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=∆=∆≡