Inductors
A coil of wire can create a magnetic field if a
current is run through it. If that current
changes (as in the AC case), the magnetic field
created by the coil will change.
Will this changing magnetic field through the
coil cause a voltage to be created across the
coil? YES!
This is called self-inductance and is the basis
behind the circuit element called the inductor.
Inductors
Recall Faraday’s Law: V = d/dt [ BdA ] .
Since the voltage created depends in this case on
the changing magnetic field, and the field
depends on the changing current, and since the
angle (dot product) and the area don’t change,
we have: Vinductor = -L dI /dt
where the L (called the inductance) depends on
the shape and material (just like capacitance and
resistance).
Inductors
Vinductor = -L dI /dt
Here the minus sign means that when the
current is increasing, the voltage across the
inductor will tend to oppose the increase, and
it also means when the current is decreasing,
the voltage across the inductor will tend to
oppose the decrease.
Units: Henry
From Vinductor = -L dI /dt
L has units of Volt / [Amp/sec] which is
called a Henry:
1 Henry = Volt-sec / Amp .
This is the same unit we had for mutual
inductance before (recall the transformer).
Solenoid type inductor
For a capacitor, we started with a parallel plate
configuration since the parallel plates
provided a uniform electric field between the
plates.
In the same way, we will start with a device that
provides a uniform magnetic field inside the
device: a solenoid.
Solenoid inductor
Recall: Bsolenoid = nI where n = N/Length .
Recall Faraday’s Law: V = d/dt [ BdA ] .
Since the I in the B is a constant with respect to
the variable of integration (dA), we have:
V = L dI/dt where L = (n) dA , and since
B is uniform over the area for the solenoid and
in the same direction as area: L = (n)A.
But the area is for each of N loops, so:
L = nNA = N2A/Length where Aeach loop = R2.
L = N2R2 / Length .
Size of a Henry
From the inductance of a solenoid:
Lsolenoid = N2R2 / Length
we see that with vacuum inside the solenoid,
becomes o which has a value of 4 x 10-7 T-m/A, a
rather small number. R (the radius) will normally be
less than a meter, and so R2 will also make L small.
However, N can be large, and can be a lot larger than
o if we use a magnetic material – see the slides later in
Altogether, a henry is a rather
this section.
large inductance.
Inductors
Lsolenoid = N2R2 / Length
As we indicated before, the value of the
inductance depends only on the shape (Radius,
Length, Number of turns) and materials ().
The inductance relates the voltage across it to
the changing current through it: V = - L dI/dt .
We again use Lenz’s Law to give us the direction
(sign) of the voltage.
Remember that the V above is really the ΔV across the terminals of the
inductor.
Energy Stored in an Inductor
We start from the definition of voltage:
V ≡ PE/q (or PE = qV). But since the voltage
across an inductor is related to dI/dt, the current
change, we might express q in terms of I:
I = dq/dt, or dq = I dt. Therefore, we have:
Estored = qi Vi = V dq = V I dt and now we
use VL = L dI/dt to get:
Estored = (L dI/dt) I dt = L I dI = (1/2)LI2.
Sorry about the multiple L’s. On this slide, all the L’s are inductances, not lengths.
Review of Energy in Circuits
There is energy stored in a capacitor (that has
Electric Field): Estored = ½CV2 .
Recall the V is related to Efield, so Estored Efield2.
(here means proportional to)
There is energy stored in an inductor (that has
Magnetic Field): Estored = ½LI2 .
Recall the I is related to B, so Estored Bfield2.
There is power dissipated (as heat) in a resistor:
Plost = RI2 .
Sorry for the multiple E’s: for energy and for electric field.
Review of Circuit Elements
Resistor: VR = R I where I = q/t
Capacitor: VC = (1/C)q (from C ≡ q/V)
Inductor: VL = -L I/t
We can make an analogy with mechanics:
q is like x; t is like t,
I = q/t is like v = x/t; I/t is like a = v/t
V is like F; C is like 1/k (spring);
R is like air resistance, L is like m.
(we do this in the following slides)
RL Circuit (DC)
What happens when we have a resistor in series
with an inductor in a circuit?
From the mechanical analogy, this should be
like having a mass with air resistance. If we
have a constant force (like gravity), the object
will accelerate up to a terminal speed (due to
force of air resistance increasing up to the point
where it balances the gravity).
F=ma bv mg = ma, or
m dv/dt + bv = -mg
If we treat down as positive, then the – sign goes away.
Mechanical Analogy: mass
falling with air resistance
M a s s fa l lin g w i th a ir re s i s ta n c e
20 0
sp e e d in m / s
15 0
10 0
50
0
1 6.0
2 4.0
3 2.0
4 0.0
4 8.0
5 6.0
6 4.0
8.0
0
ti m e in se c o n d s
RL Circuit (DC) (cont.)
If we connect the resistor and the inductor to a
battery (constant voltage similar to constant gravity)
and then turn the switch on, from the
mechanical analogy we would expect the
current (which is like velocity) to begin to
increase until it reaches a constant amount
(like velocity reaches a terminal velocity).
LR Circuit - qualitative look
From the circuit point of view, initially we
have zero current so there is no V R (voltage
drop across the resistor). Thus the full voltage
of the battery is trying to change the current,
hence VL = Vbattery, and so dI/dt = Vbattery /L.
This is like the falling mass: initially there is no
speed so there is no air resistance and the mass falls
with the full acceleration due to gravity.
LR Circuit - qualitative look
However, as the current increases, there is more
voltage drop across the resistor, V R, which
reduces the voltage across the inductor
(VL = Vbattery - VR), and hence the smaller VL
reduces the rate of change of the current!
Remember that all the V’s here - and most places - are really ΔV’s, voltage
drops across the terminals of the elements.
This is like the mass with gravity and air resistance:
as the mass falls and picks up speed there is more and
more air resistance opposing gravity so there is less and
less acceleration and so the speed levels out.
RL Circuit:
Differential Equation
To see this behavior quantitatively, we need
to get an equation.
We can get a differential equation by using
the Conservation of Energy (Vi = 0):
Vbattery - Vresistor - Vinductor = 0.
This looks like an ordinary algebraic
equation. But all the V’s are not constants:
we have relations for Vresistor and Vinductor.
RL Circuit:
Differential Equation
Vbattery - Vresistor - Vinductor = 0.
With Vbattery = constant, Vresistor = IR, and
Vinductor = L dI/dt, we have the differential
equation (for I(t)):
Vbattery - IR - L dI/dt = 0. This can be
rewritten as: IR + L dI/dt = Vbattery
which is an inhomogeneous first order differential
equation, just like we had for the mechanical
analogy: m dv/dt + bv = mg
(we drop the minus sign on mg to make down positive).
RL Circuit (cont.)
IR + L dI/dt = Vbattery
The homogeneous equation is: LdI/dt + RI = 0.
This has a dying exponential solution:
IH(t) = Io e-(R/L) t
which is similar to the Q(t) dying exponentially in the RC
circuit, with the e-1 time of =L/R instead of =RC.
The inhomogeneous equation is:
L dI/dt + RI = Vbattery .
This has the simple solution: II(t) = Vbattery / R.
RL Circuit (cont.)
I(t) = IH(t) + II(t) = Io e-(R/L) t + Vbattery / R .
To find the complete solution (that is, find Io), we
apply the initial conditions:
I(t=0) = 0 = Ioe0+ Vbattery/R so Io = -Vbattery/R .
Therefore, we have:
I(t) = [Vbattery/R]*[1 - e-(R/L) t ] .
Compare this to the charging of a capacitor in a RC circuit:
Q(t) = CVbattery*[1 - e-t/RC]
RL Circuit
I(t) = [Vbattery/R]*[1 - e-(R/L)t ] .
At t=0, I(0) = 0.
As time goes on, the current, I(t), does increase
and finally (when t is large and e-(R/L)t is tiny) reaches
the value Vbattery /R which is what it would be
without the inductor present.
The graph on the next slide shows this function
when Vbattery = 24 volts; R = 24 , and
L = 1 H. Note that the max current is 1 Amp,
and the time scale is in milliseconds.
RL Circuit: I(t) versus t
I(t) vs t for RL Circuit
1.2
1
I in amps
0.8
0.6
0.4
0.2
0
1
6
11
16
21
26
31
36
41
46
t in milliseconds
Mechanical Analogy: mass
falling with air resistance
M a s s fa l lin g w i th a ir re s i s ta n c e
20 0
sp e e d in m / s
15 0
10 0
50
0
1 6.0
2 4.0
3 2.0
4 0.0
4 8.0
5 6.0
6 4.0
8.0
0
ti m e in se c o n d s
RL Circuit
Note that this graph of I(t) versus t looks
qualitatively just like v(t) versus t for the
mass falling under constant gravity with air
resistance. According to the analogy, the
inductance acts like the mass, and the
resistance acts like the coefficient for air
resistance.
RL Circuits – cont.
If you have an inductor in a DC circuit, it only
comes into play when you turn on the DC
voltage or turn it off (with a switch).
When you turn the switch on, we saw that it
takes a little time for the current to build up.
What happens when you turn the switch off?
RL Circuits – cont.
What happens when you turn the switch off?
When you turn the switch off, there is a large
change in the current in a short time, which
means there is a large voltage across the
inductor. There is also a lot of energy stored
in the inductor in the form of the magnetic
field in the inductor. What will this large
voltage with energy behind it do?
RL Circuits – cont.
What will this large voltage in the inductor with
energy behind it do?
If you don’t have a path to dissipate that
energy, it will cause a spark to occur at the
switch!
What circuits have a large inductance?
Anything with a large magnetic field,
especially electromagnets!
DC Circuits with C and L
Because of the energy stored in the inductor, it is
good practice in lab to turn the voltage down
before you turn a power switch off - as we just
saw.
What about the energy in a capacitor when the
DC power switch is turned off?
Unless we give the capacitor a path to discharge,
it may keep its charge until the air discharges it,
and so may give someone a jolt if touched soon
after the switch is turned off!
R, L, and C in a circuit
We have already considered an RC circuit (in
Part 2) and an RL circuit (just now). We looked
at these cases for a circuit in which we had a
battery and then threw the switch.
Other than when turning a DC circuit on and off, the L has
no effect and the C acts as a break in the circuit.
However, the main reason these circuit
elements are important is in AC circuits. We
look next at the case of the three elements in
a series circuit with an AC voltage applied.
LRC Circuit: Oscillations
Newton’s Second Law: F = ma can be
written as: F - ma = 0 . With a spring and
resistance, this becomes: -kx -bv -ma = 0
This is like the equation we get from
Conservation of Energy when we have a
capacitor, resistor and inductor: V = 0 , or
(1/C)Q + RI + L dI/dt = 0 ,
where VC = (1/C)Q, VR = RI, and
VL = L dI/dt.
Resonance
If we put an inductor and a capacitor with an AC
voltage, we have the analogy with a mass connected
to a spring that has an oscillating applied force.
In both of these cases (mechanical and electrical),
we get resonance. We’ll demonstrate this in
class with a mass and spring.
This, it turns out, is the basis of tuning a radio!
The system will “resonate’ at some frequency. What
does the resonant frequency for the system
(mechanical or electrical) depend on?
LRC Circuit:
Differential Equation
Starting with Conservation of Energy
(Vi = 0) and using VR = IR, VC = Q/C,
VL = LdI/dt , and applying a sine wave
voltage (AC voltage), we get:
(1/C)Q + RI + LdI/dt = Vosin(t)
Putting this in terms of Q (I = dQ/dt):
(1/C)Q + R(dQ/dt) + L(d2Q/dt2) = Vosin(t)
we have a second order linear inhomogeneous
differential equation for Q(t).
LRC Circuit:
Differential Equation
(1/C)Q + R(dQ/dt) + L(d2Q/dt2) = Vosin(t)
Note: Once we find Q(t), we can find I(t) since I(t) = dQ(t)/dt.
We again look at the homogeneous solution
(that is, without the applied voltage), and then try
to find the inhomogeneous solution.
If the resistance is small, the homogeneous
equation becomes fairly simple
(1/C)Q + Ld2Q/dt2 = 0 and has the solution:
Q(t) = Qosin(ot) .
LC Circuit
When we substitute this expression
Q(t) = Qosin(ot)
into the diff. Eq.,
(1/C)Q + Ld2Q/dt2 = 0
we find the “natural frequency”: o = [1/LC]
And just like the case of the mass on the spring
(mechanical analogy), when the applied
frequency approaches this “natural
frequency”, o, the amplitude of the resulting
oscillation gets very large (it resonates).
LRC Circuit
By placing a resistor in the circuit, the
differential equation becomes a little harder.
We need to consider either both sines and
cosines, or we need to consider
exponentials with imaginary numbers in the
exponent. This can be done, and reasonable
solutions can be found (we do this in PHYS 380
Advanced Mechanics and in PHYS 340 Electromagnetic
Fields), but
we will not pursue that here. We
will pursue an alternate way.
LRC Circuit:
Impedance
An alternate way of considering the LRC Circuit
is to use the concept of impedance. The idea
of impedance is that all three of the major
circuit elements “impede” the flow of the AC
current.
A resistor obviously impedes (limits) the current
in a circuit. But a capacitor and an inductor
also limit the current in an AC circuit.
LRC Circuit
The basic idea we will pursue is that in a
series circuit,
a) the same current flows through all of the
elements: I = Io sin(t) (in a parallel circuit, the
same voltage is across each arm) ; and
b) the voltages at any instant add up to zero
around the circuit:
VR(t) + VC(t) + VL(t) = VAC(t) .
Resistance
Because of Ohm’s law (VR = IR), we see that
the current and the voltage due to the
current are in phase, that is, when the
oscillating voltage is at a maximum, the
oscillating current is also at a maximum.
VR = VRo sin(t)
I = Io sin(t)
VRo = IoR
where the sub zero indicates the amplitude of the oscillating
voltage or current.
Capacitive Reactance
For a capacitor: VC = (1/C)Q, [from C=Q/V]
and with I = dQ/dt, or Q = I dt, if I = Iosin(t),
then Q = I dt = Iosin(t) dt = (-Io/) cos(t),
so VC = -(1/C)Iocos(t).
Note that cosine is 90o different than sine - we
say it is 90o out of phase. This means that VC
is 90o out of phase with the current!
Note that the constant (1/C) acts just like R in
that it relates V and I. We call this the
capacitive reactance, XC = (1/C) so VCo=XCIo.
Voltage across the capacitor
VC = -VCo cos(t)
I = Io sin(t)
VCo = IoXC where XC = 1/C
For large capacitors it takes longer to fill up the capacitor with
charge, so the voltage across the capacitor is slow to rise
making it less of an impedance to the current flow.
For high frequencies the time for the current to flow to charge
up the capacitor is less so the capacitor doesn’t charge up
very much and so the voltage across the capacitor doesn’t get
very high making it less of an impedance to the current flow.
This is why both C and are in the denominator of XC.
Inductive Reactance
For an inductor, VL = L dI/dt,
if I = Iosin(t), then dI/dt = Io cos(t).
Note that cosine is 90o different than sine - we
say it is 90o out of phase. This means that
VL is 90o out of phase with the current.
Since VL = L dI/dt, VL = LIo cos(t) , and
we see that the constant L acts just like R.
We call this the inductive reactance,
XL = L so VLo=XLIo.
Voltage across the inductor
VL = VLo cos(t)
I = Io sin(t)
VLo = IoXL where XL = L
For higher inductance, L, there is a larger voltage across
the inductor.
For a higher frequency, , the dI/dt term in the voltage
is higher making the VL larger.
This is why both L and are in the numerator of XL.
LRC in series
Note that in a series combination, the current
must be the same in all the elements, while
the voltage adds. However, the voltage
must add to zero across a complete circuit
at every instant of time. But since the
voltages are out of phase with each other,
the amplitudes of the voltages (and hence
the rms voltages) will not add up to zero!
LRC in series
For: I = Io sin(t)
VR = IR = RIo sin(t)
VC = (1/C)*Q = -(1/C) Io cos(t)
VL = L*dI/dt = (L) Io cos(t)
VAC(t) = VR(t) + VC(t) + VL(t)
= Io [R sin(t) – (1/C) cos(t) + L cos(t)]
= Io Z sin(t+θ) = Vo sin(t+θ) = VAC(t)
where [R sin(t) – (1/C) cos(t) + L cos(t)] = Z sin(t+θ).
Z is some constant (independent of time), and the current and
voltage are out of phase with each other by a phase angle, θ.
Impedance
In 2-D space, x and y are 90o apart. We combine
the total space separation by the Pythagorean
Theorem: r = [x2+y2]1/2 .
If we add up the V’s, this is equivalent to adding
up the reactances. But we must take the phases
into account. For the constant, Z, which we
call impedance, we get :
Vrms = Irms Z where Z = [R2+(L-1/C)2]1/2.
(Note that we had to subtract XL from XC because of the
signs involved.)
Trig Relation
The relation
Io [R sin(t) – (1/C) cos(t) + L cos(t)] = Io Z sin(t+θ)
can also be solved using the trig relation:
sin(a+b) = sin(a)cos(b) + sin(b)cos(a)
where a=t and b=. This will give two
relations: one for sin(t) and one for cos(t),
to solve for two unknowns: Z and . We get
the same value for Z as we do on the previous
slide using the x-y scheme.
Resonance
Vrms = IrmsZ where Z = [R2 + (L-1/C)2]1/2
Note that when (L - 1/C) = 0, Z is smallest
and so I is biggest! This is the condition for
resonance. Thus when LC]1/2, we have
resonance. This is the same result we would
get using the differential equation route.
Note that this is equivalent to the resonance of a
spring when km]1/2 , since k is like 1/C
and L is like m.
V(t) versus Vrms
From Conservation of Energy, Vi = 0. This
holds true at every instant of time. Thus:
VC(t) + VR(t) + VL(t) = Vosin(t+) . But due
to the phase differences, VC will be maximum at
a different time than VR, etc.
However, when we deal with rms voltages, we
take into account the phase differences by using
the Pythagorean Theorem. Thus,
VC-rms + VR-rms + VL-rms > VAC-rms .
This is like a 3,4,5 right triangle: where [32+42]½ = 5, but 3+4 5.
Net Power Delivered
What about energy delivered?
For a resistor, electrical energy is changed into heat, so
the resistor will remove power from the circuit.
For an inductor and a capacitor, the energy is
merely stored for use later. Hence, the
average power used is zero for both of these.
Note that Power = I*V, but I and V are out of
phase by 90o for both the inductor and
capacitor, so on average there is zero power
delivered [since, on average, sin(q)*cos(q) = 0].
Net Power Delivered
Thus, even though we have V=IZ as a
generalized Ohm’s Law, power is still:
Pavg = I2R (not P=I2Z).
Computer Homework
There is a computer homework program on
Inductance on Vol. 4, #5, that gives you
practice on DC and AC behaviors of
inductors.
Determining the Inductance
In a previous lab we determined the
capacitance by timing how long it took a
charged capacitor to discharge to 1/e of its
initial charge. We used a DC power supply to
charge up the capacitor.
How can we determine the inductance of an
inductor?
RL AC Circuit
Consider an AC voltage source connected to an inductor that has
some resistance. To measure the frequency of the AC voltage,
we’ll connect an oscilloscope in parallel. The oscilloscope acts
like a capacitor (since its deflection plates act like a capacitor). We need to
be careful where we put the oscilloscope in the circuit.
If we connect the
O-scope oscilloscope across the
VAC as in the diagram,
~ L&R we have an LR circuit (R
is in the coil) in parallel
A with the oscilloscope,
and so we can ignore the capacitance of the oscilloscope since any current in
the O-scope does not go through the ammeter. We now can use V=IZ and
Z = [R2 + (L)2 ]1/2, so if we can measure V and I, we can determine Z, and knowing Z
and measuring R and f (recall =2f) we can determine L. We do this in the AC Circuits
I lab.
Parallel LC AC Circuit
Consider an AC voltage source connected to a capacitor and
inductor connected in parallel. (We’ll assume R is negligible) In the AC
Circuits lab, the oscilloscope acts like a capacitor (since its deflection
plates act like a capacitor). We need to be careful where we put the
oscilloscope in the circuit.
O-scope O-scope
~ L
A
If we connect the oscilloscope across the VAC as in the diagram, we have an LR circuit
(R is in the coil) on the left since any current in the O-scope does not go through the
ammeter; however if we connect the O-scope across the L as in the dotted position, we
have L (and R) with C in parallel which will give different ammeter readings since the
current in the ammeter is due to both the current in the inductor and the current in the
O-scope.
Parallel LC AC Circuits
How do we deal with a parallel LC circuit with an
AC voltage? We first note that in parallel, the
voltage across each leg is the same: VAC = VC = VL.
VAC = Vosin(ωt) = VC = (1/C)Q, so Q = C Vosin(ωt),
so IC = dQ/dt
~ = ωCVocos(ωt) .
A
V AC = V o sin(ωt)
= VL = L dIL/dt so
dIL = (1/L) Vosin(ωt) dt and IL = (-1/ωL) Vocos(ωt).
Therefore: Itotal = IC+IL = {(ωC – (1/ωL)}
Vocos(ωt).
Magnetism in Matter
Just as materials affect the electric fields in
space, so do materials affect the magnetic
fields in space.
Recall that we described the effect of materials
on the electric fields with the dielectric
constant, K. This measured the
“stretchability” of the electric charges in the
materials. This stretching due to applied
electric fields caused electric fields itself.
Two main effects
Atoms have electrons that “orbit” the positive
nuclei. These “orbiting” electrons act like
little current loops and can create small
magnets.
Effect 1: A diamagnetic effect is similar to the
dielectric effect which reduces the field inside
the material – a result of Lenz’s Law.
Effect 2: An aligning effect among the atoms’
individual fields tends to increase the field
inside the material.
1. Diamagnetic Effect
Effect 1: By Lenz’s law, when the applied
magnetic field changes, there is a tendency in
the circuit to resist or oppose the change. This
effect tends to create an induced magnetic
field opposing the change in the external field.
In this effect, the material acts to reduce any
applied external magnetic field.
This is similar to the dielectric effect that leads
to the dielectric constant for electric fields.
Diamagnetic Effect – cont.
In certain materials at (usually) very low
temperatures, these materials can exhibit
superconductivity in which the resistance is
not just small but zero! In these cases, the
diamagnetic effect makes the B field inside
the material zero.
Usually, though, the diamagnetic effect is
very small.
2. Aligning Effect
Effect 2: Since the normal “currents” due to
the “orbiting” electrons act like tiny magnets,
these magnets will tend to align with an
external magnetic field. This tendency to
align will tend to add to any external applied
magnetic field.
2. Aligning Effect – cont.
In ferromagnetic materials (Fe, Ni, Co, and
some alloys), this aligning effect causes areas of the
material to align. These areas of aligned atoms are
called domains.
Within the domains the alignment is strong, but
the domains themselves are usually randomly oriented.
A strong magnetic field will tend to align these
domains. A strong magnetic field when the metal is
hot and then cooled will tend to “freeze” the alignment
of the domains.
Heat or shock can cause these domains to
become randomly oriented.
Net Result
When both of these effects (Lenz’s law and
aligning) are combined, we find three
different types of results:
1. Diamagnetic: Magnetic field is slightly
reduced in some materials
2. Paramagnetic: Magnetic field is slightly
increased in some materials
3. Ferromagnetic: Magnetic field is greatly
increased in a few materials
B, M and H
We are already familiar with B. It is called the
magnetic field or the magnetic flux density
(from its use in V = -d/dt [BdA] ). This is the
total field in space or in the material.
We have H, called the magnetic field strength or
magnetic intensity. This is the field due to
external currents only.
We have M, called the magnetization. It is the
field due to the material only.
Magnetic Permeability and
Magnetic Susceptibility
To give a quantitative measure to the effects
of materials on magnetic fields, and to
relate B, M and H to one another, we define
two additional quantities:
= magnetic permeability: B = H , and
o is the value of in vacuum)
= magnetic susceptibility: M = H .
Further Relations
B = H M = H
B = H = o(H + M) = o(H + H)
= o(1+)H so = o(1+) .
Diamagnetic materials: < 0, «
Paramagnetic materials: > 0, «
(in both cases, 10-4 so o )
Ferromagnetic materials: > 0, »
and for Ferromagnetic Metals
At room temperature, only Fe, Ni and Co are
ferromagnetic.
The values of and for a particular
ferromagnetic metal piece depend on how the
metal piece is formed and the particular metal
piece’s past history [called hysteresis – see next slide].
Values of for ferromagnetic materials can be
between 100 and 100,000 ! This explains why
motors and generators are so heavy – they need the
strong B field that iron supplies.
Hysteresis
Since ferromagnetic materials have domains
where the atomic magnetic fields are aligned,
once the domains themselves are aligned, it is
sometimes hard to get these domains to
switch their orientation. This leads to a
phenomenon called hysteresis. Applying an
external magnetic field (such as the external
H field) will sometimes not be enough to
switch many of the domains.
Hysteresis
is not constant - it depends on history
hard – permanent magnet soft - transformer
B B
H
H
