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Chapter 26
Magnetic Fields
Magnets
• In each magnet there are two poles present (the ends
where objects are most strongly attracted): north and
south
• Like (unlike) poles repel (attract) each other (similar to
electric charges), and the force between two poles
varies as the inverse square of the distance between
them
• Magnetic poles cannot be isolated – if a permanent
magnetic is cut in half, you will still have a north and a
south pole (unlike electric charges)
• There is some theoretical basis for monopoles, but
none have been detected
Magnets
• The poles received their names due to the way a
magnet behaves in the Earth’s magnetic field
• If a bar magnet is suspended so that it can move
freely, it will rotate
• The magnetic north pole points toward the Earth’s
north geographic pole
• This means the Earth’s north geographic pole is a
magnetic south pole
• Similarly, the Earth’s south geographic pole is a
magnetic north pole
Magnets
• An unmagnetized piece of iron can be magnetized by
stroking it with a magnet (like stroking an object to
charge an object)
• Magnetism can be induced – if a piece of iron, for
example, is placed near a strong permanent magnet, it
will become magnetized
• Soft magnetic materials (such as iron) are easily
magnetized and also tend to lose their magnetism
easily
• Hard magnetic materials (such as cobalt and nickel)
are difficult to magnetize and they tend to retain their
magnetism
Magnetic Fields
• The region of space surrounding a moving charge
includes a magnetic field (the charge will also be
surrounded by an electric field)
• A magnetic field surrounds a properly magnetized
magnetic material
• A magnetic field is a vector quantity symbolized by B
• Its direction is given by the direction a north pole of a
compass needle pointing in that location
• Magnetic field lines can be used to show how the field
lines, as traced out by a compass, would look
Magnetic Field Lines
• A compass can be used to show the direction of
the magnetic field lines
Magnetic Field Lines
• Iron filings can also be used
to show the pattern of the
magnetic field lines
• The direction of the field is
the direction a north pole
would point
• Unlike poles (compare to the
electric field produced by an
electric dipole)
Magnetic Field Lines
• Iron filings can also be used
to show the pattern of the
magnetic field lines
• The direction of the field is
the direction a north pole
would point
• Unlike poles (compare to the
electric field produced by an
electric dipole)
• Like poles (compare to the
electric field produced by like
charges)
Magnetic Fields
• When moving through a magnetic field, a charged
Nikola Tesla
particle experiences a magnetic force
1856 – 1943
• This force has a maximum (zero) value when the
charge moves perpendicularly to (along) the magnetic
field lines
• Magnetic field is defined in terms of the magnetic force
exerted on a test charge moving in the field with
velocity v
F
• The SI unit: Tesla (T)
N
T
Am
B
N
T
qv sin 
C  (m / s)
F  q v B sin 
Magnetic Fields
• Conventional laboratory magnets: ~ 2.5 T
• Superconducting magnets ~ 30 T
• Earth’s magnetic field ~ 5 x 10-5 T
Direction of Magnetic Force
• Experiments show that the direction of
the magnetic force is always
perpendicular to both v and B
• Fmax occurs when v is perpendicular to
B and F = 0 when v is parallel to B
FB  qv  B
• Right Hand Rule #1 (for a + charge):
Place your fingers in the direction of v
and curl the fingers in the direction of B
– your thumb points in the direction of F
• If the charge is negative, the force
points in the opposite direction
Direction of Magnetic Force
• The x’s indicate the magnetic field when it is directed
into the page (the x represents the tail of the arrow)
• The dots would be used to represent the field directed
out of the page (the • represents the head of the arrow)
Differences Between Electric and
Magnetic Fields
• The electric force acts along the direction of the
electric field, whereas the magnetic force acts
perpendicular to the magnetic field
• The electric force acts on a charged particle regardless
of whether the particle is moving, while the magnetic
force acts on a charged particle only when the particle
is in motion
• The electric force does work in displacing a charged
particle, whereas the magnetic force associated with a
steady magnetic field does no work when a particle is
displaced (because the force is perpendicular to the
displacement)
Force on a Charged Particle in a
Magnetic Field
• Consider a particle moving in an
external magnetic field so that its
velocity is perpendicular to the field
• The force is always directed toward
the center of the circular path
• The magnetic force causes a
centripetal acceleration, changing
the direction of the velocity of the
particle
mv
F  q v B sin   qv B 
r
2
mv
r
qB
Force on a Charged Particle in a
Magnetic Field
• This expression is known as the
cyclotron equation
• r is proportional to the momentum
of the particle and inversely
proportional to the magnetic field
• If the particle’s velocity is not
perpendicular to the field, the path
followed by the particle is a spiral
(helix)
v qB
 
r m
2 2m

T
 qB
mv
r
qB
Particle in a Nonuniform Magnetic Field
• The motion is complex
Charged Particles Moving in Electric and
Magnetic Fields
• In many applications, charged particles move in the
presence of both magnetic and electric fields
• In that case, the total force is the sum of the forces due
to the individual fields:
F  qE  qv  B
Chapter 26
Problem 23
Microwaves in a microwave oven are produced by electrons circling in
a magnetic field at a frequency of 2.4 GHz. (a) What’s the magnetic
field strength? (b) The electrons’ motion takes place inside a special
tube called a magnetron. If the magnetron can accommodate electron
orbits with maximum diameter 2.5 mm, what’s the maximum electron
energy?
Magnetic Force on a Current Carrying Wire
• The current is a collection of many charged
particles in motion
• The magnetic force is exerted on each
moving charge in the wire
• The total force is the sum of all the
magnetic forces on all the individual
charges producing the current
• Therefore a force is exerted on a currentcarrying wire placed in a magnetic field:





 
F  L I B


 
 
F  q v d  B # carriers  q v d  B  nAL

Magnetic Force on a Current Carrying Wire
• The direction of the force is given by right hand rule #1,
placing your fingers in the direction of I instead of v
Magnetic Force on a Current Carrying
Wire of an Arbitrary Shape
• For a small segment of the wire, the force exerted on
this segment is
dFB  I ds  B
• The total force is
b
FB  I  ds  B
a
Chapter 26
Problem 28
A wire with mass per unit length 75 g/m runs horizontally at right
angles to a horizontal magnetic field. A 6.2-A current in the wire
results in its being suspended against gravity. What’s the magnetic
field strength?
Biot-Savart Law
• Biot and Savart arrived at a mathematical
expression that gives the magnetic field at
some point in space due to a current
• The magnetic field is dB at some point P;
the length element is ds; the wire is
carrying a steady current of I
Jean-Baptiste Biot
1774 – 1862
Félix Savart
1791 – 1841
Biot-Savart Law
• Vector dB is perpendicular to both ds and to
the unit vector r̂ directed from ds toward P
• The magnitude of dB is inversely
proportional to r2, where r is the distance
from ds to P
• The magnitude of dB is proportional to the
current and to the magnitude ds of the
length element
Biot-Savart Law
• The magnitude of dB is proportional to sin,
where  is the angle between the vectors ds
and r̂
• The observations are summarized in the
mathematical equation called the BiotSavart law (magnetic field due to the
current-carrying conductor):
μo I ds  ˆr
dB 
4π r 2
• µo = 4  x 10-7 T.m / A: permeability of free
space
Biot-Savart Law
• To find the total field, sum up the
contributions from all the current elements
μo I
B
4π
ds  ˆr
 r2
μo I ds  ˆr
dB 
4π r 2
Biot-Savart Law
• The magnitude of the magnetic field varies as the
inverse square of the distance from the ds element
• The electric field due to a point charge also varies as
the inverse square of the distance from the charge
• The electric field created by a point charge is radial in
direction
• The magnetic field created by a current element is
perpendicular to both the length element and the unit
vector
• The current element producing a magnetic field is part
of an extended current distribution
A Long, Straight Conductor
• The thin, straight wire is carrying a constant current

d s  rˆ  ( dx cos  ) kˆ
ad
dx  
2
cos 
x   a tan 
a
r
cos 
  0 I dx cos 
ˆ
dB 
k
2
4
r
0 I ˆ

k cos  d 
4a
2
0 I
B
cos d

4a 
1
A Long, Straight Conductor
• The thin, straight wire is carrying a constant current
2
0 I
B
cos d

4a 
0 I
sin 1  sin  2 

4a
1
• If the conductor is an infinitely
long, straight wire, θ1 = π/2 and
θ2 = – π/2 , and the field becomes
0 I
B
2a
2
0 I
B
cos d

4a 
1
A Long, Straight Conductor
• The magnetic field lines are circles
concentric with the wire
• The field lines lie in planes
perpendicular to the wire
• The magnitude of the field is
constant on any circle of radius a
• Right Hand Rule #2: Grasp the wire
in your right hand and point your
thumb in the direction of the current
and your fingers will curl in the
direction of the field
A Curved Wire Segment

Ids
0
• Find the field at point O due to the dB 
2
wire segment (I, a are constants)
4 a
0 I
0 I
0 I
0 I
B
ds 
s
a 

2 
2
2
4a
4a
4a
4a
• The field at the center of the full
circle loop
0 I
B
2
4a
0 I
B
2a
Magnetic Field of a Current Loop
Magnetic Field of a Current Loop
• The field contribution from a current
element I dl = I dx
0 I dl
dBx 
4 x 2  a 2
B   dBx 
0 Ia
4 ( x  a )
2
2 3/ 2
a
x a
2
2
 dl  2( x
0 Ia 2
2
loop
• For large distances (x >> a), this
reduces to
B
0 Ia
2x 3
2
 a 2 )3/ 2
Chapter 26
Problem 30
A single-turn wire loop is 2.0 cm in diameter and carries a 650-mA
current. Find the magnetic field strength (a) at the loop center and (b)
on the loop axis, 20 cm from the center.
Torque on a Current Loop
F 2  F 4  BIa  max
b
b
 F2  F4
2
2
b
b
 BIa  BIa  BIab  BIA
2
2
  BIA sin 
  N BIA sin 
Torque on a Current Loop
• Applies to any shape loop
• Torque has a maximum value when  =
90°
• Torque is zero when the field is
perpendicular to the plane of the loop
  N BIA sin 
Magnetic Moment
• The vector  is called the magnetic dipole
moment of the coil
• Its magnitude is given by
μ = IAN
• The vector always points perpendicular to the plane of
the loop(s)
• The equation for the magnetic torque can be written as
τ = BIAN sinθ = μB sinθ
   B
• The angle is between the moment and the field
Potential Energy
• The potential energy of the system of a magnetic
dipole in a magnetic field depends on the orientation of
the dipole in the magnetic field
 
U    B
• Umin = – μB and occurs when the dipole moment is in
the same direction as the field
• Umax = + μB and occurs when the dipole moment is in
the direction opposite the field
Chapter 26
Problem 35
A single-turn square wire loop 5.0 cm on a side carries a 450-mA
current. (a) What’s the loop’s magnetic dipole moment? (b) If the loop
is in a uniform 1.4-T magnetic field with its dipole moment vector at
40° to the field, what’s the magnitude of the torque it experiences?
Electric Motor
• An electric motor converts electrical energy to
mechanical energy (rotational kinetic energy)
• An electric motor consists of a rigid current-carrying
loop that rotates when placed in a magnetic field
• The torque acting on the loop
will tend to rotate the loop to
smaller values of θ until the
torque becomes 0 at θ = 0°
Electric Motor
• If the loop turns past this point and the current remains
in the same direction, the torque reverses and turns the
loop in the opposite direction
• To provide continuous rotation in one direction, the
current in the loop must periodically reverse
• In ac motors, this reversal
naturally occurs
• In dc motors, a split-ring
commutator and brushes are
used
Electric Motor
• Just as the loop becomes perpendicular to the magnetic
field and the torque becomes 0, inertia carries the loop
forward and the brushes cross the gaps in the ring,
causing the current loop to reverse its direction
• This provides more torque to
continue the rotation
• The process repeats itself
• Actual motors would contain
many current loops and
commutators
Magnetic Force Between Two Parallel
Conductors
0 I 2
B2 
2a
0 I 2
F1  B 2 I 1l 
I1l
2a
F1  0 I1 I 2

l
2a
Magnetic Force Between Two Parallel
Conductors
• The force (per unit length) on wire 1
due to the current in wire 1 and the
magnetic field produced by wire 2:
FB  0 I1 I 2

l
2a
• Parallel conductors carrying
currents in the same direction
attract each other
• Parallel conductors carrying
currents in the opposite directions
repel each other
Chapter 26
Problem 63
A long, straight wire carries 20 A. A 5.0-cm by 10-cm rectangular wire
loop carrying 500 mA is 2.0 cm from the wire, as shown in the figure.
Find the net magnetic force on the loop.
Ampère’s Law
• Ampère’s Circuital Law: a procedure
for deriving the relationship between
the current in an arbitrarily shaped
wire and the magnetic field produced
by the wire
• Choose an arbitrary closed path
around the current and sum all the
products of B|| Δℓ around the closed
path (put the thumb of your right
hand in the direction of the current
through the loop and your fingers
curl in the direction you should
integrate around the loop)
 
 B  ds   0 I
Ampère’s Law for a Long Straight Wire
• Use a closed circular path
• The circumference of the circle is 2
 
 B  d s  B  ds  B 2 r   0 I
0 I
B
2r
r
Ampère’s Law for a Long Straight Wire
I ' r
r
 2
I' 2 I
I R
R
 


I
'

B
2

r
B

d
s
0

2
r
 0 2 I
R
2
0 I
B
r
2
2R
2
Magnetic Field of a Solenoid
Magnetic Field of a Solenoid
• If a long straight wire is bent into a coil of
several closely spaced loops, the
resulting device is called a solenoid
• It is also known as an electromagnet since
it acts like a magnet only when it carries a
current
• The field inside the solenoid is nearly
uniform and strong – the field lines are
nearly parallel, uniformly spaced, and
close together
• The exterior field is nonuniform, much
weaker, and in the opposite direction to
the field inside the solenoid
Magnetic Field of a Solenoid
• The field lines of the solenoid resemble
those of a bar magnet
• The magnitude of the field inside a
solenoid is approximately constant at all
points far from its ends
B = µo n I
• n = N / ℓ : the number of turns per unit
length
• This result can be obtained by applying
Ampère’s Law to the solenoid
Magnetic Field of a Solenoid
• A cross-sectional view of a tightly wound
solenoid
• If the solenoid is long compared to its
radius, we assume the field inside is
uniform and outside is zero
• Apply Ampère’s Law to the blue dashed
rectangle
 
 B  ds 
 
 B  ds  B
path1
 ds
 Bl
path1
  0 NI
N
B   0 I   0 nI
l
Magnetic Effects of Electrons – Orbits
• An individual atom should act like a magnet because of
the motion of the electrons about the nucleus
• Each electron circles the atom once in about every 10-16
seconds; this would produce a current of 1.6 mA and a
magnetic field of about 20 T at the center of the circular
path
• However, the magnetic field produced by one electron
in an atom is often canceled by an oppositely revolving
electron in the same atom
• The net result is that the magnetic effect produced by
electrons orbiting the nucleus is either zero or very
small for most materials
Magnetic Effects of Electrons – Spins
• Electrons also have spin (it is a
quantum effect)
• The classical model is to consider
the electrons to spin like tops
• The field due to the spinning is
generally stronger than the field due
to the orbital motion
• Electrons usually pair up with their
spins opposite each other, so their
fields cancel each other, hence most
materials are not naturally magnetic
Magnetic Effects of Electrons – Domains
• In some materials – ferromagnetic – the spins do not
naturally cancel
• Large groups of atoms in which the spins are aligned
are called domains
• When an external field is applied, it causes the material
to become magnetized: the domains that are aligned
with the field tend to grow at the expense of the others
Domains and Permanent Magnets
• In hard magnetic materials, the domains remain
aligned after the external field is removed
• The result is a permanent magnet
• In soft magnetic materials, once the external field is
removed, thermal agitation causes the materials to
quickly return to an unmagnetized state
• With a core in a loop, the magnetic field is enhanced
since the domains in the core material align,
increasing the magnetic field
Ferromagnetism
• Some substances exhibit strong magnetic effects
called ferromagnetism (e.g., iron, cobalt, nickel,
gadolinium, dysprosium)
• They contain permanent atomic magnetic moments
that tend to align parallel to each other even in a
weak external magnetic field
Paramagnetism
• Paramagnetic substances have small but positive
magnetism, which results from the presence of
atoms that have permanent magnetic moments
• These moments interact weakly with each other
• When placed in an external magnetic field, atomic
moments tend to line up with the field and the
alignment process competes with thermal motion
which randomizes the moment orientations
Diamagnetism
• When an external magnetic field is applied to a
diamagnetic substance, a weak magnetic moment is
induced in the direction opposite the applied field
• Diamagnetic substances are weakly repelled by a
magnet
Earth’s Magnetic Field
• The Earth’s geographic north (south) pole corresponds
to a magnetic south (north) pole – a north (south) pole
should be a “north- (south-) seeking” pole
• The Earth’s magnetic field
resembles that achieved by burying
a huge bar magnet deep in the
Earth’s interior
• The most likely source of the
Earth’s magnetic field – electric
currents in the liquid part of the
core
Earth’s Magnetic Field
• The magnetic and geographic poles are not in the
same exact location – magnetic declination is the
difference between true north (geographic north pole)
and magnetic north pole
• The amount of declination varies
by location on the earth’s
surface
• The direction of the Earth’s
magnetic field reverses every
few million years (the origin of
these reversals is not
understood)
Earth’s Magnetic Field
• If a compass is free to rotate vertically as well as
horizontally, it points to the earth’s surface
• The angle between the horizontal and the direction of
the magnetic field is called the dip angle
• The farther north the device is moved, the farther from
horizontal the compass needle would be
• The compass needle would be horizontal at the
equator and the dip angle would be 0°
• The compass needle would point straight down at the
south magnetic pole and the dip angle would be 90°
Magnetic Flux
• Magnetic flux associated with a magnetic field is
defined in a way similar to electric flux
 
 B   B  dA
• SI unit of flux: Weber
• Wb = T. m²
Wilhelm Eduard Weber
1804 – 1891
Magnetic Flux
• For a flat surface with an area A in a
uniform magnetic field, the flux is (θ is the
angle between B and the normal to the
plane):
ΦB = BA = B A cos θ
• When the field is perpendicular to the
plane, θ = 0 and ΦB = ΦB, max = BA
• When the field is parallel to the plane, θ =
90° and ΦB = 0
• The flux can be negative, for example if θ
= 180°
Magnetic Flux
• The value of the magnetic flux is
proportional to the total number of
magnetic field lines passing through
area
• When the area is perpendicular to
the lines, the maximum number of
lines pass through the area and the
flux is a maximum
• When the area is parallel to the lines,
no lines pass through the area and
the flux is 0
Gauss’ Law in Magnetism
• Magnetic fields do not begin or end at any point
• The number of lines entering a surface equals the
number of lines leaving the surface
• Gauss’ law in magnetism says the magnetic flux
through any closed surface is always zero:
 
 B   B  dA  0
Answers to Even Numbered Problems
Chapter 26:
Problem 16
(a) 3.4 × 105 m/s
(b) does not change
Answers to Even Numbered Problems
Chapter 26:
Problem 20
3.9 mm
Answers to Even Numbered Problems
Chapter 26:
Problem 32
4.0 A
Answers to Even Numbered Problems
Chapter 26:
Problem 36
480 mT
Answers to Even Numbered Problems
Chapter 26:
Problem 38
24 A
Answers to Even Numbered Problems
Chapter 26:
Problem 42
(a) (−1.1iˆ + 1.5 ˆj + 1.7kˆ) × 10−3 N
(b) 0
Answers to Even Numbered Problems
Chapter 26:
Problem 58
10 m