1E6 Electrical Engineering
Electricity and Magnetism
Lecture 18: Solenoids and Relays
18.1 Magnetomotive Force and Magnetic Field Strength
When a length of conductor is wound to form multiple current-carrying
loops side-by-side, the magnetic fields associated with the individual loops
combine to create a magnetic field acting lengthways through the coil as shown
in Fig. 1. This magnetic field is present while current flows through the
conductor but collapses when the current ceases to flow if the source of
electricity is removed.
Fig. 1 A Wound Coil Generating an Internal Magnetic Field
If a piece a magnetisable material is placed within this field, it will
experience a force referred to as a Magnetomotive Force. The force experienced
will of course depend on the intensity or strength of the magnetic field. The
magnetomotive force should not be thought of as a physical Newtonian force,
(though it often results in this), but rather as analogous to the electromotive
force of a battery or electrical supply. In this case, the magnetic flux is thought
of as analogous to the current flowing in the loop of in an electric circuit.
Magnetomotive Force is formally defined as the force tending to induce
alignment of the magnetic domains within a material to form lines of flux
under the influence of a magnetic field.
1
I
current
flow
into coil
current
flow
out of coil
S
N
force experienced
by a piece of
magnetisable material
d
longitudinal
magnetic
field
distance or length of
the magnetic path
Fig. 2 A Closely Wound Coil and its Associated Magnetic Field
The coil shown in Fig. 2 consists of N closely wound turns with the
conductor fed with a current I, so that the magnetic field created within the coil
is uniform and continuous. A piece of magnetisable, or previously magnetised,
material is placed inside the region within the coil. The magnetomotive force
experienced by the material is directly proportional to the current, I,
generating it and the number of turns, N, of the coil.
Magnetomotive Force is given the symbol MMF and has SI Units of Amps (A)
Magnetomotive Force MMF = Current x No. Turns = N I (A)
Strictly, the units are those of current, i.e. Amperes, in the SI system. This,
however, tends to hide the fact that the force depends on the number of turns
of the coil and consequently the units are often stated as Ampere-turns (At),
where the number of turns is in effect dimensionless.
Magnetic Field Strength is defined as the magnetomotive force experienced per
unit length in a uniform magnetic field.
Magnetic Field Strength has the symbol H and has Units of Amps/metre (Am-1)
Magnetic Field Strength =
Magnetomot ive Force
Length of Coiled Path
H=
NI
Am -1
d
where d is the distance over which the field in the coil is generated, i.e. the
length of the wound coil.
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18.2 Permeability
There is also an intrinsic relationship between Magnetic Field Strength
and Magnetic Flux Density. This is characterised by the property known as
Permeability.
Permeability is defined as the degree of magnetisation which a material
undergoes in response to an applied magnetic field. It is the ratio of the
Magnetic Flux Density to the Magnetic Field Strength. It is analogous to
resistance in an electrical circuit.
Permeability is given the symbol µ and has units of Henrys/metre (Hm-1)
Permeability =
Magnetic Flux Density
Magnetic Field Strength
µ=
B
H
Hm -1
Permeability of Free Space: This is defined as the ratio of the Magnetic Flux
Density to Magnetic Field Strength measured at a distance of 1m from a long
straight conductor carrying a current of 1A as
shown in Fig. 3. It is designated as µ0 = 4π x 10-7 =
d = 2πr
12.57 x 10-7 Hm-1.
I = 1A
r = 1m
The permeability of different materials is usually
specified as relative permeability, µr , i.e. relative
to free space so that:
Permeability
µ = µ rµ 0
Hm -1
Fig. 3. Permeability of Free Space
From this the Magnetic Flux Density of the field within the coil can be
found as:
Magnetic Flux Density B = µH = µ
NI
d
Wbm− 2
If the wound coil has air as the core then µ = µ0. If, on the other hand, the core
is a magnetic material such as Ferrite, then the relative permeability of this
material must be used.
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18.3 Solenoids
The principle of generating a magnetic field within a coil is exploited in
the Solenoid. A solenoid is constructed by winding a coil on an insulated
former. A magnetic/magnetisable shaft is then inserted into the former so that
it is subjected to the magnetic field which is created when the coil is energised,
i.e. when a current flows through it. Usually, one end of the shaft protrudes
from the former. The magnetised shaft has North and South poles, which can
be aligned in either direction with respect to the field associated with the coil as
shown in Fig. 4.
When the coil is energised, the shaft
will move under the influence of the
magnetic field in a direction
determined by the rule for poles so
that the North pole of the shaft moves
towards the South pole of the field
associated with the energised coil.
Alternatively, the South pole of the
shaft may be attracted towards the
North pole of the field associated with
the coil, depending on the orientation
of both.
Fig. 4. Motion of the Shaft in the
Magnetic Field of a Solenoid
The end of the shaft protruding from the former can be connected to a
mechanical actuator, that is, a mechanical mechanism which causes some kind
of physical action or movement. This will be forced to move in the same
direction as the shaft when the solenoid is energised. Various mechanical
configurations can be constructed to accomplish a variety of tasks. Often the
solenoid is spring-loaded as shown in Fig. 5 so
that the shaft returns to a neutral position when
the coil is de-energised, i.e. when the solenoid is
deactivated. Solenoids come in a variety of
shapes, sizes and power ratings. They are
commonly used in door-lock mechanisms,
pressure and flow control valves, pumps and
many other industrial applications.
Fig. 5. A Spring Loaded Solenoid
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Linear Action Solenoids:
In its simplest form a shaft can be inserted into the solenoid and the protruding
end connected to a mechanism which requires a linear push or pull action. An
example of this is a door lock mechanism, and other examples can be seen in
Fig. 6.
Fig. 6 Examples of Linear Shaft Solenoids
Lever Action Solenoids:
Some mechanical devices require a lever action or even a rotary twisting action
to generate a bending force. Various mechanisms and forms of levers can be
designed as shown in Fig. 7. All of them, however, ultimately rely on being
driven by the end of the mechanism inserted into the longitudinal field of the
coil.
Fig. 7 Examples of Lever and Rotary Action Solenoids
Valves and Actuators:
The solenoid coil can be built in as part of the mechanical device which it is
used to drive, such as in the valves and actuators shown in Fig. 8.
Fig. 8 A Selection of Solenoid-Based Valves and Actuators
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18.4 Relays
The solenoid principle can be extended to construct a relay. This is a
device which, when current flows through the coil of the solenoid, causes a
mechanical arm to move and open or close electrical contacts. These contacts
are used to feed power to a load, which can be electrically isolated from the
circuit driving the relay. Hence, using a relay driven by a low-power, solid-state
switch like a transistor operating from a low-voltage supply, high voltage or
current and heavy power such as mains electricity can be switched on an off by
other electronic circuits or equipment as shown in Fig. 9.
5V
L
changeover
contacts
openclose
contacts
solenoid
logic
gate
mains supply
fan
T
lamp
M motor
heating
coil
N
Fig. 9 A Relay Driven by a Transistor Switch
A relay may have more than one set of contacts, all activated by the same
solenoid and switching circuit as shown. The contacts can be of the simple
ON/OFF type, which either open or close when the relay is activated, or they
may be of the CHANGEOVER type where one contact is closed and one is
open when the relay is inactive and this situation is reversed when the relay is
activated. A selection of relays is shown in Fig. 10.
Fig. 10. A Selection of Relays
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18.5 Case Studies
Design Example 1
The coil of the solenoid used in an automatic door-lock is wound on a
former of length 15cm and diameter 2.5cm. It is fed with a current of 100mA to
generate an internal magnetic field strength of 250Am-1. Determine the number
of turns required in the coil and the total length of wire used.
Solution:
Magnetic Field Strength = H =
Then
N=
NI
= 250 Am -1
d
Hd 250 x 0.15 2.5 x 15 37.5
=
=
=
I
0.1
0.1
0.1
so that N = 375 turns
The length of wire can be found as the number of turns times the diameter
times π. Then:
Length of wire = Nπ2r = 375 x 3.14 x 0.025
Length of wire = 29.43 m
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Design Example 2:
The energising coil of the relay driving a motor consists of a winding of
6500 turns on a former of length 2.5 cm and diameter 1.5 cm. A total flux of
0.3Wb is required to operate the lever driving the contacts of the relay which is
made of a high-µ material having a relative permeability of 20,000. If the wire
used for the coil has a resistance of 0.15 Ωm-1, determine the power drawn
from the supply by the relay.
Solution:
The total flux which must be generated by the energising coil is:
Φ = 0.3Wb
The flux density is given as:
B=
B=
But
H=
Φ Φ
0.3
= 2=
A πr
3.14 x 0.75 x 10- 2
(
)
2
0.3
−2
=
1700
Wbm
3.14 x 0.5625 x 10-4
B = µH = µ rµ 0 H
so that
H=
B
µ rµ 0
1700
1700
=
= 68 x103 Am−1
3
-7
20 x 10 x 12.57 x 10
0.025
Then H =
NI
Hd
so that I =
d
N
68 x 103 x 2.5 x 10−2 1700
I=
=
= 0.26 A = 260 mA
3
6.5 x 10
6500
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The total length of wire used in the solenoid coil is obtained as the number of
turns times the circumference of the former:
Length of wire = Nπ2r = 6500 x 3.14 x 1.5 x 10-2
Length of wire = 306.3 m
Then the resistance of the solenoid coil is found as the resistance of the wire per
unit length times the length of wire used:
Resistance of coil = R = 306.3 x 0.15 = 46 Ω
The supply voltage needed to feed the solenoid can be found from the current
determined above and the resistance so that:
Supply Voltage = V = IR = 0.26 x 46 = 11.96 V
In practice this relay would operate from a 12V supply.
Finally the power drawn from the supply by the relay is given as:
Power drawn
P = V x I = 12 x 0.26 = 3.12 W
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