# INTRODUCTION SELF INDUCTANCE. Introduction. Self inductance. Mutual inductance. Transformer. RLC circuits. AC circuits

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2 Figure 2 Self inductance of a solenoid. Figure 3 Mutual inductance. MUTUAL INDUCTANCE Φ Note the negative sign in the equation for emf which results from Faraday s law. Example: Self inductance of a solenoid. Consider that the solenoid has turns, length radius, and carries a current as shown in figure 2. Ampère s Law can be used to show that the magnetic field in a solenoid is axial with a magnitude: = 0 The flux linkage Φ, taking into account that the field is uniform across the solenoid, and that it is linked times since the coil has turns. Φ = B S = 2 2 = 0 2 Thus the self inductance is given as: = Φ 2 = 0 2 The self inductance is just a simple geometric number for any coil. NB, Many books use =turns per unit length in his formula, whereas these notes use =thetotal number of turns; be careful not to get confused. The SI unit of inductance is the Henry after the US scientist. The SI unit of magnetic flux Φ is the weber. Thus, since = Φ = = The Henry is a large unit. For example, for the above case let =10 3, =002, =01 and 0 =410 7, then =16. For this coil, if =10 3 per second, then the induced back emf will be 16 volts. Mutual inductance is the induced emf in one circuit due to a changing field produced by a second circuit. Faraday s Law can be written as: = Φ = Φ = where the mutual inductance is defined as: Φ Calculation of the mutual inductance for any pair of circuits can be a complicated integral. One uses the Biot Savart law to compute the B field at circuit due to circuit B = 0 4 I dl cr 2 Knowing the magnetic field, then one can compute the flux linkage in circuit due to the field produced by circuit. Φ = B S Φ = I 0 dl cr S 4 2 Thus the mutual inductance is given by Φ = I 0 4 dl cr S 2 This complicated double integral can be simplified mathematically using Stokes Theorem, to give that: = I I 0 l l 4 This non-trivial step gives that the double integral is a symmetric geometric factor. This relation is called Neumann s Formula. 98

3 = Φ = 0 2 You can easily compute the mutual inductance due to the magnetic flux due to in and you will obtain the same mutual inductance relation. Typical values might be = =10 3 =01 =001 then one obtains =40. Figure 4 Concentric solenoids The above proof can be repeated for the emf in due to a changing current in circuit. Thatis; = Φ = Φ = leading to mutual inductance = I I 0 l l = = 4 That is; the mutual inductance is symmetrical whether one is considering the flux in due to or vice versa. Thus we can write for the two coils that: = = where the mutual inductance is a geometrical factor expressing the degree of coupling of the magnetic flux between two separate circuits. Note the negative sign remaining from Faraday s and Lenz s laws. Mutual inductance between two concentric solenoids In general the computation of mutual inductance is non trivial. However, one can easily calculate the mutual inductance between concentric solenoids. Consider the system shown in figure 4 where the radii of the coils are such that. The magnetic field in due to circuit is = 0 The field from only extends over an area 2. Thus the flux linkage in circuit due to the magnetic flux from is: Φ = 2 = 0 2 Thus the mutual inductance is: TRANSFORMER The transformer is a nice example of use of inductance. Consider two tightly-coupled circuits such as two concentric solenoids with an alternating emf applied to the primary coil and a resistor dissipating energy connected to the secondary coil. Consider that all of the magnetic flux Φ goes through both coils. Then the flux linkage for the primary circuit is Φ = Φ while the flux linkage for the secondary is Φ = Φ If the magnetic flux is time dependent then we have Φ = Similarly for the secondary Φ = Eliminating Φ gives = That is the voltage ratio equals the turns ratio. The perfect transformer does not dissipate energy in the transformer, thus we must have power conserved, that is: = Thus: = The non perfect transformer can be solved using Kirchhoff s loop rule that the sum of emfs around the primary circuit equals zero: =0 where we have to include the induced emf due to self inductance as well as the mutual inductance term. For the secondary the voltage across the resistor = Then Kirchhoff s loop rule gives: 99

4 =0 Multiply the first equation by and the second equation by andthentakethedifference of these equations gives ( )=( 2 ) Obviously the closest coupling of magnetic flux occurs for self inductance where a coil is coupled perfectly to the magnetic flux it generates. Thus we must have that: Figure 5 The transformer. Figure 6 RLC circuit. Thus : RLC CIRCUITS 2 For perfect coupling of magnetic flux between the coupled circuits, then: 2 = In the case of perfect coupling then the right-hand side of the previous equation relating the emfs is zero, therefore: It is useful to consider the response of simple circuits involving resistance, capacitance, and inductance. The response of general LRC circuits to AC input signals is important because of many applications to technology. However, the discussion of such response requires a detailed discussion of both the amplitude and phase of the output relative to the input waveforms. The following is a brief summary of some concepts of AC circuits. Consider the series combination of, and shown in figure 6. Assume that initially the capacitor is charged with charge 0 when the switch is closed at atime =0. Using Kirchhoff s loop rule, and knowing that voltage across the capacitor = then: = For perfect coupling this equation gives the same equationsasgivenabove; = = = Note that the transformer only works for oscillating currents and emf s, otherwise =0However, the ratio of voltages is independent of frequency in this elementary theory. For perfect coupling and a resistive load, then the primary and secondary waveforms are in phase and the solution is simple. The ability of the transformer to easily and efficiently transform voltages for AC power is the reason that AC is used for power distribution. = From charge conservation, Kirchhoff s node rule, we have: + =0 Using these two equations gives a second order differential equation =0 100

5 Figure 7 Damped RLC circuit response compared with undampedsolutionwhenr=0. Since = this also can be written as: =0 If you have studied second order homogeneous differential equations you will know that the solution for 1 light damping, that is, 2 4 is: 2 Figure 8 When the motion is overdamped 2 leading to an exponetial decay. When 1 = 2 4 the 2 system is critically damped leading to the most rapid damping. Critical damping is used for meter systems to ensure that the needle reaches the correct value in the shortest time. where: () = 0 2 [ sin + cos ] 2 = The time dependence is that of a damped harmonic oscillation with angular frequency and damping time constant = 2 as shown in figure 7. Note that for =0there is no damping and one has a constant harmonic oscillation with angular frequency = 1 For the damped case the frequency is slightly reduced. On the other hand, when the relation 2 for 2 is negative leading to an imaginary value for producing a non-oscillatory over-damped motion that decays exponentially as shown in figure 8. If one applies an sinusoidal voltage from a power supply then one will have the phenomena of resonance when the applied frequency approaches the resonant frequency of the circuit as will be dicussed next lecture. Tesla Coil The Tesla coil provides a nice example of RLC circuits coupled to transformers. The first transformer raises the Hz primary voltage to 15,60Hz.The small spark gap breaks down at 15 kv stimulating the Figure 9 The Tesla coil. 101

6 Figure 11 The series resonant circuit and the corresponding phasor diagram. Capacitor C Since = and from charge conservation Figure 10 Phase relations between current (solid line) and voltage (shaded) for a resistor, capacitor, and inductor. The phasor diagram is shown on the right. LC circuit to oscillate at about 500 khz. The rate of change of current in the primary of the second transformer is 10,000 times what it would be at 60 Hz. The turns ratio for the second coil then produces 300 kv across the final spark gap. AC CIRCUITS This discussion leads naturally to the topic of AC circuits which is of considerable technical importance. This relates to the response of R,L,C circuits to an applied sinusoidal voltage. This topic is not included in this course and the examinations because of the mathematical complexity. However, for your education it is useful to recognize the basic elements of AC circuits. Consider an applied voltage that is a cosine function of time = 0 cos It is useful to define an impedance by Resistor Ohm slawgivesthat = = = 0 cos Thus the current and voltage are in phase as shown in figure 10a and the impedance is = = = = 0 sin = 0 cos( + 2 ) Thus as shown in figure 10b for a capacitor the current leads the voltage by 90 and the impedance is = 1 and 90 out of phase. This is obvious in that you can only change the voltage across a capacitor by having current flow into the capacitor to change the stored charge. Inductor L Since by Kirchhoff s rules for circuit figure 10c, =0 Thus 0 cos = By integration this gives 0 sin = This can be rewritten as = 0 cos( 2 ) Thus for an inductor the the current lags the voltage by 90 and the impedance is = and 90 out of phase. This is obvious in the the back emf opposes change of the field, that is the current when a voltage is applied. 102

7 Thus using Kirchhoff s rules for the RLC circuit in figure 11 it can be seen that the magnitude of the effective impedance can be calculate using Pythagorus Theorem to have a magnitude given by " = 2 + µ 1 # and the voltage leads the current by a phase angle µ 1 =tan 1 Figure 12 Build up of magnetic energy as the current increases in a RL circuit connected to a battery after the switch is closed at t=0. This series is an example of the fact that any combination of passive impedances can be represented as a net load having a resultant complex impedance such that V = I where has an inphase resistive component = cos and a reactive, or out of phase, component Figure 13 LC circuit. = sin It is interesting that the resistive load dissipates power where = = where the factor of 12 comes from the fact that the average of (cos ) 2 over one complete cycle is 12. However a reactive load does not dissipate power since the voltage and current are out of phase then the product of gives a (cos sin ) term which averages to zero over one complete cycle of oscillation. Because of the mathematical complexity of this topic this discussion will not be pursued further. It is suggested that you skim over chapter 31 of Giancoli to get an broader impression of this topic. MAGNETIC ENERGY Energy stored in Inductor Since forces occur between magnetic circuits, energy must be stored in the magnetic field. Consider the system shown in figure 12. Using Kirchhoff s loop rule we have: 0 = + Consider that in a time dt, a charge = flows. The work done by the battery is given by 0 This equals: 0 = 2 + This is equivalent to the statement that the energy provided by the battery equals the energy dissipated in the resistor plus the energy stored in the self inductance. Thus we have that the energy stored in the inductance is: = Integrating the energy from =0to the final value gives the magnetic energy stored in the self inductance as: = 0 = Consider a simple LC circuit, shown in figure 13, with oscillating current and charge. The total energy is distributed between the capacitor and the inductor as: = + = Note that the energy oscillates between the capacitor, when Q is maximum and =0 to the inductor when =0and I is a maximum. This is analogous to 103

8 Ampère s law gives the magnetic field inside the toroid is () = 0 2 Integrating over the rectangular cross section inside the toroid windings, gives the magnetic flux inside the windings to be Φ = () = 0 2 Figure14 Nturntoriodwithinnerradius, outer radius, andthickness. harmonic oscillations of a pendulum where the energy oscillates between kinetic energy and potential energy. The inertia in the inductance is analogous to moment of inertia in the kinetic energy term for angular motion of the pendulum. The energy stored in the capacitor is analogous to the gravitational potential energy stored at the extreme positions of the pendulum oscillation. Energy Density in a Magnetic Field It is more useful to express the stored magnetic energy density in terms of the magnetic field B just as the electric energy density was expressed in terms of the electric field E Inthecaseoftheelectricfield, the stored electric energy for a capacitor, of = was used to show that the electric energy can be expressed as the integral of the electric energy density in vacuum = Thus the total stored energy in the electric field in vacuum 1 = where the integral is taken over all space For the magnetic fielditwillbeshownlaterthat the magnetic energy = can be expressed in terms of the magnetic energy density in vacuum = Thus the total stored energy in the magnetic field in vacuum is 1 2 = 2 0 The equivalence of this expression and = can be illustrated by considering the toriod shown in figure 14. Φ = 0 2 ln( ) Thus the flux linkage for the N turns wrapped around the toroid is Φ = Φ = ln( ) This gives that the self inductance = Φ = ln( ) Therefore the stored magnetic energy = = 0 4 ()2 ln( ) Consider the integral of = Knowing that () = 0 2 and that the volume element of a ring inside the torus is d =2 gives = 1 µ That is = 0 4 ()2 = = 0 4 ()2 ln( ) which is the same relation obtained using = That is, the two expressions for magnetic energy give the same answer for this case. In fact it can be proven, using vector differential calculus, that this is always true. As a result, the most general expression for the total electromagnetic energy can be written in terms of the electric and magnetic fields as given by the integral over all space of the energy density = =( ) 2 0 ( ) 2 0 This is especially useful for discussions of electromagnetic waves. 104

9 SUMMARY The concepts of self inductance and mutual inductance and some applications have been discussed. It was shown that the induced emf in an isolated circuit can be written as: = where the self inductance L is defined as: Φ Similarly, the coupling between two circuits is given by = where the mutual inductance is defined as: Φ Elements of the response of circuits with L, R, and C were discussed as well as applications to the transformer and the induction coil. It was shown that the energy stored in an inductor is given by = Thus the total energy stored in a circuit having both inductors and capacitors is: = + = It is especially useful to express the total energy stored in an electromagnetic field in terms the energy density of the E and B fields. = ( ) This form will be used in discussing electromagnetic radiation. Reading assignment: Giancoli, Chapter 30 plus skim through Chapter

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