volts. A metal slab is inserted into the capacitor. Now ΔV = VB
–
VA
=
1) + 1000 volts2) +500 volts3) 0 volts4) –500 volts5) –1000 volts
1 2 3 4 5
With a plastic slab in the capacitor: Now ΔV = VB
–
VA
=
1. between –500 and –1000 volts2. between +500 and+ 1000 volts3. –500 volts4. +500 volts5. not enough information to tell
1 2 3 4 5
Potential along the axis of a ringPotential 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
14
qVR zπε
=+
Adding up the potential contributed by all the point charges:
ring 2 2 2 2 2 20 0 0
1 1 1 1 4 4 4
q QV qR z R z R zπε πε πε
= = =+ + +
∑ ∑
M&I16.8
Potential along the axis of a uniformly charged disk
ΔE
j
i
R
rΔ
... with radius R and total charge Q
k
r
z
ring 2 2 2 20 0
1 1 24 4
q Q r rVAz r z r
ππε πε
Δ Δ= =
+ +
2 200
12
RQ rdrVA z rε
=+∫
2 2
00
12
RQ z rAε⎡ ⎤= +⎣ ⎦
( )2 2
0
12
QV z R zAε
∴ = + −
Check:2 2
0
1 12z
V Q zEz A z Rε
⎛ ⎞∂= − = −⎜ ⎟∂ +⎝ ⎠
M&I16.12
What is VB
−
VA
?
1. 270 V2. 18 V3. 6 V4. –6 V5. –18 V6. –270 V
1 2 3 4 5
What is VB
−
VA
?
1. +1350 V2. –1350 V3. +3375 V4. –3375 V5. none of the above
1 2 3 4 5
VP
−
VQ
is:
1. positive2. negative3. zero4. not enough information to tell
1 2 3 4 5
Along the semicircular path, VB
−
VA
is:
1. positive2. negative3. zero4. not enough information to tell
1 2 3 4 5
Which of the following quantities are zero?
1. VC
–
VA2. VD
–
VC3. VB
–
VD4. VC
–
VA
and VB
–
VD5. VC
–
VA
and VB
–
VD
and VD
–
VC6. none of the above
1 2 3 4 5
Along the straight path through the metal sphere, VB
−
VA
is:
1. positive2. negative3. zero4. not enough information to tell
1 2 3 4 5
M&IChapter 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&I17.1
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&I17.2
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)
02
ˆ4
qr
μπ
×=
v rB
0
-7 -1
permeability of free space
= 4 10 T m A
μ
π
=
×
0
4μπ
= 10-7
T m A-1 exactly
+r
v
B
θq
r
M&I17.3
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 sinθAB=×
θ A
The cross product
B
A
B
θ
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
iˆ−ikˆ−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
iˆ−ikˆ−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
jˆ−jkˆ−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
jˆ−jkˆ−k
1 2 3 4 5
Relativistic effectsM&I17.4
02
ˆ4
qr
μπ
×=
v rB v : velocity of source or observer?
+
vor
+v
Retardation
+ v
B
+ 0=v
0?=B
+ 0=v
0=B
t1 t2
t3
(no t in here)
Electron current i
Distance traveled by electron sea in time = v tΔtΔ
Drift speed of mobile electrons = v
Number of mobile electrons in shaded cylinder =
A
v tΔ
vE
nAv tΔ
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&I17.5
Metal 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&I17.5
I enAv=
The Biot-Savart
law for currents
02
ˆ 4
Ir
μπ
Δ ×Δ =
l rB
0
4μπ
= 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 lΔ
lΔ
Then the total contribution to =qv ( ) ( ) nA l q v q nAv l I lΔ = Δ = Δ
Now can write
where is a vector with magnitude pointing in the direction of the conventional current I
ΔllΔ
r
Δl
B
θI
r
M&I17.6
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
12
3
IP
A B
C
1. into the page2. out of the page3. 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 page2. out of the page3. zero
D E
1
2
3
Magnetic field of a straight wire
Magnitude of : r
Then( )22
ˆ ˆˆ x y
r x y
−= =
+ −
r i jr
( )22r x y= + −
ˆ ˆx y= −r i j
ΔB
y r
j
i
Δy
x
k
r
I
ˆyΔ = Δl j
02 2
ˆ ˆ 4
I yx y
μπ
Δ ×+j rThen magnetic field
due to small piece only =ΔB
( )0
2 2 22
ˆ ˆ ˆ 4
I y x yx y x y
μπ
Δ −= ×
+ + −
j i j
... of length L, carrying current I
M&I17.7
Magnetic field of a straight wire …2
( )0
2 2 22
ˆ ˆ ˆ 4
I y x yx y x y
μπ
Δ −Δ = ×
+ + −
j i jB ( )ˆ ˆ ˆ ˆx y x× − = −j i j k
( )32
0
2 2
ˆ4
I x y
x y
μπ
Δ∴Δ = −
+B k
Let and integrate over entire length L of wire (only Bz is non-zero):
( )32
20
2 2
2
4
L
zL
dyB Ixx y
μπ
+
−
=+∫
Tables of integrals
( )
2
2
0 02 2 2 22
... 4 4 2
L
L
zy LIB Ix
x x y x x L
μ μπ π
+
−
= = =+ +
0yΔ →
Magnetic field of a straight wire …3
( )0
22
4 2wire
LIBr r L
μπ
=+
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
IBr
μπ
=
Another special case :r L 02
4wireI lBr
μπ
Δ→
Direction of ? ... use right hand rulewireB
I
wireB
Magnetic field of a straight wire …4
wireB curls around the wire
( )0
22
4 2wire
LIBr r L
μπ
=+
I
B
Go to worksheets ...Electric currents produce magnetic fields (1 & 2)
VPython scriptBwire_with_r.py
Magnetic field along the axis of a circular loop of wire
... with radius R and current I
ˆ ˆR z= − +r 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
Ir
μπ
Δ ×l r ( )( )
32
0
2 2
ˆ ˆ ˆ
4
R R zI
R z
θμπ
− Δ × − +=
+
i j k
ˆR θΔ = − Δl i
M&I17.8
ΔB
j
iθ
k
r
z
Δl
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
θμπ
− Δ × − +Δ =
+
i j kB
( )2
ˆ ˆ ˆ
ˆ ˆ
R R z
zR R
θ
θ θ
− Δ × − +
= + Δ + Δ
i j k
j k
( )32
20
2 2
ˆ ˆ +4
zR RIR z
μ θ θπ
Δ + Δ∴Δ =
+
j kB
( )32
20
2 2
4z
IRBR z
μ θπ
ΔΔ =
+
ΔB
R
zzBΔ
r
Only Bz will be non-zero:
B_loop_with_r_dB.pySee:
( )32
20
2 2
2 = 4loop
I RBR z
μ ππ +
Check the result ... units? ... direction?
Special case: centre of the loop, z = 0
BI
0 2 = 4loop
IBR
μ ππ
Another special case :z R ( ) ( )3 32 22 2 2 3R z z z+ ≈ =
20
3 2 =
4loopR IB
zμ ππ
( ) ( )( )3 3
2 2
22 2
0 0
2 2 2 20
= 2 4 4z
I R I RB dR z R z
πμ μθ ππ π
=+ +∫
Now let and integrate around loop0θΔ →loop 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
iˆ−ikˆ−k
1 2 3 4 5
Which components of at the observation location are nonzero?
1. z2. y3. x4. y & z5. x & y6. x & z7. all components
1ΔB
1 2 3 4 5
Which components of at the observation location are nonzero?
1. z2. y3. x4. y & z5. x & y6. x & z7. all components
2ΔB
1 2 3 4 5
What is the direction of magnetic field at location A? ... and B?
1) +x2) –x3) +y4) –y5) +z6) –z7) 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 page2. out of the page3. 4. 5. zero
Magnetic dipole moment, μ
I
B Bμ
20
3 2 =
4axisR IB
rμ ππ
for r R
Write 03
2 = 4axisB
rμ μπ
Where the magnetic dipole moment
μ = IA
In an applied magnetic field, a current-carrying loop rotates so as to align the magnetic dipole moment with the field. μ
M&I17.9
A
The magnetic field of a bar magnet
… what about magnetic monopoles?
S N
M&I17.10
Be careful of pictures like this ...
03
2 = 4axisB
rμ μπ
… 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
A I
Say that a circuit containing a ¾
loop of wire 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
θ
d l
R
r
A
k axis out5
Earthˆ6 10 T −= ×B k
straight wires 0=B
03/4 loop 2
ˆ 4
I ddr
μπ
×= =∫ ∫ l rB B
Put the origin at the centre of the loop.
R=r (constant)
At all locations thereforeˆd ⊥l r ˆ sind dl dl Rdθ θ× = = =l r
( )2 2
20 0 03/4 loop 2 2 2 2
2 2
= =
4 4 4I Rd IR IRd
R R R
π π
ππ
π π
θμ μ μθ θπ π π
∴ = ∫ ∫B
503/4 loop
3 ˆ ˆ = ... = 4.7 10 T 4 2
IR
μ ππ
−= ×B k k
Antarctic loop …2
5Earth
ˆ6 10 T −= ×B k
53/4 loop
ˆ= 4.7 10 T −×B k
Therefore want
4Earth 3/4 loop
ˆ1.06 10 T −+ = ×B B k
( )4magnet
ˆ1.06 10 T −= × −B k
40magnet 3
2 1.06 10 T 4 zμ μπ
−≈ = ×B
( )( )
1130 7 -1 2 3
4magnet
2 1 10 T m A (2)(1.2 A m )4 0.13 m 1.06 10 T
zB
μ μπ
−
−
⎛ ⎞⎡ ⎤×⎜ ⎟
∴ = = =⎢ ⎥⎜ ⎟ ×⎢ ⎥⎣ ⎦⎜ ⎟⎝ ⎠
Antarctic loop …3
The atomic structure of magnets
M&I17.11
Each atomic current loop contributes an amount of magnetic field:
20 0
3 3 2 2=
4 4R I
r rμ μμ ππ π
B B
2 2e e evI
RT Rvπ π
= = =⎛ ⎞⎜ ⎟⎝ ⎠
and 2A Rπ=
2 12 2evIA R eRv
Rμ π
π⎛ ⎞∴ = = =⎜ ⎟⎝ ⎠
Bohr atomic model ...
Estimating the magnetic dipole moment: a simple model of the atom
+R
2
netd vm m Fdt R
= =v
2 2
20
14
v emR Rπε
∴ =
2
20
14
evR mπε
∴ = … 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 )2
eRvμ − −= = × ×Then
23 21.3 10 A m per atomμ −∴ ≈ ×
Estimating the magnetic dipole moment: quantized angular momentum
Orbital angular momentum: L Rmv=
( )1 1 12 2 2
e eeRv Rmv Lm m
μ = = =Then
L is quantized in units of = 1.05 ×
10-34
J s
23 20.9 10 A m per atomμ −∴ ≈ ×
191342
31
1 (1.6 10 C) (1.05 10 Js)2 (9 10 kg)
em
μ−
−−
×∴ ≈ = ×
×
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)
12
em
μ ≈
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 R 0z
NIBL
μ=
IL
N loops
RI
solenoid_drag.py
M&I17.13
M&IChapter 18
A Microscopic
View of Electric Circuits
electron current i = no. of electrons per second passing a pointelectron current flows in direction opposite to conventional current I = no. of coulombs per second = conventional current flows in the direction of
EE
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&I18.2
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&I18.3
The 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
Δ= =
Δp F
If an electron loses all its momentum in a collision, 0p p eE tΔ = − = Δ
If speed of electron << c, writee e
p eE tvm m
Δ= =
Averaging over all collisions:e
eE tv uEmΔ
= =
where is the electron “mobility”e
e tumΔ
=
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 thickthin
Av vA
=
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
ABCDAA
V dΔ = −∫ E li
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&I18.4
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 wire
Ebattery
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&I18.5
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&I18.6
Feedback
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
< i2i2
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.
0≠E
0=E
Surface charge and resisitorsM&I18.7
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”
iEwire
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.
Eresistorthick thick thin thinnA uE nA uE=
or thickthin thick
thin
AE EA
=
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&I18.8
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 lsot
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 VΔ + Δ + Δ + = 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 he 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 Ce= −F E
of platesCE
NCF
C NC= −F 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 is
NCbattery C
F sV E se
Δ = =
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 Ce= −F ENCF
v
C NCF F=
However, in any real battery there is “internal resistance.”The drift velocity in the battery: NC
CFv u Ee
⎡ ⎤= −⎢ ⎥⎣ ⎦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
0batteryVΔ =
emfbatteryVΔ =
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 VΔ + Δ =
or emfemf ( ) 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: +E2
L2
along L2Potential rise: +E3
L3
along L3
Potential drop: −emf
through the battery
And along path through L1
and L3
:
For the round trip: +E2
L2
+E3
L3
−
emf
= 0
For the round trip: +E1
L1
+E3
L3
−
emf
= 0This implies that E1
L1
= E2
L2
... 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
i1i2
i3 = i1 + i2
i3i3E3
E2E1
General use of the loop rule
Consider one loop of a multi-loop circuit:
loop 1
loop 1
loop2
B C D
A F E
B C
A F
ΔV2 = VC −
VB
ΔV1 = VB −
VA
ΔV3 = VF −
VC
ΔV4 = VA −
VF ≈
0
ΔV1 + ΔV2
+ ΔV3
+ ΔV4 = 0
Any round trip potential difference must be zero:
Hence: (VB −
VA
) + (VC
−
VB
) + (VF
−
VC
) + (VA
−
VF
) = 0
Energy conservation circuits
M&I18.9
L1 L3
L2
A D
B C
E1
E1E2
E3
E4E3
E1
V VB −
VA
VC −
VB
VA −
VD
EE1
E2
E3
E4
Consider the circuit shown which contains a (thin) resistor:The electric field is the (negative) gradient of the potential.
ΔV1 + ΔV2
+ ΔV3
+ ΔVbattery
= 0
(−E1
L1) + (−E2
L2
) + (−E3
L3
) + emf
= 0
VD −
VC
Going around the circuit:
Applications of the theoryM&I18.10
• 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 VΔ + Δ + Δ + =
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 change2. 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 change2. 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 wire2. On the surface of the wire3. 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 wire3. Only charges on the battery4. Only charges on the surface of the wire
1 2 3 4 5
Circuit 1: 1 battery, NiCr
wire length Lcross-sectional area A electric field E1
inside wire
Circuit 2: 1 battery, NiCr
wire length (3L)cross-sectional area Aelectric field E2
inside wire.
Which statement is correct?
1. E1
= E22. E1
= 3*E23. E1
= E2
/3
1 2 3 4 5
Circuit 1: 1 battery, NiCr
wire length Lcross-sectional area A electric field E1
inside wire
Circuit 2: 1 battery, NiCr
wire length Lcross-sectional area (4A)electric field E2
inside wire.
Which statement is correct?
1. E1
= E22. E1
= 4*E23. E1
= E2
/4
1 2 3 4 5
Which statement is correct?
1. i1
= i22. i1
= 4*i23. i1
= i2
/4
Circuit 1: 1 battery, NiCr
wire length Lcross-sectional area A
Circuit 2: 1 battery, NiCr
wire length Lcross-sectional area (4A)
1 2 3 4 5
Energy conservation (loop) equation:
1. +emf
–
E*(2L1
+ L2
) = 02. +emf + E*(2L1
+ L2
) = 03. +emf –
2E1
L1
–
E2
L2 = 04. +emf
+ 2E1
L1
–
E2
L2
= 05. None of the above
1 2 3 4 5
Current conservation (node) equation:
1. i1
= 2*i22. 2i1
= i23. i1
= i24. i1
= (A2
/A1
)*i25. None of the above
1 2 3 4 5
What is E2
?
1. 50.4 V/m2. 12.86 V/m3. 3.15 V/m 4. 0.788 V/m5. None of the above
emf
= 1.5 Vn = 9 ×
1028
electrons/m3,
u = 7 ×
10-5
(m/s)/(V/m)
L1
= 0.2 m, L2
= 0.05 mA1
= 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 emfE
L=
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