electromagnetics - 23.1
23. ELECTROMECHANICAL SYSTEMS
23.1 INTRODUCTION
• Magnetic fields and forces are extremely useful. The fields can allow energy storage,
or transmit forces.
•
23.2 MATHEMATICAL PROPERTIES
• Magnetic fields have direction. As a result we must pay special attention to directions,
and vector calculations.
23.2.1 Induction
• Magnetic fields pass through space.
Topics:
Objectives:
electromagnetics - 23.2
• resistivity of materials decreases with temperature
• Amperes Circuit Law
• Flux density can be calculated for low H values. As the value climbs the relationship
becomes non-linear.
• Permeability,
I= °∫Hdl
where,
I = current flowing along a line (A)
H = magnetic field intensity (A/m)
l = A perpendicular path around the current flow (m)
For example, at a fixed radius (r) around a wire,
I Hrdθ
0
2π
= ∫
∴I = 2πHr
∴H I
2πr
= ---------
B=μH=μrμ0H
where,
B = Flux density (Wb/m*2 or T)
μ = permeability of material
μ0 permeability of free space 4π –7 ×10
Henry
m
= = ----------------
μr = relative permeability
electromagnetics - 23.3
• Permeability is approximately linear for smaller electric fields, but with larger
magnetic fields the materials saturate and the value of B reaches a maximum value.
Figure 23.1 Saturation for a mild steel (approximately)
μ0 4π10–7H
m
= ----
μ=μrμ0
where,
μ0 = permeability of free space
μ = permeability of a material
μr = relative permeability of a material
μ B
H
= ----
H Magnetic field intensity
A
m
----
=  
B
H
linear saturation
4000
1.5
electromagnetics - 23.4
Figure 23.2 Magnetization curves (Sen, 1989)
• Flux density about a wire
0 500 1000
0
0.5
1.0
1.5
H (At/m)
B (T)
silicon sheet steel
cast steel
cast iron
electromagnetics - 23.5
• Flux and flux density,
• When a material is used out of the saturation region the permeabilities may be
written as reluctances,
B
I
2πr
= --------- For an infinitely long straight conductor
where,
B Flux density
Wb
m2
-------- or Tesla
=  
I = Current in the conductor (A)
r = radial distance from the conductor
where,
Φ = Flux density (Wb)
A = Cross section area perpendicular to flux
Φ = ∫BdA
R
L
μA
= -------
where,
R = reluctance of a magnetic path
L = length of a magnetic path
A = cross section area of a magnetic path
electromagnetics - 23.6
• Electric circuit analogy
• Example,
• Faraday’s law,
Φ
R
F Φ Ni
l
μA
-------
 
------------ F
R
= =---
lc
lg
I Rc
lc
μcAc
= -----------
Rg
lg
μgAg
= ------------
Φ Ni
Rc + Rg
= ------------------
electromagnetics - 23.7
• Field energy,
• Force can be derived from the energy,
• The basic property of induction is that it will (in the presence of a magnetic field)
convert a changing current flowing in a conductor to a force or convert a force to a current
flow from a change in the current or the path.
e N
d
dt
= ----Φ For a coil
where,
e = the potential voltage across the coil
N = the number of turns in the coil
W
B2
2μ
------V
B2
2μ
= =------Al
F
B2
2μ
= ------A
electromagnetics - 23.8
Figure 23.3 The current and force relationship
• We will also experience an induced current caused by a conductor moving in a
magnetic field. This is also called emf (Electro-Motive Force)
F=(I×B)L
F
I, L
B
F=(L×B)I
NOTE: As with all cross products
we can use the right
hand rule here.
L = conductor length
where,
F = force (N)
I
M
F
The FBD/schematic equivalent is,
electromagnetics - 23.9
Figure 23.4 Electromagnetically induced voltage
• Hysteresis
23.3 EXAMPLE SYSTEMS
• These systems are very common, take for example a DC motor. The simplest
motor has a square conductor loop rotating in a magnetic field. By applying voltage the
wires push back against the magnetic field.
em =(v×B)L
where,
em = electromotive force (V)
φ = magnetic flux (Wb - webers)
v = velocity of conductor
e
v
B
+
-
e
v
M
The FBD/schematic equivalent
electromagnetics - 23.10
Figure 23.5 A motor winding in a magnetic field
x
y
z
I
1 I
2
3
4
5
a
b
B
axis of
rotation ω
electromagnetics - 23.11
Figure 23.6 Calculation of the motor torque
For wire 3,
P3
r cos(ωt)
r sin(ωt)
b
2
--to
b
2
– --
= V3
–rωsin(ωt)
rωcos(ωt)
0
=
dem3 =(V×B)dL
B
0
–B
0
=
em3
–rωsin(ωt)
rωcos(ωt)
0
0
–B
0
× dr b
2
–--
b
2
--
= ∫
em3
0
0
Brωsin(ωt)
dr b
2
–--
b
2
--
= ∫
em3 B
r2
2
----ωsin(ωt)
b
2
–--
b
2
--
Bω (ωt) b
2
--
  b
2
–--
 2
= = sin – = 0
For wires 1 and 5,
By symmetry, the two wires together will act like wire 3. Therefore they both
have an emf (voltage) of 0V.
em1=em5=0V
electromagnetics - 23.12
Figure 23.7 Calculation of the motor torque (continued)
• As can be seen in the previous equation, as the loop is rotated a voltage will be
generated (a generator), or a given voltage will cause the loop to rotate (motor).
• In this arrangement we have to change the polarity on the coil every 180 deg of
rotation. If we didn’t do this the loop the torque on the loop would reverse for half the
motion. The result would be that the motor would swing back and forth, but not rotate
fully. To make the torque push consistently in the same direction we need to reverse the
For wire 2 (and 4 by symmetry),
P2
b
2
-- cos(ωt)
b
2
-- sin(ωt)
0toa
= V2
b
2
–--ωsin(ωt)
b
2
--ωcos(ωt)
0
=
dem2 =(V×B)dL
B
0
–B
0
=
em2
b
2
–--ωsin(ωt)
b
2
--ωcos(ωt)
0
0
–B
0
× dl
0
a
= ∫
em2
0
0
B
b
2
--ωcos(ωt)
dl
0
a
∫ aB
b
2
= = -- cos(ωt)
em4 = em2
For the total loop,
em = em1+em2+em3+em4+em5
em 0 aB
b
2
--ωcos(ωt) 0 aB
b
2
= + + + --ωcos(ωt)+0
em = aBbωcos(ωt)
electromagnetics - 23.13
applied voltage for half the cycle. The device that does this is called a commutator. It is
basically a split ring with brushes.
• Real motors also have more than a single winding (loop of wire). To add this into
the equation we only need to multiply by the number of loops in the winding.
• As with most devices the motor is coupled. This means that one change, say in
torque/force will change the velocity and hence the voltage. But a change in voltage will
also change the current in the windings, and hence the force, etc.
• Consider a motor that is braked with a constant friction load of Tf.
Figure 23.8 Calculation of the motor torque (continued)
• We still need to relate the voltage and current on the motor. The equivalent circuit
for a motor shows the related components.
em = aBbωcos(ωt)
em = NaBbωcos(ωt)
Fw=(I×B)L=IBa
Tw 2 F
b
2
 × --
=  =Fb = IBab
ΣM=Tw – Tf=Jα
IBab – Tf = Jα
electromagnetics - 23.14
Figure 23.9 Calculation of the motor torque (continued)
• Practice problem,
RA
+-
+
-
IA
eA
LA
em
where,
IA, eA = voltage and current applied to the armature (motor supply)
RA, LA = equivalent resistance and inductance of windings
ΣV eA – IARA LA
d
dt
= – ----IA–em=0
we can now add in the other equations,
eA – IARA LA
d
dt
– ----IA–NaBbωcos(ωt) =0
and recall the previous equation,
IBab – Tf = Jα
electromagnetics - 23.15
Figure 23.10 Drill problem: Electromotive force
• Consider a motor with a separately excited magnetic field (instead of a permanent
magnet there is a coil that needs a voltage to create a magnetic field). The model is
similar to the previous motor models, but the second coil makes the model highly nonlinear.
R
Ks
Kd
x
N
S
a
M
Write the transfer function relating the displacement ’x’ to the
current ’I’
I
electromagnetics - 23.16
23.4 SUMMARY
•
23.5 PRACTICE PROBLEMS
1.
23.6 PRACTICE PROBLEM SOLUTIONS
23.7 ASSIGNMENT PROBLEMS