Electromagnetic Induction and Motional EMF

Electromagnetic Induction and Motional EMF
An EMF (Electromotive Force – a source of electrical potential) created by a
changing magnetic field is known as an induced EMF.
Induced EMF’s generate currents
Consider a conductor, length l moving to the right in a magnetic field as shown
below.
⊗
⊗
l ⊗
⊗
v v
⊗
⊗
v
F
v v
= qv × B
• A charge, +q in the conductor experiences an electric force directed upward
(be sure to verify this for yourself) of magnitude qvB, due to the motion of
the conductor through the field
• The magnetic force causes the + charges in the conductor to move upward
and the negative charges to move downward
+ +
+ +
+ +
E
- -
- -
- -
• The result of this process is a pileup of + charge at the upward end of the
conductor and – charge at the downward end creating a potential difference
between the ends of the conductor and an E field pointing downward within
the conductor.

• The E field within the conductor produces a force, qE, also directed
downward on positive charges. .
• The force qvB causes polarization of the conductor and establishes a
potential difference between the top and bottom of the conductor.
• Due to this polarization an electric field is established within the conductor
and any positive charge in this field experiences a force qE directed
downward - which grows as charges accumulate at the ends of the conductor
(it also continues to experience a force of qvB directed upwards as long as
the conductor continues to move to the right).
+ +
+ +
+ +
v v v
F = qvB
- -
- -
- -
v v
F = qE
• The accumulation of charges at the ends of the conductor continues until
v v v
qE = qvB . When this occurs the forces within the conductor are in
equilibrium and no more accumulation of charges at the ends of the bar
occurs.
• The magnitude of the potential difference at equilibrium is:
q / vB = qE / Q E = vB
Recalling that V = Ed → V = El = vBl
• The potential difference between the top and bottom of the conductor is
vBl

What if we connect this polarized conductor, a source of electrical potential, to an
external circuit?
⊗
⊗
a
∆s
⊗
⊗
⊗
I
⊗
⊗
+
⊗
⊗
v v
l
⊗
⊗
b
⊗
moving conductor and source of EMF
• A counterclockwise current is established due to the motion of the moving
portion of the circuit.
• The moving element is a source of EMF.
• Charges move from lower to higher potential (as in other sources) in the
moving segment, and from higher to lower throughout the rest of the
circuit.
• Motional EMF:
ξ = vBl
• If resistance (R) is negligible in the sliding segment ξ = V ab .
• In real conductors ξ = V ab – IR where the potential drop is due to the
resistance in real conductors.
• The quantitative result here is applies generally to any source of electrical
potential though the fundamental source of the EMF may be different.
• In this particular example we had the moving conductor moving with a
velocity v ⊥ to B (which was constant). In general the conductor could move
in any plane (or B could vary) and still produce a motional EMF as long as a
component of that plane is perpendicular to B.
• The general expression for motional EMF is:
ξ = vBlsinθ

Example 1
Compute the motional EMF and the current in the circuit for the following
⊗
⊗
l = 0.1m
⊗
l
⊗
v
v = 0.1m
⋅ s
−1
⊗
⊗
R
= 0. 01Ω
B = 1. 0T
The motional EMF is: ξ = vBlsinθ = (0.1m/s)(1.0T)(0.1m) = 0.01 volts
The current in the loop is: I = ξ/R = 1 Amp
This is essentially (schematically) how all electrical current is generated.
Force and Work on a Current Loop
In the previous example the current flowing through the circuit moves in the
presence of an external B field that generates a force on the conducting elements
of the circuit as shown below.
F = Il × B = IBlsinθ = 0.1N (on the moving conductor)
F
I
F
⊗ ⊗ ⊗
F
F
This magnetic force opposes
the motion of the conductor
In order to move the bar to the right at a constant velocity we must continue to
apply a force to it that is equal in magnitude to IlB force. In other words we must
do work on the system.

Recalling that work per unit time is power:
or, equivalently:
−1
( 0.1m
⋅ s ) = 0. watts
P = Fv = (0.1N
)
01
P = ξI = ( 0 .01V )( 1A) = 0. 01 watts
The rate of energy conversion, ξI, equals the rate of mechanical energy input to
the system.
• We have conjured up a 0.01 watt electrical generator.
• We would have to supply 0.01 watts of mechanical energy to move the
conductor to produce 0.01 watts of electrical energy assuming 100%
conversion.
• Is this realistic? What is an obvious source of loss we have not computed?

Example 2
Consider the induced EMF from a conducting loop rotating as shown below with an
angular velocity ω.
z
y
v
θ
B
ξ = v × B
B
θ
x
B
ξ = v × B
a
v
ω
b/2
First let’s examine the motional EMF produced by each side “a “of the loop.
B
⎛ b ⎞
v = rω = ⎜ ⎟ ω
⎝ 2 ⎠
The velocity of side a.
a
ξ
v
b/2
ξ = vBlsinθ = vBasinθ
b
= ω sinθ
2 Ba

Since there are two a sides for this loop these two EMF’s add and the total EMF is:
ξ = ωabB
sinθ
Notice that this EMF is directed in such a manner that it will cause current to flow
through the a sides of the loop in the direction shown.
On the “b” sides of the loop (top and bottom):
B
ξ
v
The magnetic forces on b sides are transverse to the conductor and do not
contribute an EMF in a direction that produces current flow down the length of
either side b
The total EMF of the loop is therefore produced solely from the sides a.
ξ total
= ωabB
sinθ
Noting that ab = the area of the loop (A) and thatθ = wt :
ξ total
= ωAB sinωt
= ωAB
sinθ
Note:
• ξ varies sinusoidally with time.
• ξ = ξ max when sin ωt = 1 this occurs when B
is parallel to the plane of the loop and is
perpendicular to µ and A
• ξ max = ωAB
• ξ = ξ max sin ωt
B
A
µ r
Basis for the construction of an electrical generator or alternator

The Alternator
a
slip ring
B
V ab
+
b
ring rotates with
the loop
Notice current reverses direction in the loop as it turns. To see this look only at
one segment of the loop as it rotates through 360 0 . The angles given are between
the B and µ vectors.
• θ = 0 0 → 90° I starts at zero and is increases to I max + (ccw)
I
• θ = 90 0 → 180 0 I begins at its maximum value and
decreases to zero, still + (ccw)
t
v
I
B
B
ξ
v

• θ = 180 0 → 270 0 I starts at zero and is increasing – (cw)
• θ = 270 0 → 360 0 I starts – I max and decreases to zero – (cw)
v
B
B
ξ
v
I
So the direction of the current in side a reverses each half cycle as it does in the
entire loop.
Let’s modify this device as follows:
V ab
a
B
t
b
ring rotates with loop
gap
commutator
At the angular positions where the current reverses itself, the connections to the
external circuit are reversed. The EMF is always is the same direction but varies
from 0 to some maximum value. This “half wave” may be easily rectified or
converted from alternating to direct current.

Faraday’s Law
Faraday’s Law relates changes in magnetic flux to EMF
d
a
∆s
⊗
A
•
⊗
l +
v
⊗
⊗
⊗
⊗
c
b
• We have already looked at this circuit using the concept of magnetic force
to produce a motional EMF.
• We can also examine the change in magnetic flux through this circuit due to
the increasing area. A change in magnetic flux produces EMF.
• Changes in magnetic flux may be due to changes in the magnetic field
strength or direction or a change in the area being penetrated by the
magnetic field lines (which applies here?).
• When the conductor moves to the right at distance ds, the area enclosed by
r r
abcd increases by: ∆ A = l∆s
v v r r
• The change in magnetic flux through abcd is: ∆Φ
m
= B ⋅ ∆A
= Bl∆s
cosθ
r r
Since B and A are parallel: ∆Φ
m = Bl
∆ s
• Noting that l and B are constant, the time rate of change of the magnetic
∆Φ
m ∆s
flux is:
= Bl = Blv
∆t
∆t
r r
∆Φ
m ∆( B ⋅ A)
• Faraday’s Law: ξ = -
∆t
= −
• Faraday’s Law of induction is valid for any circuit for which there is time
varying magnetic flux, even circuits in which there is no evident motion.
• The change in flux may be caused by a change in area with respect to time, a
change in magnetic field with respect to time, or both
• The negative sign is a matter of convention.
∆Φ
m
• For multiple loops: ξ = -n
∆t
∆t

Consider a Transformer – a device that uses a change in magnetic flux to induce a
current.
B
I
Magnetic forces
only influence
charges in
motion
If I varies with time B also varies with time and an induced current appears in the
winding on the right.
Transformers are widely used for switching, electronic isolation and stepping
voltages up or down.

Example 3
Consider a solenoid; n = 500 loops, r = 4 cm as shown below. The magnetic field is
changing at a rate of 0.2T per second. Find the induced EMF in this solenoid
B
B
B is increasing at
B
B
−1 ∆B
0.2T
⋅ s = while the area remains unchanged.
∆t
The change in magnetic flux here is due to a changing magnetic field rather than a
change in area penetrated by the magnetic field.
Φ
m
=
BA
∆Φ
∆t
m
=
∆B
A
∆t
A = π(.04m)² = 0.00503m²
∆Φ
Q
∆t
m
=
∆B
A
∆t
= (.00503m
)(0.2T
⋅ s
2 −1
)
.00101
2 −1
= T ⋅ m ⋅ s
= .00101Wb
⋅ s
−1
∆Φ
ξ = -n
∆t
m
= (500)(.00101T
⋅ m ⋅ s
2 −1
) = 0.505 Volts
Note that we are not really worried about the sign. We were not really told that
∆B
was positive or negative so worrying about the significance of the sign in the
∆t
answer makes no sense.
Which direction does the induced current flow in the coils? (bottom to top)

What if the situation had looked like this:
θ = 30°
B
We want ⊥ component of B so B(cos 30°) = B ⊥ = B. So we would multiply the
result above by cos 30°

B v v v w
Example 4
Let’s look again at the induced EMF produced by the same rotating loop we looked
at earlier.
z
y
v
θ
B
ξ = v × B
B
θ
x
B
Recall:
ξ = v × B
a
v
ω
b/2
B
θ
v
B
r
F m
r v
= qv × B
Notice that the cross product in this expression “picks off” components of v
r v
perpendicular to B. In this particular configuration v × B = 0 (F m and I at minimum
values) but when the loop rotates 90 0 both I and F m are at a maximum value.
Based on evaluating the time varying cross product (with a few twists and turns) we
arrived at:
b
ξ = vBl sinθ → ξ = 2 ω a Bsinθ
= ωABsinθ
= ωABsinωt
2

Now consider evaluating the induced EMF with Faraday’s Law.
A v
r
B v
w
B v
v v
The magnetic flux through the loop is Φ
m
= B ⋅ A = BAcosθ
. Notice that the dot
product in this expression “picks off” parallel components of B and A. In this
particular configuration the maximum area of the loop is exposed to the B field so
flux through the loop is at a maximum value.
The instantaneous flux through the loop is
v v
Φ = BA cos θ = B ⋅ A
But the loop is rotating with an angular velocity ω so the flux is changing as a
function of ω . Recalling that θ = ωt
:
Φ
m
= BAcosωt
Faraday’s Law yields (with a little arithmetic massage):
∆Φ
ξ = −
∆t
m
= ωBAsinωt
= ωBAsinθ
Notice that this is the same result as derived earlier.

Lenz’s Law
Lenz’s Law is an alternative method for determining the direction of an induced
current.
The direction of an induced current is such as to oppose the cause producing it.
“Causes” may include:
• motion of a conductor in a B field
• change in flux
Lenz’s Law states that any induced EMF opposes either the motion causing the
induced current or the change in flux causing the induced current. Conservation of
energy is the underlying principle.
Examples
Moving conductor in a B field
x
I
x
x
x
x
x
x
x
v
F
v v
= qv × B
F
The induced EMF opposes the motion of
the conductor creating the current

Rotating loop in a B field.
B v
v
v v × B
I
v v
× B
ω
The velocity here is
of the current not the
tangential velocity of
the loop.
Notice that the induced EMF
creates a current which produces a
force that in turn opposes rotation
of loop.
Two solenoids in motion with respect to each other:
a
R
b
+
ε
_
1) ξ increasing
N
repulsive
N
Current must be CCW i.e.
a → b
2) ξ decreasing
S
Current must be CW
attractive
N

Summary - Induced EMF
• ξ = vBl
Works for a conductor moving through a stationary B field
− ∆Φ
m
• ξ = Works for B fields that change with time
∆t
Since motion is relative the second method always works
For coils of N turns:
∆Φ
m
ξ = −N
∆t

Applications of Induced EMF’s
Transformers are devices that exploit the constantly changing magnetic flux of
alternating currents to induce currents in devices that do not physically touch.
Transformers are widely used for both electronic isolation (for low electronic noise
and for safety) and to step up or step down AC voltages
Some advantages of AC over DC:
• Easier to step up and down
• Easier to transmit
• Can use high voltage & low current to reduce power losses in
transmission lines
Most transmission lines carry about 500kV which must be stepped down (converted
to lower voltages) for household or office operation.
Primary
Secondary
V
V
S
P
=
N
N
S
P
⇒
The transformer equation
N
N
S
S
> N
< N
P
P
⇒
⇒
Step up transformer
Step down transformer
• It is important to note that power in = power out.
• Transformers trade voltage for current
V
P S P
V
P
I
P
= VS
I
S
→ = → =
VS
I
P
N
S
I
N
I
I
S
P

A transducer is a device that converts mechanical motion to an electrical signal or
vice versa. Transducers generally work by exploiting the behavior of conductors in
magnetic flux.
The Electric Guitar
String
Nickel
To preamplifier
Permanent
Magnet
To preamplifier
Coil Windings
• Virtually all electric guitars use electromagnetic pickups
• Most guitars have at least two pickups for each string and some have three.
• Pickups are positioned at different locations under the string so that each is
sensitive to different harmonics produced by the vibrating string.
• Electric guitar strings are generally made of nickel – a material easily
magnetized.
• The pickup consists of a coil of wire with a permanent magnet inside the coil.
• The magnet in the pickup induces a magnetic field in the string as shown
above
• The magnetic field of the string vibrates with the string as it is plucked
• The change in magnetic flux due to the vibration of the field lines across the
coil windings produces an oscillating change if EMF in the coil
• An oscillating current is produced at the same frequency as that of the
vibrating string
• The EMF produced is very weak (a few microvolts) and must be amplified to
a level of a few millivolts (line level) in order to drive any subsequent
electronic devices. The device that does this is referred to as a preamp.
• Preamps are also responsible for modifying the tonal characteristics of the
signal.
• In order to drive a loudspeaker the signal must be amplified again by a power
amplifier to a level of typically 10 – 70 volts.
• A loudspeaker is another transducer that converts electricity to mechanical
motion which in turn produces an acoustic wave

Loudspeakers and Microphones
• We have previously learned how a vibrating object such as a guitar string
produces an acoustic wave. We have also seen how acoustic waves propagate
through the air and how the energy in acoustic waves can be transferred to
an object some distance from the source of the wave causing it to vibrate at
the same frequency as the source.
• Sound waves consist of very small displacement amplitudes and minute
fluctuations of pressure. When sound waves impinge upon a surface they
impart very small energy pluses to the incident material because their
displacement amplitudes are small.
• Because sound waves suffer losses from several sources they must be
amplified to travel long distances and still arrive at the source with
sufficient amplitude (volume) for clarity. This is the primary purpose of a
sound system.
• In order to amplify an acoustic wave we must find a way of converting it to
an electrical signal. Transducers are devices that may be used to convert
acoustic waves to electrical signals.
• A very common type of transducer used in converting acoustic signals to
electrical signals is known as a linear electromagnetic motor. The guitar
pickup is one example of a linear electromagnetic motor.
• Linear electromagnetic motor are based on electromagnetic induction. The
diagram below contains a cross-sectional view of a simplified linear
electromagnetic motor.

• This particular L.E.M. consists of an iron magnet and a coil of wire (normally
copper) attached to a diaphragm.
• As a sound wave impinges upon the diaphragm it vibrates at the same
frequency as the incoming wave.
• The greater the SPL (amplitude) of the incident wave, the greater the
amplitude of vibration imparted to the diaphragm and attached wire coil.
• As the coil moves back and forth within the magnetic field of the fixed
magnet an alternating current is induced that changes direction each time
the coil changes direction.
• The larger the amplitude of oscillations, the stronger the induced voltage.
• Notice that as long as the movement of the diaphragm does not exceed the
elastic limits of the mounting system it behaves exactly like a simple
harmonic oscillator. The waveform generated is therefore sinusoidal.
• This is generally how a dynamic microphone operates. In order to contain the
infrastructure needed to create this transformation in a compact package it
is necessary to limit the size of the magnet and wire coil. Large diaphragms
and massive wire coils, due to their large inertia, will not vibrate as easily as
smaller diaphragms and light coils unless the SPL of the incoming wave is
extremely high.
• Because of inertial limitations most dynamic microphones produce extremely
low-voltage signals, typically a few microvolts, which must be amplified by
other electronic devices in order to gain enough strength to drive a
loudspeaker.
• There are other devices capable of converting an acoustic signal to an
alternating current including capacitor, piezoelectric and ribbon transducers.
• Loudspeakers are linear electromagnetic motors that are essentially
microphones in reverse, i.e., loudspeakers convert electrical signals to
acoustic waves.
• In most loudspeakers alternating current flowing through a wire coil in the
presence of a fixed magnetic field causes the coil to oscillate at the
frequency of the source signal. If the coil is attached to a diaphragm,
acoustic waves will be produced as the diaphragm vibrates back in forth in
air.