MAGNETIC FORCE & MAGNETIC FIELD (Chapter 22)
Read historical background in 22.1
 Magnetism known of for over 2000 years.
 Magnets known to have two poles
o North points to Earth’s North Pole (which is a south magnetic pole!)
o South points to Earth’s South Pole (which is a north magnetic pole!)
 Poles cannot be separated – no magnetic monopoles (so far)
 Connection between magnetism and electricity dates to 1819 (Oersted)
 Important contributions in early 1820’s from Ampére, Faraday, Henry
 Relationship between magnetic field and changing electric fields and vice
versa formulated by Maxwell in 1861
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MAGNETIC FIELD  B
 Magnetic field is a VECTOR (important)
 Source can be permanent magnet OR nearby electric current
Can use MAGNETIC FIELD LINES to represent how magnetic field depends
on location in space
N
S
 Magnetic Field Lines – RULES
N
S
o Lines start and stop on poles of a magnet OR lines form closed loops
around current-carrying wires
 No such thing as a magnetic monopole
 For magnet – lines start at North pole and end at South pole

o MAGNITUDE of magnetic field | B | indicated by DENSITY of lines
o DIRECTION of magnetic field at a point is tangent to field lines at point
 Direction of magnetic field at some location is the direction that a
compass needle would point at that location.
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MAGNETIC FORCE FB on a moving charged particle (charge q, velocity v )
Experimental observations show:

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1 : For magnitude, find | FB |  q and | FB |  v BUT
st



FB  0 if v parallel to B



2 : For direction, find that if v and B not parallel, then FB perpendicular to both
nd
FB
B
θ
B
v
θ
v
FB
(negative charge)
(positive charge)



|
F
|

sin

3 : Find B
where  is angle between v and B
RD
Trickier than electric force on static charge

MAGNETIC FORCE FB on a moving charge – VECTOR CROSS PRODUCT

 
Observations above amount to: FB  q v  B
(a vector cross product)
VECTOR CROSS PRODUCT (review page 316)
 Operation on two vectors – result is a VECTOR
 
 Result of A  B is a VECTOR


o perpendicular to BOTH A AND B


o i.e. perpendicular to plane formed by A AND B
AxB
B
θ
 
 Direction of vector A  B
o Given by RIGHT-HAND RULE

 

o If fingers go from A → B , then thumb shows direction of vector A  B
 
 
 Does NOT COMMUTE: A  B   B  A
A
VECTOR CROSS PRODUCT – using unit vectors
 Use rules to get vector products of unit vectors:
ˆj  kˆ  iˆ
iˆ  ˆj  kˆ
kˆ  iˆ  ˆj
ˆj  iˆ  kˆ
kˆ  ˆj  iˆ
iˆ  kˆ   ˆj
iˆ  iˆ  0
ˆj  ˆj  0
kˆ  kˆ  0
AxB
B
θ
  
 Use these to evaluate C  A  B using components:


  
C  A  B  Ax iˆ  Ay ˆj  Az kˆ  Bx iˆ  B y ˆj  Bz kˆ

o Multiply out - get nine terms
 each contains one of the unit vector cross products above
 three of these are zero
  
ˆ
 Result: C  A  B  Ay Bz  Az B y  iˆ   Ax Bz  Az Bx  ˆj  Ax B y  Ay Bx  k
o Gives components of vector resulting from cross product
A
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MAGNETIC FORCE FB on a moving charge

 
FB  q v  B
What do we learn from this?
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
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 Magnitude is | FB |  | q | vB sin  where  is the angle between v and B
 NO magnetic force on a stationary (v = 0) charge


 NO magnetic force on a moving charge if v and B parallel

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 MAXIMUM magnetic force is when v and B are perpendicular


F
v
 Because B is perpendicular to
o Magnetic force does NO work on moving charge

o Magnetic force can change direction of v but not kinetic energy!
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MAGNETIC FIELD B
 SI Unit is Tesla: 1T = 1 N·s/(C·m)
o So Magnetic force on a 1 Coulomb charge moving perpendicular to a 1
Tesla magnetic field with a speed of 1 m/s is 1 Newton
 Magnetic force problems are 3-dimensional

o Need a convention for representing B in 3rd dimension
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B out of board/page
B into board/page
MOTION OF A CHARGED PARTICLE IN A UNIFORM MAGNETIC FIELD
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1 case: velocity perpendicular to magnetic field ( vi  B )
st

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 Resulting magnetic force FB perpendicular to v
o A force perpendicular to velocity is a CENTRIPETAL force

o Charge moves in a circle in a plane perpendicular to B
v
+q
FB


o v remains perpendicular to B as charge moves

 Path is a circle - FB points toward centre – called cyclotron motion
Cyclotron motion:
Charge moving with velocity perpendicular to magnetic field
 will move on circular path in plane perpendicular to
magnetic field – cyclotron orbit
What is the radius of the cyclotron orbit?
v
+q
r
FB
v2
 Magnitude of centripetal acceleration is ar  r
mv 2
 By Newton’s 2nd law: FB  m ar  r
 
o
 But for v  B , magnetic force is F B q v B (because   90 and sin   1 )
mv 2
qv B 
 So:
Can solve for radius of cyclotron orbit.
r
mv
r

Can measure charge to mass ratio if field and speed known.
Gives:
qB
v
Cyclotron motion continued:
Charge with velocity perpendicular to magnetic field moves on
circular path (cyclotron orbit) in plane perpendicular to magnetic field
Radius of orbit is
r
r
mv
qB
For each orbit, line from centre to charge moves through 2 radians
 can calculate angular frequency for motion (cyclotron frequency)
v qB
 
r
m
 Period of circular motion is distance (circumference) over speed
T
m
circumference 2  r

 2
v
v
qB
 Period independent of speed:
o important for particle accelerators – can “pump” energy into orbiting
particles with constant frequency as particles gain kinetic energy
+
F
2nd case: velocity has component perpendicular to magnetic field AND
component parallel to magnetic field
 Look at case with magnetic field along x

axis (i.e. B  B iˆ )
 No magnetic force parallel to magnetic
field
o SO no acceleration in x direction
o x-comp. of vel. constant vx  vxi
 particle does cyclotron motion around direction of magnetic field
o result is a HELICAL path
o view along x-axis (at right)
mv

r
o radius of helix is
qB
z
v  v y2  v z2
y
EXAMPLE: problem 8 from page 773
A proton (charge = +e = 1.6 x 10-19 C ) moves with velocity

v  2 iˆ  4 ˆj  kˆ m/s in a region where the magnetic field is

B  iˆ  2 ˆj  3 kˆ T . What is the magnitude of the magnetic force on this particle?
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
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RECOMMENDED: Work through example 22.3 on page 750. Calculating
magnetic field from bending of an electron beam.