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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… 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 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 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… 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… 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… 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 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… 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… 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 Introduction Mutual Inductance Energy in a coupled circuit Introduction 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 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… 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 d12 dt Continued… As the fluxes are caused by the current i1 flowing in coil 1, hence V1 = N1 d1 di1 = L1 di1 di1 dt dt Similarly v2 can be written as; V2 = N2 d12 di1 = M21 di1 di1 dt dt where M21 = N2 d12 dt Where M21 = N2 d12 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 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… 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… 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) = 2.9114.04 ------------------------------ (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) = 1,868.274.5 = 20.33.5A 30 + j87 92.3071 I2 = 2 = j800 = 80090 = 8.69319A 30 + j87 92.3071 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… 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… 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 6030, 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 = 6030 ---------(1) For mesh 2, j10I1 + (j16 – j4)I2 = 0 -------------------(2) or I1 = -1.2I2 Continued… Substitute eqn (2) into (1), yields I2(-12 – j14) = 6030 I2 = 3.254160.6A 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…