Chp. 29 Magnetic Fields slide 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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WORKPLAN Halliday, Resnick Walker 5 ed.
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• Chapter 29 Magnetic Fields
• The science of magnetism had its
  origin in ancient times. It grew from the
  observation that certain naturally occurring
  stones would attract one another and would also
  attract small bits of one metal, iron, but not
  other metals, such as gold or silver. The word
  magnetism comes from the name of the districts
  (Magnesia) in Asia Minor, one of the locations
  where the stones were found.

AIMS Magnetic Field (B) and field lines Force
(F) on a charged particle conductor Crossed
fields Circulating charged particle
cyclotrons and synchrotrons
29-1 The Magnetic Field (B) Historical
Perspective China (13th century
BC) Compass Greeks (800 BC) Fe3O4 attracts
iron P. de Maricourt (1269) Orientation of
compass needle W. Gilbert (1600) Earth is a
magnet J. Michell (1750) Magnetic forces H.
Oersted (1819) Magn. and elec. M. Faraday
(1820) Magn. and elec. J. Henry (1820)
Magn. and elec. J.C. Maxwell (1860) Equations
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29-1 The Magnetic Field (B) cont.

• Magnitude
• Moving charged particle (ie. Electron)
• Direction (B-field is a vector quantity)
• Strength of B ? density of field lines
• N-pole to S-pole
• ???

29-2 The Definition of B

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The Vector Product (cross product) pg. 46,
Sample problem 3-7
Charged particle - different velocities
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• Problem
• What is the net force (magnitude and direction)
  on the electron moving in the magnetic field in
  the figure if B  2 T, v  4 x 104 m/s, and
  ? 30o?

Right-hand rule
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SUMMARY

• SI unit for B

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Opposites attract

• Sample Problem 29-1
• Hints
• Conversion of kinetic energy of 5.3 MeV to J
• (5.3 ? 106 eV)(1.6 ? 10-19) 8.48 ? 10-13 J
• angle between v and B is 900
• small force acting on a small mass? large
  acceleration.
• Checkpoint 1
• Application of right hand-rule

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Summary (lecture 1)
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J.J. Thomson (1897) m/q
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29-3 Crossed Fields Discovery of the Electron

• Crossed fields 2 fields ? to each other

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Checkpoint 2 (pg. 706) (a) FB 0, when v is
?? to B (b) net force 0 FB FE
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• Step 1 B-field direction
• ? B is out of page
• (a)
• Upperplate
• (b)
• Lower plate

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Question 10 (pg. 720)

• What is the particles charge?

NEGATIVE (electron)
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Question 6 (pg. 720)
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Summary (lecture 2) Thomsons experiment
(1897) m/q ratio of charged particles E ?
B Cross - fields When v E/B, no
deviation Right-hand rule
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• 29-4 Crossed Fields The Hall Effect
• Hall Effect (E.H. Hall (1879))
• Carrier charge
• Carrier density

Sample problem 29-2, Checkpoint 3 (pg. 708)
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29-5 A circulating charged particle
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Question 12 (pg. 721)
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• Summary (lecture 3)
• Hall effect (1879)
• Type, number and mobility (charge carriers)
• When v ? B ? circular motion
• Mass spectrometer (m/q)
• Right hand rule

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Helical Paths
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29-6 Cyclotrons and Synchrotrons Cyclotron
Accelerates beams of charged particles particle
energies 0-50 MeV Synchrotron Accelerates
beams of charged particles particle energies up
to 1TeV
http//lynx.uio.no/cycdescr.html
http//www2.slac.stanford.edu/vvc/accelerators/cir
cular.html
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Solved example (41E) A physicist is designing a
cyclotron to accelerate protons to one tenth the
speed of light. The magnet used will produce a
field of 1.4 T. Calculate (a) the radius of the
cyclotron and (b) the corresponding oscillator
frequency. Relativity considerations are not
relevant. qproton  1.602 x 10-19 C,
mproton  1.67 x 10-27 kg c 3 x 108 m.s-1
29-7 Magnetic Force on a Current-Carrying
Wire Previously charged particles (ie.
electrons and protons) Exp. Cond positive
carriers, ? 90o, v vd
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Solved Example (51P) A long, rigid conductor,
lying along the x-axis, carries a current of 5.0
A in the negative direction. A magnetic field B
is present, given by B 3.0i 8.0x2j, with x in
meters and B in milli-Teslas. Calculate the
force on the 2.0 m segment of the conductor that
lies between 1.0 m and 3.0 m.
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ANSWERS (SELF-EVALUATION)
1. d 2. c 3. a 4. b 5. a 6. d 7. a
8. a 9. c 10. e
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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