Hf. 29 Magnetiese Velde transperant 1

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FSK 126 Semester 2, 1999

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http//www.up.ac.za/science/phys/study126.htm
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WERKPLAN Halliday, Resnick Walker 5 ed.
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• 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
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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

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Die vektor produk (kruis produk) bl. 46,
Probleem voorbeeld 3-7
Gelaaide deeltjie - verskillende snelhede
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• 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
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Opsomming

• SI eenheid vir B

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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)
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J.J. Thomson (1897) m/q
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29-3 Kruis velde Ontdekking van die elektron

• Kruis velde 2 velde ? aan mekaar

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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

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Vraag 10 (bl. 720)

• Wat is die lading van die deeltjie?

NEGATIEWE (elektron)
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Vraag 6 (bl. 720)
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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
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• 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)
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29-5 n Gelaaide deeltjie in sirkel beweging
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Vraag 12 (bl. 721)
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• Opsomming (lesing 3)
• Hall effek (1879)
• Tipe, aantal en mobiliteit (lading draers)
• Wanneer v ? B is ? sirkel beweging
• Massa spektrometer (m/q)
• Regterhand reel

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Spiraal bane
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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
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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
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  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    
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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?  
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  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