Transcript chapter 4: magnetism/electromagnetism

CHAPTER 4:
MAGNETISM/ELECTROMAGNETISM
Electromagnetism
 Magnetic Field
 Direction on magnetic field
 Magnetic field due to an electric current
 Force determination
 Electromagnetic Induction
 Direction of induced emf
Electromagnetism

Magnetic Field
- A permanent magnet on the table, cover it over with a sheet of
smooth cardboard and sprinkle steel filings uniformly over the sheet.
- Slight tapping of the latter causes the fillings to set themselves in
curved chains between the poles as shown in figure below:
A suspended permanent magnet
Use of steel fillings for
determining distribution of
magnetic field
Continued…
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The shape and density of these chains
enable one to form a picture of the magnetic
condition of the space or ‘field’ around a bar
magnet and lead to the idea of lines of
magnetic flux.
Noted that these lines of magnetic flux have
no physical existence, they are purely
imaginery.
Direction of Magnetic Field
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The direction of a magnetic field is taken as that in
which the north-seeking pole of magnet points when
the latter is suspended in the field.
Thus, if bar magnet rests on a table and 4 compass
needles are placed in positions indicated in figure
below, it is found that the needles take up positions
such that their axes coincide with the corresponding
chain of fillings and their N poles are all pointing
along the dotted line from the N pole of the magnet
to its S pole.
Continued…

The lines of magnetic flux are assumed to pass
through the magnet, emerge from the N pole
and return to the S pole.
Use of compass needles for determining direction of magnetic field
Magnetic Field due to an
electric current

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When a conductor carries an electric current,
a magnetic field is produced around the
conductor – phenomenon discovered by
Oersted at Copenhagen in 1820.
He found that when a wire carrying an
electric current was placed above a magnetic
field (figure below) and in line with the normal
direction of the latter, the needle was
deflected clockwise or anticlockwise,
depending the direction of the current.
Continued…

Thus, if the current is flowing away from us,
as shown in figure below, the magnetic field
has a clockwise direction and the lines of
magnetic flux can be represented by
concentric circles around the wire.
Magnetic flux due to current in a straight conductor
Continued…
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In Figure below, we have a conductor in
which drawn an arrow indicating direction of
conventional current flow.
If the current is flowing towards us, we
indicate this by a dot equivalent to the
approaching point of the arrow and if the
current is flowing away then it is represented
by a cross equivalent to the departing tail
feathers of the arrow.
Continued…
Current Conventions
Right-hand screw rule
Continued…
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A convenient method of representing relationship
between directions of current and its magnetic field
is to placed a corkscrew or a woodscrew alongside
the conductor carrying the current.
In order that the screw is in the same direction as
the current, towards right, it has to be turned
clockwise when viewed from the left-hand side.
Similarly, direction of the magnetic field, viewed from
the same side, is clockwise around the conductor as
indicated by the curved arrow F.
Force determination
The force on the conductor can be measured for
various currents and various densities of the
magnetic field. It is found that
Force on conductor proportional to current x (flux
density) x (length of conductor)

F(N)  flux density x l(m) x I(A)
* The unit of flux density is taken as the density of a
magnetic field such that conductor carrying I ampere
at right angles to that field has a force of I N/m
acting upon it.
Continued…
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Magnetic flux density ----- Symbol : B-------Unit:
Tesla (T)
For a flux density of B teslas,
force on conductor = Bl I (N)
F=BlI
For a magnetic field having cross-sectional area of A
square metres and a uniform flux-density of B
teslas, the total flux in weber (Wb) is represented by
Greek capital letter  (phi).
Magnetic flux ------- Symbol:  -------- Unit: weber
(Wb)
 = BA ---------- 1 T = 1 Wb/m2
Electromagnetic Induction
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In 1831, Michael Faraday discover electromagnetic
induction namely method of obtaining an electric
current with the aid of magnetic flux.
He wound two coils A & C on a steel ring R as figure
below, and found that when switch S was closed,
deflection was obtained on galvanometer G and when S
was opened, G was deflected in reverse direction.
Then, he found that when permanent magnet NS was
moved relatively to coil C, galvanometer G was deflected
in one direction when the magnet moved towards the coil
and in reverse direction when the magnet was
withdrawn. And this experiment convinced Faraday that
an electric current could be produced by the movement
of magnetic flux relative to a coil.
Continued…
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Alternatively, we can say that when a conductor
cuts or is cut by magnetic flux, an e.m.f. is
generated in the conductor and the magnitude of
e.m.f. is proportional to the rate at which the
conductor cuts or is cut by the magnetic flux.
Electromagnetic Induction
Electromagnetic Induction
Direction of induced e.m.f…

Two methods available for deducing the direction or generated e.m.f.
namely, a) Fleming’s right-hand rule and b) Lenz’s Law
a) Fleming’s right-hand rule
If the first finger of the right is pointed in the direction of the magnetic
flux, as figure below and if the thumb is pointed in the direction of the
motion of the conductor relative to the magnetic field, then the second
finger held at right angles to both the thumb and the first finger
represents the direction of the e.m.f.
Flemming’s right-hand rule
Continued…

Field or Flux with First finger, Motion of the
conductor relative to the field with the M in thuMb
and e.m.f. with the E in sEcond finger.
b) Lenz’s Law
In 1834, Heinrich Lenz, a German physicist
enunciated a simple rule, known as Lenz’s Law: The
direction of an induced e.m.f. is always such that it
tends to set up a current opposing the motion or the
change of flux responsible for inducing that e.m.f.
Continued…
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Consider the application of Lenz’s Law. We find that
when S is closed and the battery has the polarity
shown, the direction of the magnetic flux in the ring
is clockwise.
The current in C must try to produce a flux in an
anticlockwise direction tending to oppose the growth
of the flux due to A namely the flux which is
responsible for the e.m.f. induced in C.
But an anticlockwise flux in the ring would require
the current in C to be passing through the coil from
X to Y. Hence, this must also be the direction of the
e.m.f induced in C.
MAGNETIC CIRCUITS
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Introduction
Mutual Inductance
Energy in a coupled circuit
Introduction
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When two loops with or without contacts
between them affect each other through the
magnetic field generated by one of them, are
said to be magnetically coupled.
TRANSFORMER is an electrical device
designed on the basis of the concept of
magnetic coupling.
Mutual Inductance
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Mutual inductance happens when two
inductors or coils are in close proximity to
each other, the magnetic flux caused by
current in one coil links with other coil,
thereby inducing voltage in the latter.
Consider a single inductor, a coil with N turns.
When current i flows through the coil, a
magnetic flux  is produced around it.
Continued…

According to Faraday’s Law, the voltage v induced
in the coil is proportional to the number of turns N
and the time rate of change of the magnetic flux :
v = N d
dt
Continued…
The flux  is produced by the current i so that any
change in  is caused by a change in current. Hence
it can be written as
V = L di
dt
which is the voltage-current relationship for the
inductor.
 The inductance L of the inductor is thus given by
L = N d
dt
This inductance commonly known as self-inductance

Continued…

Consider two coils with self-inductances L1 & L2
that are in close proximity with each other.

Coil 1 has N1 turns, while coil 2 has N2 turns.
Assume that second inductor carries no current.
Continued…
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Magnetic flux 1 emanating from coil 1 has 2
components: one component 11 links only
coil 1 and other component 12 links both
coils. Hence,
1 = 11 + 12

Although the two coils are separated, they
are said to be magnetically coupled.
Continued…

Since flux 1 links coil 1, the voltage induced
in coil 1 is:
V1 = N1 d
dt

Only flux 12 links coil 2, the voltage induced
in coil 2 is:
V2 = N2 d12
dt
Continued…

As the fluxes are caused by the current i1
flowing in coil 1, hence
V1 = N1 d1 di1 = L1 di1
di1 dt
dt

Similarly v2 can be written as;
V2 = N2 d12 di1 = M21 di1
di1 dt
dt
where M21 = N2 d12
dt
Where M21 = N2 d12
di1
Continued…

M21 is known as the mutual inductance of coil 2
with respect to coil 1. Subscript 21 indicates that
the inductance M21 relates the voltage induced
in coil 2 to the current in coil1. Thus the opencircuit mutual voltage (induced voltage) across
coil 2 is
V2 = M21 di1
dt
Continued…

Thus the open-circuit mutual voltage (induced
voltage) across coil 2 is
V1 = M12 di2
dt
Continued…
M12 = M21 = M
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M is the mutual inductance between two
coils.
Both self-inductance L and mutual inductance
M is measured in henrys (H).
We can conclude that mutual inductance is
the ability of one inductor to induce a voltage
across a neighbouring inductor.
Continued…
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Since it is convenient to show the construction of
coils on a circuit, we apply the dot convention in
circuit analysis.
A dot is placed in the circuit at one end of each two
magnetically coupled coils to indicate the direction of
magnetic flux if current enters that dotted terminal of
the coil.
Dot convention stated that if the current enters the
dotted terminal of one coil, reference polarity of the
mutual voltage in the second coil is +ve at the dotted
terminal of the second coil.
Continued…

Alternatively, if a current leaves the dotted
terminal of one coil, the reference polarity of the
mutual voltage in the second coil is -ve at the
dotted terminal of the second coil
Continued…
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The sign of mutual voltage
v2 is determined by the
reference polarity for v2
and the direction of i1.
Since i1 enters the dotted
terminal of coil 1 and v2 is
+ve at the dotted terminal
of coil 2, the mutual
voltage is +Mdi1/dt
The current i1 is enters the
dotted terminal of coil 1
and v2 is -ve at the dotted
terminal of coil 2. Hence,
the mutual voltage is –
Mdi1/dt
Continued…

The current i2 is leaves the
dotted terminal of coil 2
and v1 is +ve at the dotted
terminal of coil 1. Hence,
the mutual voltage is –
Mdi2/dt

The sign of mutual voltage
v1 is determined by the
reference polarity for v1
and the direction of i2.
Since i2 leaves the dotted
terminal of coil 2 and v1 is ve at the dotted terminal of
coil 1, the mutual voltage
is +Mdi2/dt
Continued…
L = L1 + L2 + 2M

Dot convention for coils in series, the sign indicates the
polarity of the mutual voltage (series-aiding connection)
L = L1 + L2 - 2M

Dot convention for coils in series, the sign indicates the
polarity of the mutual voltage (series-opposing
connection)
Continued…
Time-domain analysis of a circuit containing coupled coils
As in figure above, applying KVL to coil 1 gives
V1 = i1R1 + L1di1/dt + M di2/dt
Coil 2, KVL gives
V2 = i2R2 + L2di2/dt + M di1/dt
Continued…
Frequency-domain analysis of a circuit containing coupled coils
As in figure above, applying KVL to coil 1 gives
V = (Z1 + jωL1)I1 – jωMI2
Coil 2, KVL gives
0 = – jωMI1 + (ZL + jωL2)I2
Example 1
Solution:
For coil 1, KVL gives
-12 + (-j4 + j5)I1 – j3I2 = 0 ---------> jI1 -3jI2 = 12 ----(1)
For coil 2, KVL gives
-j3I1 + (12 + j6)I2 = 0 ----------------> I1 = (2 – j4)I2 ----(2)
Substitute eqn (2) into (1)
(j2 + 4 – j3)I2 = (4 - j)I2 = 12
Or I2 = 12/(4 - j) = 2.9114.04 ------------------------------ (3)
From eqn (2) & (3)
I1 = (2 – j4)I2 = 13.01-49.39 A
Exercise 1
Determine the voltage Vo in the circuit of Figure above
Solution:
For mesh 1
For mesh 2
For the matrix form
Example 2
Calculate the mesh currents in the circuit of figure
above
SOLUTION:
1st: know the polarity of the mutual voltage. Need to apply the dot
rule.
2nd: By refer to the figure above, suppose coil 1 is the reactance
6 and coil 2 is the reactance 8.
3rd: Figure out the polarity of the mutual voltage in coil 1 due to
current I2, we observe that I2 leaves the dotted terminal of coil 2.
Since we are applying KVL in clockwise direction, it implies
that mutual voltage is –ve, that is –j2I2.
The best way to figure mutual
voltage is by redraw the
portion of the circuit as
shown on the right
Continued…
Thus, mesh 1, KVL gives
-100 + I1(4 – j3 + j6) – j6I2 – j2I2 = 0
or 100 = (4 + j3)I1 – j8I2 --------(1)
Similarly, to figure out mutual voltage in coil 2 due to
current I1, consider figure below

Applying dot convention
gives the mutual voltage
as V2 = -2jI1. Also current
I2 sees the two coupled
coils in series since it
leaves the dotted terminals
in both coils.
Continued…
Therefore mesh 2, KVL gives
0 = -j2I1 – j6I1 + (j6 + j8 + j2 x 2 + 5)I2
or 0 = -j8I1 + (5 + j18)I2
In matrix form, by using cramer’s rule
we obtain mesh current as

I1 = 1 = 100(5 + j18) = 1,868.274.5 = 20.33.5A

30 + j87
92.3071
I2 = 2 = j800
= 80090 = 8.69319A

30 + j87
92.3071
Exercise 2
Determine the phasor currents I1 and I2 in the circuit of figure above
Solution:
Since I1 enters the coil with reactance 2 and I2 enters the coil with
reactance 6, the mutual voltage is +ve. Hence, for mesh 1;
Energy in a coupled Circuit
The energy stored in an
inductor is;
W = ½ Li2


Consider the circuit above, assume current i1 and i2 are
zero initially, so that energy stored in coils is zero.
If we let i1 increase from 0 to I1 while maintain i2 = 0, the
power in coil 1 is
P1(t) = v1i1 = i1L1 di1/dt
Continued…
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The energy stored in the circuit is:
W1 = ½ L1I12
Now, we maintain i1 = I1, i2 increase from 0 to I2, the
mutual voltage induced in coil 1 is M12 di2/dt while the
mutual voltage induced in coil 2 is zero since i1 does not
change. The power in the coil is;
P2(t) = I1M12 di2/dt + i2L2 di2/dt
The energy stored in the circuit is
W2 = M12I1I2 + ½ L2I22
So, the total energy stored in the coils is
W = w1 + w2 = ½ L1I12 + ½ L2I22 + M12I1I2
Continued…
Since the total energy stored should be the same, we can conclude
that
M12 = M21 = M
and
W = 1/2 L1I12 + ½ L2I22 + MI1I2
(This equation based on assumption that the coil currents both entered
the dotted terminals)
If one current enters one dotted terminal while the other current leaves
the dotted terminal, the mutual voltage is negative, so that the
mutual energy MI1I2 is also negative.

So, w = ½ L1I12 + ½ L2I22 – MI1I2
Hence, instantaneous energy stored in the circuit as below;
W = ½ L1i12 + ½ L2i22 ± Mi1i2
Continued…
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To establish an upper limit for Mutual Inductance, M.
The energy stored cannot be negative because the
circuit is passive.
The extent to which the mutual inductance M
approaches the upper limit is specified by the
coefficient of coupling k, given by
M = k√L1L2
where 0 ≤ k ≤ 1
The coupling coefficient k is a measure of the
magnetic coupling between two coils.
Continued…

If entire flux produced by one coil links another coil, then
k = 1 & we have 100% coupling or the coils are said to
be perfectly coupled. For k < 0.5, coils are said to be
loosely coupled and for k > 0.5, they are said to be tightly
coupled.
a)
b)
Windings loosely
compound
Windings tightly
coupled
Example 3
Consider the circuit above, determine the
coupling coefficient. Calculate the energy
stored in the coupled inductors at time t = 1s
if v = 60 cos (4t + 30) V
Solution:
The coupling coefficient is
k=
M
= 2.5
= 0.56
√L1L2
√20
This indicates that the inductors are tightly coupled. To
find the energy stored, we need to calculate the current.
To find the current, we need to obtain the frequencydomain equivalent of the circuit
60 cos(4t + 30)
5H
2.5H
4H
1/16F
 6030, ω = 4 rad/s
 jωL1 = j20
 jωM = j10
 jωL2 = j16
 1/jωC = -j4
Continued…
The frequency domain equivalent circuit as
shown in figure below. Apply mesh analysis. For
mesh 1,
(10 + j20)I1 + j10I2 = 6030 ---------(1)
For mesh 2,
j10I1 + (j16 – j4)I2 = 0 -------------------(2)
or I1 = -1.2I2
Continued…
Substitute eqn (2) into (1), yields
I2(-12 – j14) = 6030  I2 = 3.254160.6A
and I1 = -1.2I2 = 3.905- 19.04 A
In time domain,
i1 = 3.905 cos(4t – 19.4) A, i2 = 3.254 cos(4t + 160.6) A
At time t = 1s, 4t = 4 rad = 229.2 and
i1 = 3.905 cos(229.2 - 19.4) = -3.389 A
i2 = 3.254 cos(229.2 + 160.6) = 2.824 A
The total energy stored in the coupled inductor is
W = ½ L1i12 + ½ L2i22 + Mi1i2 = 20.73 J
Exercise 3:
Consider the circuit above, determine the
coupling coefficient and the energy stored in
the coupled inductors at t = 1.5s
Solution:
Continued…
Continued…