Chapter 23
Hans Christian Oersted
 1777 – 1851
 Best known for
observing that a
compass needle deflects
when placed near a wire
carrying a current
 First evidence of a
connection between
electric and magnetic
phenomena
Magnetic Fields – Long Straight
Wire
 A current-carrying wire
produces a magnetic
field
 The compass needle
deflects in directions
tangent to the circle
 The compass needle
points in the direction
of the magnetic field
produced by the
current
Direction of the Field of a Long
Straight Wire
 Right Hand Rule #1
 Grasp the wire in your
right hand
 Point your thumb in
the direction of the
current
 Your fingers will curl in
the direction of the
field
Magnitude of the Field of a Long
Straight Wire
 A straight current-carrying wire generates a
cylindrical magnetic field in the space surrounding
it.
 The magnitude of the field at a distance r from a
wire carrying a current of I is
B 
 µo = 4  x 10-7 T.m / A
0 I
2 r
 µo is called the permeability of free space
 The magnetic field of a long straight wire in a
vacuum is
Example 1
 The overhead power cable for a street trolley is
strong horizontally 10 m above the ground. A long
straight section of it carries 100 amps dc due west.
Describe the magnetic field produced by the
current, and determine its value at ground level just
under the wire. Compare that to the strength of the
Earth's field.
 Given: r = 10 m and I = 100 A
 Find: B
Example 1
 Solution: We've got a straight current-carrying wire and
that produces a known B-field.
 See diagram on last slide with the current assumed
heading west. At ground level (at a point beneath the
westerly current), the Right-Hand-Current Rule tells us
that B points due south. We already have an expression
for the field of a long straight current-carrying wire in
terms of I and r.
 0 = 4 x 10-7 T-m/A.
B 
( 4   10
7
T  m / A )( 100 A )
2  (10 m )
 2 . 0  10
 Which is only 4% of the Earth's field.
6
T
Force Between Two Conductors
 The magnetic fields created by moving
charges can interact and create forces much
like the electric forces between those charges.
 Parallel conductors carrying currents in the
same direction attract each other
 Parallel conductors carrying currents in the
opposite directions repel each other
Magnetic Force Between Two
Parallel Conductors
 The force on wire 1 is due to
the current in wire 1 and the
magnetic field produced by
wire 2
 The force per unit length is:
FM
l

 0 I1 I 2
2 d
Example 2
 What is the force per unit
length experienced by each of
two extremely long parallel
wires carrying equal 1.0-A
currents in opposite directions
while separated by a distance
of 1 m in vacuum?
 Given: I1 = I2 = 1.0 A and
d
=1m
 Find: FM
Example 2
 Make a drawing. The defining equation is
FM
l

 0 I1 I 2
2 d

( 4   10
7
T  m / A )( 1 . 0 A )( 1 . 0 A )
2  (1m )
FM/l = 2 x 10-7N/m, repulsion
André-Marie Ampère
 1775 – 1836
 Credited with the
discovery of
electromagnetism
 Relationship between
electric currents and
magnetic fields
 Mathematical genius
evident by age 12
Ampère’s Law
 André-Marie Ampère found a procedure for
deriving the relationship between the current in an
arbitrarily shaped wire and the magnetic field
produced by the wire
 Ampère’s Circuital Law
 B l    I
||
0
 Sum over the closed path
***THE MORE CURRENT….THE STRONGER
THE MAGNETIC FIELD***
Magnetic Field of a Current
Loop
 The strength of a
magnetic field
produced by a wire can
be enhanced by
forming the wire into a
loop
 All the segments, ∆x,
contribute to the field,
increasing its strength
Magnetic Field of a Current Loop
– Total Field
Magnetic Field of a Current Loop
– Equation
 The magnitude of the magnetic field at the center of
a circular loop with a radius R and carrying current I
is
0I
B 
2R
 With N loops in the coil, this becomes
B  N
0I
2R
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
Magnetic Field of a Solenoid, 2
 The field lines inside the solenoid are
nearly parallel, uniformly spaced, and close
together
 This indicates that the field inside the
solenoid is nearly uniform and strong
 The exterior field is nonuniform, much
weaker, and in the opposite direction to the
field inside the solenoid
Magnetic Field in a Solenoid, 3
 The field lines of the solenoid resemble those of a bar
magnet
Example 3
 A 20-cm-long solenoid with a 2.0-cm inside
diameter is tightly wound on a hollow quartz
cylinder. There are several layers with a total of 20 x
103 turns per meter of a niobium-tin wire. The
device is cooled below its critical temperature and
becomes superconducting. Since the wire is then
without resistance, it can easily carry 30 A and not
develop any I2R losses. Compute the approximate
field inside the solenoid near the middle. What is
its value at either end?
 Given: n = 20 x 103 m-1, I = 30 A, and D = 2.0 cm
 Find: Bz
Example 3
 Solution: We've got a current-carrying solenoid and
that produces a known B-field.
 The solenoid is long and narrow and will obey the
approximations that led to the last equation
 Using 0 = 1-257 X 10-6 T-m/A,
Bz ≈ 0nI =(1.257X 10-6 T-m/A)(20 x 103 m-1)(30A)
Bz ≈ 0.75 T
 Which is a formidable field, over 104 times that of the
Earth. The field at either end is about half this, 0.38 T
Moving Charges and Magnetism
 Moving charge creates B
 The orientation (direction) of B depends upon the
orientation (direction) of I
 WITHIN atoms, electrons move in different (often
opposing) directions, thus individual B usually cancel
out
 BETWEEN atoms that DO have B fields, those FIELDS
may oppose each other and cancel out
**Only in materials that have BOTH alignments do we
see magnetic properties***
FERROMAGNETIC
Right Hand Rule #2
 Place your fingers in the
direction of
 Curl the fingersv in the direction
of the magnetic field,
 Your thumb points in the

direction of the force, B , on a

positive charge
F
 If the charge is negative, the
force is opposite that
determined by the right
hand rule
 Maximum force is
F M  qvB sin 
Finding the Direction of
Magnetic Force
 Experiments show that
the direction of the
magnetic force is always

perpendicular
to bothv

and B which form a
plane.

 Fmax occurs when v is

perpendicular to B

 F=0
 when v is parallel
to B
Example 4
 A conventional water-cooled electromagnet
produces a 3.0-T uniform magnetic field in the 4-in.
gap between its flat pole pieces. The field is aligned
horizontally pointing due north. A proton is fired
into the field region at a speed of 5.0 x 106 m/s. It
enters traveling in a vertical north-south plane,
heading north and downward at 30° below the
horizontal. Compute the force vector acting on the
proton at the moment it enters the field.
 Given: A proton with v = 5.0 X 106 m/s, at 30° below
the horizontal in the northerly direction, and B =
3.0 T, north
 Find: FM
Example 4
 Solution: Here a charged particle is moving in a B-field and




that should call to mind v x B.
First, make a drawing
The proton has a positive charge of +1.60 X 10-19 C and so v x
B is due east, FM is due east
The basic force-on-a-moving-charge relationship is
FM = qvB sin 
The angle between v and B is  = 30° and so with q = qe
FM = qe vB sin 
FM = (+1.6 x 10-19 C)(5.0 x 106 m/s)(3.0 T)(sin 30°)
Fm= 1.2 x 10-12 N
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
Force on a Charged Particle
 Charged particle entering perpendicular to uniform
magnetic field and experiences centripetal
acceleration
 Equating the magnetic and centripetal forces:
 Solving for R:
mv 2
F  qvB 
R
mv
R
qB
 R is proportional to the momentum of the
particle and inversely proportional to the
magnetic field
 Sometimes called the cyclotron equation
Particle Moving in an External
Magnetic Field
 If the particle’s velocity is
not perpendicular to the
field, the path followed
by the particle is a spiral
 The spiral path is
called a helix
23.2
Electromagnets
 Electromagnets are created by an electric current
travelling through a solenoid. As such, their strength
and direction can be controlled.
 Strength can be increased by:
 Increasing current (I)
 Increasing the number of coils (N) – although this
means a longer total wire length and thus more
resistance.
 Increasing the core material’s permeability.
Electromagnets
Magnetic Force on Current
Carrying Conductor
 A force is exerted on a current-carrying wire placed
in a magnetic field
 The current is a collection of many charged
particles in motion
 The direction of the force is given by right hand rule
#2
Force on a Wire
 The blue x’s indicate the magnetic
field is directed into the page
 The x represents the tail of the
arrow
 Blue dots would be used to
represent the field directed out of
the page
 The • represents the head of the
arrow
 In this case, there is no current, so
there is no force
Force on a Wire
 B is into the page
 The current is up the page
 The force is to the left
Force on a Wire
 B is into the page
 The current is down the
page
 The force is to the right
Force on a Wire, equation
 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
FM = B I l sin 
 is the angle between

B
and the direction of I
 The direction is found by the right hand rule,
placing
 your fingers in the direction of I instead
of v

Example 5
 A flat, horizontal rectangular
loop of wire is positioned, as
shown in Fig. 19.32a, in a 0.10-T
uniform vertical magnetic field.
The sides of the rectangle are FC equal to 30 cm and C- equal to
20 cm. Determine the total
force acting on the loop when it
carries a current of 1.0 A.


Given: B = 0.10 T, F-C = 30 cm, C-D = 20 cm, and
I = 1.0 A
Find: FM.
Example 5
 Solution: We've got a current-
carrying loop in a S-field – FM =
IlB sin 
 First, make a drawing

Current travels from the positive terminal of the battery
clockwise around the circuit. The directions of the forces
on each segment are arrived at via v x B and are indicated
in the diagram. Because the forces on segments F-C and
D-E are equal and opposite, they cancel. The total force
acting on the loop is the force acting on segment C-D
Example 5
FM = IlB sin  = (1.0 A) (0.20m) (0.10 T) (sin 90°)
FM = 0.020 N
Torque on a Current Loop
 Applies to any shape loop
 N is the number of turns in
the coil
 Torque has a maximum
value of NIAB
 When f = 90°
 Torque is zero when the field
is parallel to the plane of the
loop
  NIAB sin f
f
f
Electric Motor
 An electric motor
converts electrical energy
to mechanical energy
 The mechanical
energy is in the form
of rotational kinetic
energy
 An electric motor
consists of a rigid
current-carrying loop
that rotates when placed
in a magnetic field
Electric Motor, 2
 The torque acting on the loop will tend to rotate
the loop to smaller values of  until the torque
becomes 0 at  = 0°
 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
Electric Motor, 3
 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

Actual motors would contain many current loops and
commutators
Electric Motor
Electric Motor, final
 Just as the loop becomes perpendicular to the
magnetic field and the torque becomes zero,
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
***This is obviously much easier if the
CURRENT itself periodically reverses – AC
electricity instead of DC***
23.3
Electromagnetic Induction
 Moving charge(s) creates (induces) a magnetic field
(B)
 A moving (changing) magnetic field (B) creates
(induces) a current (I)
 Whether B is increasing or decreasing near a wire
determines the current’s direction.
Michael Faraday
 1791 – 1867
 Great experimental
scientist
 Invented electric motor,
generator and
transformers
 Discovered
electromagnetic
induction
 Discovered laws of
electrolysis
Faraday’s Experiment – Set Up
 A current can be produced by a changing magnetic
field
 First shown in an experiment by Michael Faraday
 A primary coil is connected to a battery
 A secondary coil is connected to an ammeter
Faraday’s Experiment
 The purpose of the secondary circuit is to detect
current that might be produced by the magnetic
field
 When the switch is closed, the ammeter reads a
current and then returns to zero
 When the switch is opened, the ammeter reads a
current in the opposite direction and then returns to
zero
 When there is a steady current in the primary
circuit, the ammeter reads zero
Faraday’s Conclusions
 An electrical current is produced by a changing
magnetic field
 The secondary circuit acts as if a source of emf were
connected to it for a short time
 It is customary to say that an induced emf is
produced in the secondary circuit by the changing
magnetic field
Magnetic Flux (Ф)
 The emf is actually induced by a change in
the quantity called the magnetic flux
rather than simply by a change in the
magnetic field
 Magnetic flux (Ф) is defined in a manner
similar to that of electrical flux – the density
of the magnetic field lines in a given area
ofspace
 Magnetic flux is proportional to both the
strength of the magnetic field passing
through the plane of a loop of wire and the
area of the loop
Magnetic Flux, 2
 You are given a loop of wire
 The wire is in a uniform
magnetic field B
 The loop has an area A
 The flux is defined as
ΦM = BA = B A cos θ
 θ is the angle between B and
the normal to the plane
Magnetic Flux, 3
 When the field is
perpendicular to the plane
of the loop, as in a, θ = 0
and ΦB = ΦB, max = BA
 When the field is parallel
to the plane of the loop, as
in b, θ = 90° and ΦB = 0
 The flux can be negative,
for example if θ = 180°
 SI units of flux are
T. m² = Wb (Weber)
Magnetic Flux, final
 The flux can be visualized with respect to magnetic
field lines
 The value of the magnetic flux is
proportional to the total number of lines
passing through the loop
 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
Electromagnetic Induction –
An Experiment
 When a magnet moves toward a
loop of wire, the ammeter shows
the presence of a current (a)
 When the magnet is held
stationary, there is no current (b)
 When the magnet moves away
from the loop, the ammeter
shows a current in the opposite
direction (c)
 If the loop is moved instead of
the magnet, a current is also
detected
Electromagnetic Induction –
Results of Experiment
 A current is set up in the circuit as long as there
is relative motion between the magnet and the
loop (changing magnetic field)
 The same experimental results are found whether
the loop moves or the magnet moves
 The current is called an induced current
because is it produced by an induced emf
Faraday’s Law and
Electromagnetic Induction
 The average induced emf for a one turn loop
Eav


M
t
 A single loop of wire will experience an induced
voltage that equal the time rate of change of
magnetic flux through it at any given instant.
 For N turns of wire, the average induced emf is
Eav
 N

M
t
 Which is known as Faraday’s Induction Law
Example 6
 A circular flat coil of 200 turns of wire encloses an
area of 100 cm2. The coil is immersed in a uniform
perpendicular magnetic field of 0.50 T that
penetrates the entire area. If the field is shut off so
that it drops to zero in 200 ms, what is the average
induced emf? Given that the coil has a resistance of
25 W, what current will be induced in it?
 Given: N = 200, A = 100 cm2, Bi = 0.50 T, Bf = 0,
∆t
= 200 ms, and R = 25 W
 Find: The induced emf and current I
Example 6
 Solution: Faraday's Law problem
 The time rate-of-change of the flux equals the emf.
 The B-field links the entire area perpendicularly,
the initial flux is
M = BA = (0.50T)(0.0100 m2) = 0.0050 T-m2
 and so ∆M = - 0.005 0 T- m2 since the final flux is
zero
Example 6
 Thus,
E
 N

(  0 . 0050 T  m )
2
M
t
  200
0 . 200 s
 From Ohm's Law:
I 
V
R

(5 .0 V )
( 25 W )
 0 . 20 A
 5 .0 V
Applications of Faraday’s Law
Electric Guitar
 A vibrating string induces an
emf in a coil
 A permanent magnet inside
the coil magnetizes a portion
of the string nearest the coil
 As the string vibrates at some
frequency, its magnetized
segment produces a changing
flux through the pickup coil
 The changing flux produces
an induced emf that is fed to
an amplifier
Applications of Faraday’s Law
Apnea Monitor
 The coil of wire attached to
the chest carries an
alternating current
 An induced emf produced
by the varying field passes
through a pick up coil
 When breathing stops, the
pattern of induced voltages
stabilizes and external
monitors sound an alert
Application of Faraday’s Law
Motional emf
 A straight conductor of
length ℓ moves
perpendicularly with
constant velocity through a
uniform field
 The electrons in the
conductor experience a
magnetic force
FM = qvB sin  = qvB
 The electrons tend to move
to the lower end of the
conductor
Motional emf
 As the negative charges accumulate at the base, a
net positive charge exists at the upper end of the
conductor
 As a result of this charge separation, an electric field
is produced in the conductor
 Charges build up at the ends of the conductor until
the downward magnetic force is balanced by the
upward electric force
 There is a potential difference between the upper
and lower ends of the conductor
Motional emf, cont
 The potential difference between the ends of the
conductor can be found by
E=vBl
 The upper end is at a higher potential than the lower
end
 A potential difference is maintained across the conductor
as long as there is motion through the field
 If the motion is reversed, the polarity of the potential
difference is also reversed
 Since emf = E l, the electric field in the wire, which
exactly counters the motional emf is
E=vB
Example 7
B = 2.0 x 10-5 T
W
S



 A 1.0-meter-long wire
1.0 m
N
E
held in a horizontal eastwest orientation is
dropped at a place where
the Earth's magnetic field
is 2.0 x 10 -5 T, due north.
Determine the induced emf 4.0 s after release.
Given: l = 1.0 m, t = 4.0 s. and B = 2.0 X 10 -5 T
Find: The resulting emf
Example 7
 Solution: Because a wire is moving through a
magnetic field and we want the emf, this problem
involves the expression E = vBl. We have B and land
need v
 The speed at t = 4.0 s
v = v0 + gt = 0 + (9.81 m/s2)(4.0 s) = 39.2m/s
E = vBl = (39.2 m/s)(2.0 x 10 -5T)(l.0m) = 0.78 mV
Motional emf in a Circuit
 Assume the moving bar
has zero resistance
 As the bar is pulled to
the right with a given
velocity under the
influence of an applied
force, the free charges
experience a magnetic
force along the length of
the bar
 This force sets up an
induced current because
the charges are free to
move in the closed path
Motional emf in a Circuit
 The changing magnetic flux
through the loop and the
corresponding induced emf
in the bar result from the
change in area of the loop
 The induced, motional emf,
acts like a battery in the
circuit
E=vBl
and
I 
vBl
R
Generators
 Alternating Current (AC) generator
 Converts mechanical energy to electrical energy
 Consists of a wire loop rotated by some external
means – a Turbine
 There are a variety of sources that can supply the
energy to rotate the loop

These may include falling water, heat by burning coal to
produce steam
AC Generators
 Basic operation of the
generator
 As the loop rotates, the
magnetic flux through it
changes with time
 This induces an emf and a
current in the external
circuit
 The ends of the loop are
connected to slip rings that
rotate with the loop
 Connections to the
external circuit are made
by stationary brushes in
contact with the slip rings
AC Generators, final
 The emf generated by the
E
rotating loop can be found by
E = 2 N B l v sin θ
Emax
 If the loop rotates with a
constant angular speed, ω,
and N turns
E = N A B ω sin ω t
 E = E max when loop is
parallel to the field
 E = 0 when when the loop is
perpendicular to the field
AC Generators – Detail of
Rotating Loop
 The magnetic force on the
charges in the wires AB
and CD is perpendicular
to the length of the wires
 An emf is generated in
wires BC and AD
 The emf produced in each
of these wires is
E= B l v= B l sin θ
AC Generators
 Can also consist of a rotating turbine which passes
constantly-reversing (and changing) magnetic fields
past an electromagnet.
 This changes the electromagnet’s magnetic field and
induces AC current in its coil
DC Generators
 Components are
essentially the same as
that of an ac generator
 The major difference is
the contacts to the
rotating loop are made
by a split ring, or
commutator
DC Generators
 The output voltage always
has the same polarity
 The current is a pulsing
current
 To produce a steady
current, many loops and
commutators around the
axis of rotation are used
 The multiple outputs
are superimposed and
the output is almost
free of fluctuations
E
Example 8
 A simple single-coil dc generator rotates at a
constant frequency of 60 Hz in a 0.40-T magnetic
field. Given that the coil has 10 turns and
encompasses an area of 1200 cm2, what will be its
maximum emf?
 Given: f = 60 Hz, A = 1200 cm2, B = 0.40 T, and N =
10
 Find: Em.
Example 8
 A coil is rotating in a B-field with a specified frequency –
sinusoidal emf
 The basic formula for the emf of an ac generator is Eq.
(20.7)
 The maximum emf (Em) is the amplitude of the
oscillating voltage given by
E = NAB w sinwt
 namely
E m= NAB w
w = 2f= 2(60 Hz) = 376.99 rad/s
Em = (10)(1200X 10-4m2)(0.40T)(376.99 rad/s) = 0.18 kV
Motors
 Motors are devices that convert electrical energy
into mechanical energy
 A motor is a generator run in reverse
 A motor can perform useful mechanical work
when a shaft connected to its rotating coil is
attached to some external device
Transformers
 Use EM Induction to control voltages by “Stepping-
Up” or “Stepping-Down” the voltage.
 Using different numbers of coils of wire wrapped
around a common core, the induced emf in one part of
the coil is transferred to the other coil.
V2/V1 = N2/N1
 Larger-N coil has more voltage than the…
 Smaller-N coil (for the same B-Field generated in the
core
Transformers