Post on 17-Feb-2019
PHYS-A0130 ElectromagnetismLectures follow Chaps. 14-19 of
1
Schedule for the period 20.2-31.3.2017:
Lectures: 12 hours; a quiz exam 28.3 (215, Otakaari 4) gives 20 % of the final grade.
Exercise sessions: H01 and H02; some exercises are solved in the class and some aregiven as homework; the homework submission deadline is each Tuesday 17:00 (viaMyCourses); 40 % of the final grade from the home exercises.
Project sessions: P01-P04; the 3d, 4th and 5th weeks of the course (the first session 9.3);each group prepares a plan before each experiment, conducts the experiment andsubmits a report; the reports comprise 40 % of the final grade. 2
Monday Tuesday Wednesday Thursday Friday
8-9 H02, Y228b P03, U020
9-10 H02, Y228b P03, U020
10-11 28.3 : Quiz, 215 H01, Y228b P03, U020
11-12 28.3: Quiz, 215 H01, Y228b
12-13 P04, U020
13-14 P02, U020 P04, U020
14-15 Lecture, A1 P01, U020 P02, U020 P04, U020
15-16 Lecture, A1 P01, U020 P02, U020
16-17 P01, U020
Electric currents (§ 14)
3
• Concept and properties of electric current• Joule’s and Ohm’s laws• Connections of resistors and power supplies• Kirchhoff’s rules• Resistivity and electric current density
Concept and properties of electric current
4
Electric current produces• Magnetic field (right screw rule)• Force between wires• Heating of the wire
I
e
+−H
Force balance:
⇒ An aluminum wire of 1-mm diameter can levitate at r = 1 mm at I = 3.3 A
rIIk
lF 21=
rlIkmg
2
=
270
AN102
2−×==
πµk,I1
I2r
l
RconstIP
==2
5
1840) law, s(Joule' 2RIP =⇒
Concept and properties of electric current• Heating
heat capacity resistance [ohms]
increased using a higher voltage
1
2
3
123
cH2O ≈ 4 kJ/(lK)CH2O = cH2OV
dtdTCP =
6
Electromotive force of a power supply. Ohm’s law
E
increased by shortenning the heating wire (changing R)
I
circ2 ⇒= RI== const
IP
emf [volts]E IP =⇒ E circIR=E
In general, the potential difference (voltage)
between points A and B is (Ohm’s law) IRV =A B
I R
“web of naked fancies” in 1830; full recognition in 1840P = IV
coilcirc RR ≈
7
Emf versus voltage
A B
I
R
r
E
IRV =BARrI
+= and
rRRV+
=⇒ BA E
E
8
Connections of resistors and power suppliesIn series:
In parallel:
In series:
In parallel: For identical units
R1 R2 R3
R
IRRRRIIRIRIRVVVV
=++=++=++=
)( 321
321321
R1
R2
R3
I1
I2
I3
I I
RV
RRRV
RVRVRVIIII
=
++=
++=++=
321
321321
111
///
321 RRRR ++=
321
1111RRRR
++=
E = E1 + E2 + E3
E
E1 E2 E3
I
321ext rrrRI
+++= E1 + E2 + E3
E1 = E2 = E3= E
E1
E2
E3
I1
I2
I3
I I
3/311113/
iiiii
i
rrrrrrr
II
=⇒=++=
=
r, internalresistance
I
9
Kirchhoff’s rulesKirchhoff’s junction rule:The sum of the currents flowing into the junction is equal to the sum of the currents leaving the junction
Kirchhoff’s loop rule:The sum of emfs in any loop is equal to the sum of all the IR drops around the loop
R1
R2
R3
I1
I2
I3
E1
E2
213 III +=
11221 RIRI +−=− E
also
E E 113321 RIRI +=+−
R1
R2
R3
I1
I2
I3
E1
E2
charge decreases across the battery in the direction of
current is opposite to
10
What is R of a piece of a wire?
R = ?AlR ∝
R1 R2 R3
321 RRRR ++=
R1
R2
R3
321
1111RRRR
++=
The proportionality coefficient is a parameter of the material. It is called the resistivity, ϱ [Ωm].
AlR =
The reciprocal of resistivity is conductivity σ = 1/ϱ. Both ϱ and σ depend on T.
Examples of a conductor and an insulator: ϱAg = 1.6×10-8 Ωm, ϱglass = 1010 - 1014 Ωm
ϱ
At any point,
The voltage drop
J is a vector, and for its arbitrary direction, the differential Ohm’s law is
where is operator nabla
If J is known, the total current through any
surface A is
11
Current density and differential Ohm’s law
I∆ nA∆
z
z∆
ρρρ JdzdVzJ
AzI −=⇒∆=
∆∆
∆=
RIV ∆∆=∆
ρJ−=∇V
zyx ∂∂
+∂∂
+∂∂
=∇ zyx ˆˆˆ
.∫∫∫∫ ⋅=⋅=AA
ddAI AJnJ
J⋅n = Jnn = J cosθ
V∇
An
J
nJn
θ
ϱ ϱ ϱ
ϱ
dAdI
AI
AnnJ =
∆∆
=→∆ 0
lim
12
Electric fields (§ 15)
• Electric charge• Electric field and Coulomb’s law• Electric flux and electric flux density• Electric field of point charges• Gauss’ law for electrostatics• Electric potential• Dielectric media• Capacitors
13
Electric charge• Electric charge (Q) can be positive as that of proton
or negative as that of electron
• Opposite charges attract (→←) each otherand similar repel (←→) each other
• Charge is quantized: e = ±1.6×10-19 C
• In electric conductors, electrons are free to migrate, which leads to conductivity (in insulators they stay with their own atoms or molecules)
• Electric current = electric charge per unit time:
In terms of charge density , the current density is
+ −+ − + + + − −−
I
dtdQI =
dVdQ
=ρ vAIJ ρ=≡ /
dQvdt
electron
protonneutron
14
Electric field and Coulomb’s lawElectric field is a force field created by and acting on electric charges
• Electric field strength E is the force per unitcharge:
• Electric field lines are along the force imposedon a positive charge:
qFE =
+ +
+ +
qF
E
+
normal to the surface of a conductor
+++ + ++ + + + + +++E = 0
Coulomb’s law:
rErF ˆ4
)( ˆ4
12
112
21
rqq
rqq
πεπε=⇒=
15
Electric flux and electric flux density
2QQ
• Total flux of the electric field is
In SI, the proportionality coefficient is 1 and we have
.QE ∝Φ
QE =Φ
• Electric flux density D (local flux per unit surface area) must be proportional to E. It is a vector
, where A is perpendicular to D, andε is the electric permittivityε0 = 8.85×10-12 CV-1m-1
E
• Total flux through any surface area A is
∫∫∫∫ =⋅=ΦAA
E dADd αcos AD
E2
EdD ε=Φ
≡ ˆdA
d E
Superposition principle:
For a continuous distribution of ρ :
16
Electric field of point charges
q
ED,
2222
4
4
4 4
rqE
rqD
rDrA E
πεπππ =⇒=⇒
Φ=⇒=
independent of ε depends on ε
i
ii
ii
i
i
ii rr
q rrrEEE ==⇒= ∑∑ ˆ ,ˆ4
1 2πε
∆qi
ri
iE∆
ρ
r1
q1
1E
q2
q3
2E3E
r2
r3
dVdqqi ρ=→∆
∫∑ →Vi
rrE ˆ4
1)( 2∫∫∫=⇒V r
dVρπε
1E2E
3EE
17
Gauss’ law for electrostatics
∑∫∫ ==⋅=ΦS inside
i
iS
QqdAD
q1 q2
q3
q7
q4
q6q5
S
72 qqdS
+=⋅∫∫ AD
a>r a<rExample: Spherical charge distribution
20
20
4
4
rQE
QrE
πε
πε
=⇒
= 3
32
0 4arQrE =πε
304 a
QrEπε
=⇒
18
Electric potentialqEF =
ABAB qVxEqW −=∆=
xEV ∆−=⇒ AB
sF ⋅=ABW , if F is constant as in the picture.
Path 1:1l∆ 2l∆α β
x∆q
is independent of the path.
In general:
The electric potential V(r) is
∫ ⋅−=r
r
sEr0
)( dV
and r0 corresponds to V(r0) = 0.
The mechanical work done by the field is
x∆
QrA
rB
rQdr
rQdVQ πεπε 44
)( 2
A
B
=−=⋅−= ∫∫∞
=
∞→
rrr
r
rEr
AB21ACB )coscos( qVllEqW −=∆+∆= βαPath 2:
∫ ⋅−=−≡A
BABAB )()( sE drVrVV
x
E
qVtVItP BA=∆=∆≡