Chapter 21: Electromagnetic induction
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about
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electromagnetic
electromagnetic
electromagnetic
electromagnetic
electromagnetic
electromagnetic
electromagnetic
electromagnetic
electromagnetic
electromagnetic
induction
induction
induction
induction
induction
induction
induction
induction
induction
induction
What will we learn in this chapter?
Induction
Magnetic flux
Faraday’s law
Lenz’s law
Eddy currents
Self inductance
Transformers
Magnetic field energy
R–L and L–C cuircuits
generators in a power plant
Induced currents and emf
Induced currents play an important role in everyday life:
Power generation (power plants, dynamos, …)
Magnetic recording and data storage (tapes, hard disks, …)
Wireless energy transmission (toothbrush charger, cooktop, …)
Goal:
Understand how these phenomena work.
When are currents induced in a conductor by a magnetic field?
When are currents induced?
Study the following circuit:
A coil connected to a galvanometer.
Because there is no emf source connected to
the system, we expect to see no current in
the meter.
What happens, if a magnet is
placed close to the coil?
Still nothing.
There is still no emf
source in the circuit.
Induced currents contd.
When the following experiments are performed, a current can be
measured in the meter:
move a magnet
toward/away from the coil
move a second current-carrying
coil toward/away from the magnet
vary the current in the
second coil (open/close switch)
Changing the magnetic flux in the coil causes an induced current.
The corresponding emf is called induced emf.
Magnetic flux
Analogous to electric flux ΦE .
Derivation:
Divide an arbitrary area into small
segments ∆A .
For each segment determine the
component of the B-field normal
to the surfaceB⊥:
B⊥ = B cos φ
The magnetic flux trough the area element is
∆ΦB = B⊥ ∆A = B∆A cos φ
The total flux is then the sum of all ∆ΦB .
��
Mathematically:
� S
�
ΦB =
Bd
S
Special case: planar surface w. constant field
Magnetic flux trough a planar surface in a constant magnetic field: For a
planar surface with area A in a uniform field B the magnetic flux is given
by
Φ = B A = BA cos φ
B
⊥
The SI unit for magnetic flux is called weber (Wb):
1Wb = 1Tm2 = 1Nm/A
ΦB = BA
ΦB = BA cos φ
ΦB = 0
Faraday’s law
Faraday (UK, 1791 - 1867) discovered the relationship between an
induced emf and a changing magnetic field (flux, to be precise).
Faraday’s law of induction: The magnitude of the induced emf in a
circuit equals the absolute value of the time rate of change of the
�
�
magnetic flux trough that circuit:
� ∆ΦB �
�
E = ��
∆t �
In this definition the emf is positive. For a coil of N turns E → N E .
Example: voltage induced in a loop of area
A = 120 cm2 with a field increasing at a rate
of 0.02 T/s.
�
�
� ∆ΦB � ∆B
�=
E = ��
A = 2.4 × 10−4 V
�
∆t
∆t
Note that the sign of the voltage is not determined.
Example: railgun
Goal:
accelerate the crossbar to 300 m/s
using a 9V battery and two rails
2cm apart.
How large must the B-field be?
Solution:
�
�
� ∆ΦB � B∆A
�=
E = ��
∆t �
∆t
E = BLv
B=
∆A = L∆s
∆s = v∆t
E
9V
=
= 1.5T
Lv
0.02m · 300m/s
A neodymium magnet of 500g should do the trick (~ 1.5T field)
Note: this is an academic exercise and not meant for
experimenting at home.
Generators revisited
Principle of a generator:
A coil of area A rotates in an external magnetic field B thus
producing an emf.
Alternators (ac-generators)
Why is the induced emf sinusoidal?
The generator loop starts perpendicular to the field, i.e.,ΦB = BA .
At any given point in time during the rotation at a frequency ω:
ΦB = BA cos φ = BA cos(ωt)
In general,
∂ΦB
∂t
this means that the induced emf is the slope (derivative, or rate of
change) of the flux at any given point.
It follows:
E = ωBA sin(ωt)
The resulting current is alternating, we call this generator an
alternator or ac-generator.
E =−
Commutators (dc-generators)
Can we generate an emf which is always positive? Yes!
A commutator uses a split ring to transfer the emf to a circuit. At
the points where the emf is zero the polarity of the connections is
reversed.
Commercial dc motors have a large number of commutators and
coils thus smoothing out the bumps in the emf (patent by Tesla).
Lenz’s law
Lenz’s law: The direction of any magnetically induced current or emf
is such as to oppose the direction of the phenomenon causing it.
Note:
So far we have only discussed the magnitude of an induced emf.
Lenz’s law determines the direction.
Simple example: A ring in an increasing field.
The current in the loop produces an additional
magnetic field.
Lenz’s law states that this induced field
must oppose the direction of the
original field.
It follows that the current must be clockwise
when seen from above.
Currents induced in a loop by a magnet
Moving down…
As Lenz’s law requires,
the induced field
opposes the change
in flux trough the
circuit and not the
flux itself.
Recall: a current-carrying
conductor is acted
upon a force when
placed in a magnetic
field. According to Lenz
this force opposes the
motion that induces
the current.
Currents induced in a loop contd.
When the magnet
moves up, one can
use similar arguments
to determine the
direction of the
current in the loop.
Example: railgun revisited.
What happens if we move the rod
by hand at a given velocity v? In which
direction does the current go?
Solution:
The rod’s motion increases the flux
trough the loop.
According to Lenz, the current must
be as to oppose the motion.
Therefore the field points out of the
page.
Using the RH rule for the induced
field it follows that the current is
counterclockwise.
setup
Motional electromotive force
When a conductor moves in a magnetic field, we can gain further
insights into the induced emf by considering the forces on charges in
the conductor.
Consider a rod moving in a field:
A charge q will experience a force FB = qvB .
If B points into the plane and v is to the right,
the force points upward (positive charge).
Inside the bar a charge imbalance occurs since
positive charges move to the top thus producing
an E-field from a to b which produces a downward
force on the charges.
This continues until FB = FE = qE.
The potential Vab is given by LE, where E = vB, i.e.,Vab = vBL .
Motional electromotive force contd.
Suppose the rod slides along a U-shaped conductor:
The circuit is complete.
No magnetic force acts on the
charges in the conductor, but there
is an electric field caused by the
accumulation of charges at a and b.
Under the action of this field a
current is established counterclockwise.
The moving rod has become a source of emf with charges moving
from higher to lower potential.
The emf is called motional emf. When the velocity, magnetic field
and rod are mutually perpendicular:
E = vBL
Note: this is analogous to a battery, but the emf source is different.
Déjà vu: the railgun
What is the force on the rod?
Assume the loop has a resistance
of 0.03Ω.
Recall: B = 1.5T, L = 2cm, V = 9V.
Solution:
I = E/R = 300A
Find = ILB = 9N
The induced current is quite large!
receiver
transmitter
So far the currents induced by magnetic fields
have been confined to well-defined paths (e.g., wires).
Some electric equipment has large metallic pieces
moving in a field. Currents induced there behave
like eddies in a river, i.e., eddy currents.
The downward currents
at O–b induce a force
opposed to the direction
of the rotation. The
return currents lie
outside the field so no
force occurs.
The eddy currents cause a breaking action; electrical
energy is converted into heat (resistance of the material).
Applications: breaking devices for public transport,
ergometers, metal detectors, …
eddies
Eddy currents
Mutual inductance
Q: !
! !
A: !!
How can a 12V car battery provide thousands of Volts
needed to produce sparks in the engine chamber?
Mutual inductance.
Concept: A changing current in a coil
induces an emf in an adjacent coil.
A current i1 in coil 1 produces a
flux which passes trough coil 2.
�
�
emf in coil 2:
� ∆ΦB2 �
�
E2 = N2 ��
∆t �
The flux in coil 2 is proportional
to the current i1:
N2 |ΦB2 | = M21 |i1 |
Mutual inductance contd.
It follows:
�
�
�
�
� ∆ΦB2 �
� ∆i1 �
� = M21 �
�
N2 ��
� ∆t �
∆t �
Thus:
�
�
� ∆i1 �
�
E2 = M21 ��
∆t �
We can repeat this calculation for the
case where a current i2 induces a flux
in coil 1:
�
�
�
�
� ∆ΦB1 �
� ∆i2 �
�
�
E1 = N1 ��
E1 = M12 ��
∆t �
∆t �
The proportionality constant M12 turns out to be equal to M21, we
call it mutual inductance M:
M12 = M21 = M
Mutual inductance contd.
Mutual inductance: The mutual inductance of two coils is given by:
�
� �
�
� N2 ΦB2 � � N1 ΦB1 �
�=�
�
M = M12 = M21 = ��
i1 � � i2 �
From the preceding analysis we can also write:
�
�
�
�
� ∆i1 �
� ∆i2 �
�
�
E2 = M ��
E1 = M ��
∆t �
∆t �
The SI unit of mutual inductance is called henry (1 H):
1H = 1Wb/A = 1V · s/A = 1Ω · s
Note:
If the coils are in vacuum, M depends only on their geometry.
If magnetic material is present, M will depend on the properties of
the material.
Self-inductance
So far we have studied the effects of one circuit on another.
In general, a circuit carrying a time-dependent current has an induced
emf resulting from the variation of its own magnetic field.
This is called self-inductance.
Example: ! coil with N turns carrying a varying
! ! ! current i.
A magnetic flux passes trough the coil.
Define self-inductance L of the circuit:
�
�
� N ΦB �
�
L = ��
N |ΦB | = L|i|
i �
If the flux and the current change with time we can write:
�
�
� �
� ∆ΦB �
� �
� = L � ∆i �
N ��
� ∆t �
∆t �
Self-inductance contd.
Using Faraday’s law the magnitude of the emf due to the self
inductance is E = N |∆ΦB /∆t| . Hence we can define L as follows:
Self-inductance: The self-inductance L of a circuit is the magnitude of
the self-induced emf E per unit rate of change of current, so that
� �
� ∆i �
E = L �� ��
∆t
From the definition, the units are the same as for the mutual inductance
(Henry).
Note:
Part of a circuit designed to have a particular inductance is called
an inductor (or choke).
Inductors are indispensable in electric circuits.
Circuit symbol:
Direction of the self-inductance
The direction of the self-induced emf can be determined from Lenz’s
law (the emf opposes the rate of change in the current).
Constant current: There is no
self-induced emf, hence the
voltage Vab is zero.
Increasing current: In this case
∆i/∆t > 0 . The emf opposes
the current and so Vab > 0.
Decreasing current: In this case
∆i/∆t < 0 . The emf points in
the same direction as the
current and so Vab < 0.
Self-inductance: Final considerations
The inductance of a circuit depends on its size, shape, and number of
windings.
For a coil with N turns close together, the inductance is proportional
to N2.
The inductance depends on the magnetic properties of the
material enclosed by the circuit:
If the flux is concentrated in a region filled with magnetic
material, then in any calculation of the B field we must
replace
µ →µ=K µ
0
m 0
where Km is the permeability of the material.
For example, for a soft iron core Km = 5000! This is why iron and
ferrite cores are used in technological applications.
Transformers
No, not this kind…
Transformers
Q: !
A: !!
What is the advantage of ac over dc current?
It is easier to step voltage levels up and down.
Examples:
Long-distance power transmission: Increase
the voltage to 500kV to reduce the dissipated
power P = I2R.
America uses 120V, Europe uses 240V.
Cars use 12V batteries, but require high voltages for the sparks.
Note:
In the following discussions ‘voltage’ represents the maximum of
sinusoidal-varying voltages.
Ac voltages are usually described by their rms√values. This means
the peak voltage in a household will be 120V 2 = 170V.
Construction of a transformer
A transformer consists of 2 coils (called windings) electrically
insulated from each other but wound on the same core.
The winding to which the
power is supplied is called
primary.
The winding to which the
power is delivered is called
secondary.
Circuit symbol:
How it works:
An ac current in the primary sets up an alternating flux in the
core.
According to Faraday’s law, this induces an emf in the secondary.
Transformers contd.
Assume:
The flux is confined to the core.
The flux is the same in the primary and secondary coil.
Neglect the resistance of the windings.
Induced emfs in coils 1 and 2:
�
�
� ∆ΦB �
�
Ei = Ni ��
∆t �
Because the flux is the same
it follows
E2
N2
=
E1
N1
Since we assumed that the resistances are
zero, we obtain for the terminal voltages…
Transformers contd.
Voltage-winding turns relation for a transformer: For an ideal transformer
with zero resistance
V2
N2
=
V1
N1
By choosing the appropriate turnus ratio N2/N1, we may obtain any
desired secondary voltage from a given primary voltage. If V2 > V1 we
have a step-up transformer, if V2 < V1 we have a step-down transformer.
Note:
The V’s can either be both amplitudes or both rms values.
If the secondary circuit is completed by a resistance R, then
I2 = V2/R. From power considerations V1 I1 = V2 I2 . We thus
obtain
V1
R
R
=
R→
2
I1
(N2 /N1 )
(N2 /N1 )2
This means that the transformers also transforms resistances.
Real transformers
In real transformers, power is dissipated:
Power is dissipated due to the resistance of the windings.
Power is dissipated due to eddy currents. Solution: laminated
cores or the use of ferrites.
in the laminated cores the eddy currents are smaller.
Magnetic field energy
Goal: ! !
! ! !
! ! !
Calculate the energy input U needed to establish a final
current I in an inductor with inductance L if the initial
current is zero.
Solution:
Let the current at some instant be i and its rate of change be
∆i/∆t . Then the terminal voltage at that instant is
E = Vab = L∆i/∆t
and the average power P supplied by the current source during
the small time interval ∆t is
∆i
P = Vab i = Li
∆t
The energy ∆U supplied to the inductor during this time interval
is approximately ∆U = P ∆t = Li∆i .
To obtain the total energy while the current increases from zero
to I we need to integrate the equation.
Magnetic field energy contd.
Energy stored in an inductor: The energy stored in an inductor with
inductance L carrying a current I is
1
U = LI 2
2
Note:
We can think of the energy U as analogous to a kinetic energy
associated with the current. This energy is zero when there is no
current.
The energy of an inductor is stored in the magnetic field within
the coils (similar to the energy in a capacitor is stored in the
E-field between the plates).
Magnetic energy density
We can compute the energy density of the magnetic field inside an
inductor.
Simple case: toroidal solenoid (simple geometry, field constrained)
Calculation of the self-inductance:
�
�
� N ΦB �
I
�
L = ��
ΦB = BA
B = µ0 N
�
I
2πr
µ0 N 2 A
L=
2πr
Calculation of the magnetic energy:
1
1 µ0 N 2 A 2
U = LI 2 =
I
2
2 2πr
Volume of the torus: V = 2πrA
We can compute now the energy density u = U/V…
Magnetic energy density contd.
Energy density:
1
N 2I 2
u = U/V = µ0
2 (2πr)2
Recall that
B = µ0 N
I
2πr
We obtain:
Magnetic energy density: The energy density inside an inductor with a
1 2
field B is
u=
B
2µ0
Note:
This expression is generally valid for any geometry.
Recall for the energy density due to an E-field u = (1/2)�0 E 2.
The R–L circuit
Suppose the switch is closed. The current
will not instantly reach a given value since
this would mean an infinite induced emf
in the inductor.
The rate of growth of the current
depends on the induction L.
Let ∆i/∆t be the rate of change of the
current i. The potential difference vbc is
then
∆i
vbc = L
∆t
Across the resistor, the potential difference is vab = iR .
Apply Kirchhoff’s rule:
∆i
E − iR − L
=0
∆t
∆i
E
R
The rate of change of the
= − i
current is thus given by
∆t
L
L
R–L circuit contd.
At the instant the switch is closed (t = 0) we have i = 0 and thus
�
∆i ��
E
in =
�
∆t
L
The greater the inductance L the slower the current increases.
When the current reaches a steady-state value I the rate is zero, i.e.,
∆i
E
R
=0= − I
→
∆t
L
L
If we solve the differential equation, we
obtain for the current as a function of
time
�
E �
−(R/L)t
i(t) =
1−e
R
I=
At time τ = L/R (time constant) the
current will have risen to 1 - (1/e).
Current decay in an R–L circuit
Suppose switch S1 has been closed long
enough to reach the steady state I0.
Close switch S2 at time t = 0 to bypass the
battery (open S1 to save the battery…).
Applying Kirchhoff’s rule we obtain
∆i
R
=− i
∆t
L
and the current varies as
I = I0 e−(R/L)t
The time constant τ = L/R
is the time required for the current
to decrease to 1/e.
One can show that the energy
stored in the inductor decreases
at a rate equal to the dissipation
rate i2R.
E
R
The L–C circuit
A circuit comprised of an inductor L and a
capacitor C shows a completely different behavior:
In a R–L and R–C circuit we have exponential
damping.
In a L–C circuit the stored energy oscillates
between the capacitor and the inductor. We obtain an electrical
oscillation.
Note that the total energy of the circuit is constant. This is
reminiscent of kinetic/potential energy conservation in a
pendulum.
The oscillation occurs with an angular frequency
1
ω=√
LC
L–C circuits are used to tune radios to a given frequency.
The L–C circuit contd.