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FSK 126 Semester 2, 1999

- Formaat van lesings
- Interaktiewe leer materiaal
- Notas
- Hardekopie
- Web
- Klastoetse
- Semestertoets
- Praktiesetoets
- Eksamen

http//www.up.ac.za/science/phys/study126.htm

WERKPLAN Halliday, Resnick Walker 5 ed.

- Hoofstuk 29 Magnetiese Velde
- Die wetenskap van magnetisme het n oorsprong in

die antieke tyd. Dit het gegroei vanaf die

waarrneming dat sekere klippe wat natuurlik

voorkom mekaar aantrek asook klein stukkies van

een metaal nl. Yster aantrek, maar nie ander

metale soos goud of silver aantrek nie. Die woord

magnetisme kom van die naam van die omgewings

(Magnesia) in Asia Minor, een van die plekke waar

die klippe gevind is.

Doelstellings Magnetiise Veld (B) en veld

lyne Krag (F) op n gelaaide deeltjie

geleier Kruis velde Gelaaide deeltjies in

sirkel beweging Siklotrons and

sinchrotrons

29-1 Die magnetiese veld (B) Historiese

Perspektief China (13th century

BC) Kompas Grieke (800 BC) Fe3O4 trek yster

aan P. de Maricourt (1269) Orienteer kompas

naald W. Gilbert (1600) Aarde is n magneet J.

Michell (1750) Magnetiese kragte H. Oersted

(1819) Magn. and elek. M. Faraday

(1820) Magn. and elek. J. Henry (1820)

Magn. and elek. J.C. Maxwell (1860) Vergelyking

s

29-1 Die magnetiese veld (B) verv.

Me

- Grootte
- Bewegende gelaaide deeltjie (ie. Elektron)
- Rigting (B-veld is n vektor hoeveelheid)
- Sterkte van ? digtheid van veld lyne
- N-pool na S-pool
- ???

29-2 Die Definisie van B

Die vektor produk (kruis produk) bl. 46,

Probleem voorbeeld 3-7

Gelaaide deeltjie - verskillende snelhede

- Probleem
- Wat is die netto krag (grootte en rigting) van

die elektron wat in die magneetveld in die figuur

beweeg as B 2 T, v 4 x 104 m/s, en ? 30o?

Regterhand reel

Opsomming

- SI eenheid vir B

Teenoorgesteldes trek aan

- Probleem voorbeeld 29-1
- Wenk
- Omskakeling van kinetiese energie van 5.3 MeV na

J - (5.3 ? 106 eV)(1.6 ? 10-19) 8.48 ? 10-13 J
- Hoek tussen v en B is 900
- klein krag werk in op n klein massa? groot

versnelling. - Toetspunt 1
- Toepassing van die regterhand reel

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Opsomming (lesing 1)

J.J. Thomson (1897) m/q

29-3 Kruis velde Ontdekking van die elektron

- Kruis velde 2 velde ? aan mekaar

Toetspunt 2 (bl. 706) (a) FB 0, wanneer v

?? aan B is (b) netto krag 0 FB FE

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- Stap 1 B-veld rigting
- ? B is uit die bladsy uit
- (a)
- boonste plaat
- (b)
- onderste plaat

Vraag 10 (bl. 720)

- Wat is die lading van die deeltjie?

NEGATIEWE (elektron)

Vraag 6 (bl. 720)

Opsomming (lesing 2) Thomson se eksperiment

(1897) m/q verhouding van gelaaide deeltjies E

? B Kruis velde Wanneer v E/B, geen

uitwyking Regterhand reel

- 29-4 Kruis Velde Die Hall Effek
- Hall Effek (E.H. Hall (1879))
- Draer lading
- Draer digtheid

Probleem voorbeeld 29-2, Toetspunt 3 (bl. 708)

29-5 n Gelaaide deeltjie in sirkel beweging

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Vraag 12 (bl. 721)

- Opsomming (lesing 3)
- Hall effek (1879)
- Tipe, aantal en mobiliteit (lading draers)
- Wanneer v ? B is ? sirkel beweging
- Massa spektrometer (m/q)
- Regterhand reel

Spiraal bane

29-6 Siklotrons and Sinchrotrons siklotron

Versnel strome van gelaaide deeltjies deeltjie

energie 0-50 MeV Sinchrotron Versnel strome

van gelaaide deeltjies deeltjie met energie

tot 1TeV

http//lynx.uio.no/cycdescr.html

http//www2.slac.stanford.edu/vvc/accelerators/cir

cular.html

Voorbeeld (41E) n Fisikus ontwerp n siklotron

om protone tot een tiende van die snelheid van

lig te versnel. Die magneet sal n veld van 1.4 T

produseer. Bereken (a) die radius van die

siklotron en (b) die ooreenstemende ossilator

frekwensie. Relativiteit oorwegings is nie

nodig. qproton 1.602 x 10-19 C,

mproton 1.67 x 10-27 kg c 3 x 108 m.s-1

29-7 Magnetiese krag op n stroom draende

draad Voorheen gelaaide deeltjies (bv.

Elektrone en protone) Eks. Voor positiewe

draers, ? 90o, v vd

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Voorbeeld (51P) n Lang, starre geleier, wat

langs die x-as lê, dra n stroom van 5.0 A in die

negatiewe rigting. n Magneet veld met grootte

B 3.0i 8.0x2j is teenwoordig, met x in meter

en B in milli-tesla. Bereken die krag op n 2.0

m segment van die geleier wat tussen 1.0 m and

3.0 m lê.

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Tutorial 1 2E. An alpha particle travels at a

velocity v of magnitude 550 m/s through a

uniform magnetic field B of magnitude 0.045 T.

(An alpha particle has charge 3.2 x 10-19 C and

a mass of 6.6 x 10-27 kg.) The angle between v

and B is 52?. What are the magnitudes of (a) the

force FB acting on the particle due to the field

and (b) the acceleration of the particle due to

FB? Does the speed of the particle increase,

decrease or remain 550 m/s? Solution (a) FB

?q?vB sin ? (3.2 x 10-19 C)(550 m/s)(0.045

T)(sin 52?) 6.2 x 10-18 N. (b) a FB / m

(6.2 x 10-18 N) / (6.6 x 10-27 kg) 9.5 x 108

m/s2. (c) Since it is perpendicular to v, FB

does not do any work on the particle. Thus from

the work-energy theorem both kinetic energy and

the speed of the particle remain

unchanged. 8E. A proton travels through

uniform magnetic and electric fields. The

magnetic field is B ?2.5i mT. At one instant

the velocity of the proton is v 2000j m/s. At

that instant, what is the magnitude of the force

acting on the proton if the electric field is

(a) 4.0k V/m, (b) ?4.0k V/m, and (c) 4.0i

V/m? Solution The net force on the proton is

given by (a) F FE FB qE qv x B (1.6 x

10-19 C)(4.0 V/m) k (2000 m/s) j x (?2.5 mT)

i (1.4 x 10-18 N) k (b) In this case F

FE FB qE qv x B (1.6 x 10-19 C)(?4.0

V/m) k (2000 m/s) j x (?2.5 mT) i (1.6 x

10-19 N) k In this case (c) F FE FB qE

qv x B (1.6 x 10-19 C)(4.0 V/m) i (2000

m/s) j x (?2.5 mT) i (6.4 x 10-19 N) i (8.0

x 10-19 N) k

The magnitude of FB is now 12P. An

electron is accelerated through a potential

difference of 1.0 kV and directed into a region

between two parallel plates separated by 20 mm

with a potential difference of 100 V between

them. The electron is moving perpendicular to the

electric field when it enters the region between

the plates. What magnetic field is necessary

perpendicular to both the electron path and the

electric field so that the electron travels in a

straight line? Solution Let F q(E v x B)

0. Note that v ? B so ?v x B? vB.

Thus 19E. An electron is accelerated from

rest to a potential difference of 350 V. It then

enters a uniform magnetic field of magnitude 200

mT with its velocity perpendicular to the field.

Calculate (the speed of the electron and (b) the

radius of its path in the magnetic

field. Solution (a) In the accelerating

process the electron loses potential energy eV

and gains the same amount of kinetic energy.

Since it starts from rest, ½ mev2 eV and

(b) The electron travels with constant

speed around a circle. The magnetic force on it

has magnitude FB evB and its acceleration is

v2/R, where R is the radius of the circle.

Newtons second law yields evB mev2/R, so

24E. Physicist S. A. Goudsmit devised a method

for measuring accurately the masses of heavy ions

by timing their periods of revolution in a known

magnetic field. A singly charged ion of iodine

makes 7.00 rev in a field of 45.0 mT in 1.29 ms.

Calculate its mass in unified atomic mass units.

Actually, the mass measurements are carried out

to much greater accuracy than these approximate

data suggest. Solution The period of

revolution for the iodine ion is T 2?r/v

2?m/Bq, which gives 33P. Two types of

singly ionized atoms having charge q but masses

that differ by a small amount ?m are introduced

into the mass (see figure 2). (a) Calculate the

difference in mass in terms of V, q, m (of

either), B, and the distance ?x between the spots

on the photographic plate. (b) Calculate ?x for a

beam of singly ionized chlorine atoms of masses

35 and 37 u if V 7.3 kV and B 0.50

T. Solution (a) From m B2qx2/8V we have ?m

(B2q/8V)(2x?x). Here x 8Vm/B2q1/2, which we

substitute into the expression for ?m to obtain

(b) 46E. A wire of 62.0 cm length

and 13.0 g mass is suspended by a pair of

flexible leads in a magnetic field of 0.440 T

(Fig. 29-43). What are the magnitude and

direction of the current required to remove the

tension in the supporting leads?

46E The magnetic force on the wire must be

upwards and have a magnitude equal to the

gravitational force on the wire. Apply right-hand

rule to show that the current must be from left

to right. Since the field and the current are

perpendicular to each other the magnitude of the

magnetic force is given by FB iLB, where L is

the length of the wire. The condition that the

tension in the supports vanish is iLB mg, which

yields

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ANTWOORDE (SELF-EVALUASIE)

1. d 2. c 3. a 4. b 5. a 6. d 7. a

8. a 9. c 10. e

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