Advanced electromagnetism
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Advanced electromagnetism
E.m.f. in a conductor
• Moving a conductor through a magnetic field can induce an emf.
• The faster the conductor moves through the field the greater the emf and hence the greater the current
N S
E.m.f. in a conductor
• The bigger the length of the conductor moving through the field the greater the emf and hence the greater the current
N
S
E.m.f. in a conductor
• The greater the flux density (B) of the field, the greater the emf and hence the greater the current
N
S
•Magnitude of induced e.m.f. in a conductor •E = Blv volts
(B = flux density in Teslas, l = length in metres and v = velocity in m/s)
E.m.f. in a conductor
•It is the rate at which the conductor “cuts” through the magnetic field
= dФ/dt volts
E.m.f. in a conductor
Right hand rule
The direction of the current can be found using Fleming’s right hand rule
Example
A conductor, 800 mm long, is moved at a uniform speed at right-angles to a magnetic field
of density 0.8Tesla. Calculate the velocity required to generate an e.m.f. in the conductor of 10 V.
Example
A conductor, 800 mm long, is moved at a uniform speed at right-angles to a magnetic field
of density 0.8Tesla. Calculate the velocity required to generate an e.m.f. in the conductor of 10 V.
Example
From: E = Bℓv
V = e / Bℓ
= 10 / (0.8 x 800 x 10-3) = 15.65 m/s
E.m.f induced in a rotating coil
E = - d/dt(NΦ)
example
A coil of 3000 turns, when energised, produces a magnetic flux of 3.5 mWb. If the energising
current is reversed in 0.3 seconds, determine the direction and average value of e.m.f. induced in
the coil.
example
From: E = -N x (dΦ) / dt
E = -3000 x (-2 x 3.5 x 10-3) / 0.3 = 70V
Flux reversal hence +ve
Flux linkage
Flux relates to the number of field lines
Φ = BA
Flux linkage takes into consideration the number of turns in a coil
Flux linkage = NΦ
Induction in terms of Flux linkage (NΦ)
L = NΦ/I
= N.dΦ/dI
Rate of change in flux linkage
Example
•A coil of 250 turns is wound on a non-magnetic ring. If a current of 5 A produces a magnetic flux
of 0.4mWb, calculate:
•a) Inductance of coil
•b) Average e.m.f. induced in the coil when switching on if the current takes 3 ms to rise to
its final value.
Example
•From: L = NΦ / I
L = 250 x (0.4 x 10-3) / 5= 20 mH
b) From: e.m.f.av = - NΦ ./ t
•e.m.f.av = - 250 x (0.4 x 10-3) / (3 x 10-3) = - 33.3V.
N S
N S
Pushing a magnet into a coil induces a current in the coil wire
Pulling the magnet out of the coil induces a current in the opposite direction
Current growth in an inductive circuit
I = Io(1-eRt/L)
I
t
Current decay in an inductive circuit
I = Io(e-Rt/L)
I
t
120V
120 Ω
40 Ω
Example
•A non-reactive resistor of 120 Ω is connected in parallel with a coil of inductance 4 H and
resistance 40 Ω. Calculate the current flowing in the coil 0.06 seconds after the circuit is disconnected from a 120 V dc supply.
Example
• Initial current through coil = V/RI= 120 / 40 = 3 A
•From i = I x e -Rt/L
I = 3 x e-(120 x 0.06) / 4= 3 x e- 1.8
e-1.8 = 0.1653
I= 3 x 0.1653 = 495.9 mA
Energy stored in an inductor
W(energy) =1/2LI2
Example
A current of 5 A flows through a coil of 3 H. Calculate the amount of energy stored in the
coil.
Example
From: energy stored = 1/2 x LI2
Energy stored = 0.5 x 3 x (52) = 37.5 Joules
Mutual Inductance
E = - MdI/dt = -N2dΦ/dt
M = N2Φ2/I1
Example
•Two coils have a mutual inductance of 300 μH. Calculate the e.m.f. produced in one coil when the current in the other coil changes at the rate
of 20 x103 A/s.
Example
From: E = -M x dI / dt
•E = -300 x 10-6 x 20000 = -6 V
Magnetism
The relationship between magnetic field strength and magnetic flux density is:
B = H × µ
where µ is the magnetic permeability of the substance
Magnetic field strength equation in a coil
H = (NI) / l
where: H = magnetic field strength (ampere per metre)
I = current flowing through coil (amperes) N = number of turns in coil
l = length of magnetic circuit
Magnetism
Magnetism
The magnetomotive force in an inductor or electromagnet consisting of a coil of wire is given
F = NI
where N is the number of turns of wire in the coil and I is the current in the wire.
Magnetism
Permeability
Is a measure of how easily a magnetic field can set up in a material
It is the ratio of the flux density of the magnetic field within the material to its field strength
µ =B/H
Permeabilty of free space µo is 4π x10-7 H/m
Magnetism
Relative Permeablity µr
• This is how much more permeable the material is compared to free space (a vacuum). The permeability
of the material can be calculated by multiplying its relative permeability by the permeability of free
space.
•µ = µo x µr
Magnetism
Magnetic Flux
The rate of flow of magnetic energy across or through a (real or imaginary) surface. The unit of
flux is the Weber (Wb)
Magnetic Flux Density
A measure of the amount of magnetic flux in a unit area perpendicular to the direction
of magnetic flow, or the amount of magnetism induced in a substance placed
in the magnetic field.
The SI unit of magnetic flux density is the Tesla, (T).
One Tesla, (1T), is equivalent to one weber per square metre (1 Wb/ m2).
Magnetism
• To summarise
• The magnetic flux density , B, multiplied by the area swept out by a conductor, A, is called the magnetic
flux, Φ.
•Φ = BA
• . Unit of flux: weber, Wb
• Unit of flux density: Tesla, T
Example
•A coil of400 turns is wound uniformly over a wooden ring of mean circumference 200 mm and cross-
sectional area 150 mm2. If the coil carries a current of 2 A, calculate:
• a) Magneto-motive-force
•b) Magnetic Field Strength
• c) Flux Density
•d) Total Flux
Example
• a) From: m.m.f = NI = 400 x 2 = 800 AT
•b) From: H = m.m.f. / ℓ = 800 / 200 x 10-3 = 4 x 103 A/m
• c) From: B / H = μ0 . B = μ0 x H = 4π x 10-7 x 4 x 103 = 5.027 x 10-3 Tesla
•d) From: Φ = B x area = 5.027 x 10-3 x 150 x 10-6 = 0.754 μWeber
‘Hard’ magnetic materials
Hard magnets, such as steel, are magnetised, but afterwards take a lot of work to de-magnetise. They're
good for making permanent magnets.
.
‘Soft’ magnetic materials
•Soft magnets are the opposite. With an example being iron, they are magnetised, but easily lost their magnetism, be it through vibration or any other means. These are best for things that only
need to be magnetised at certain points, egmagnetic fuse/trip switch
Magnetism
Solenoid
Magnetism
The magnetomotive force in an inductor or electromagnet consisting of a coil of wire is given by:
F = NIwhere N is the number of turns of wire in the coil
and I is the current in the wire.The unit is amp.turns
(AT)
Magnetism
Magnetic field strength in a coil = mmf/ length of the coil
Magnetic field strength equation in a coi
H = (NI) / lwhere:
H = magnetic field strength (ampere per metre) I = current flowing through coil (amperes)
N = number of turns in coil l = length of magnetic circuit
Retentivity –
A measure of the residual flux density corresponding to the
saturation induction of a magnetic material. In other words, it is a
material's ability to retain a certain amount of residual magnetic field
when the magnetizing force is removed after achieving
saturation
•Residual Magnetism or Residual Flux - the magnetic flux density that remains in a material when the
magnetizing force is zero.
•Coercive Force - The amount of reverse magnetic field which must be applied to a magnetic material
to make the magnetic flux return to zero. (The value of H at point c on the hysteresis curve
Magnetism
• Example
• Starting with the concept of molecular magnets in a magnetic material, explain
• a) Relative permeability of a material
• b) Loss of magnetisation in a ‘soft’ material
• c) Magnetic saturation
Magnetism
• a) Relative permeability of a material, how easily molecular magnets align with applied field
• b) Loss of magnetisation in a ‘soft’ material, how easily molecular magnets take up random alignment
• c) Magnetic saturation, molecular magnets all aligned in field direction
B (Teslas)
1.7
1.6
1.5
1.4
1.3
1.2
1.1
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
1 2 3 4 5
H (At/m) x 1000
Example
• A coil, uniformly wound over a mild steel ring, produces a magnetic field strength of 2000 A/m when energised. Using an appropriate
curve calculate:
•
• i) Flux density
• ii) Relative permeability of mild steel under the stated conditions
•
1.7
1.6
1.5
1.4
1.3
1.2
1.1
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
1 2 3 4 5
H (At/m) x 1000
Example
• From Curve - Flux Density v Magnetic field strength
• Flux density = 1.3Tesla
• ii) From B / H = μ0 x μr . Then μr = B / H μ0 = 1.3 / (2000 x 4π x 10-7 )
• = 517
Example
m(relative) = m(material) / m(air)
Example
Reluctance of a magnetic circuit (S)
S = mmf/Ф
S = ℓ / μ0 μr a
Relative 850
permiability 800
(μr) 750
700
650
600
550
500
450
400
350
300
250
200
150
100
50
0
1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.2 2.3 2.4 2.5
B (Teslas)
Example
A mild steel ring has a cross- sectional area of 400 mm2 and mean circumference of 300 mm. If a coil of 250 turns is wound uniformly over
the ring, calculate:
i) Flux density in the ring assuming a total flux of 600 μWb
ii) Reluctance of the ring assuming μr
iii) Current to produce the required flux density
e
Example
• i) From Φ = B x area. Flux density B = 600 x 10-6 / 400 x 10-6 = 1.5 Tesla
ii) From curve – μr = 500
From: S = ℓ / μ0 μra
= 300 x 10-3 / (4π x 10-7 x 500 x 400 x 10-6)
= 11.9 x 105A/Wb
Example
•From: m.m.f. = ΦS = 600 x 10-6 x 11.9 x 105
•= 714 Ampere-turns.
•Current m.m.f/ turns =714/ 250
•= 2.86 Amperes
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