Physics
Energy Stored in Inductor
The energy stored in an inductor is a form of potential energy that is stored in the magnetic field surrounding the inductor when a current flows through it. This energy is proportional to the square of the current and is released when the current decreases, causing the magnetic field to collapse. The energy stored in an inductor can be calculated using the formula 1/2 * L * I^2, where L is the inductance and I is the current.
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7 Key excerpts on "Energy Stored in Inductor"
eBook - PDF- David Halliday, Robert Resnick, Kenneth S. Krane(Authors)
- 2019(Publication Date)
- Wiley(Publisher)
823 36-1 INDUCTANCE An inductor is a circuit element that stores energy in the magnetic field surrounding its current-carrying wires, just as a capacitor stores energy in the electric field between its charged plates. An inductor is characterized by its induc- tance, which depends on its geometrical properties; in a similar way, in Chapter 30 we characterized a capacitor by its capacitance, which also depends on its geometrical properties. Figure 36-1 shows an inductor, such as an ideal sole- noid, carrying a current i that produces a magnetic field in its interior. If we change the current, thereby changing and the magnetic flux through the solenoid, Faraday’s law shows that there is an emf generated in the inductor. The in- ductance L is defined* to be the proportionality constant that relates the rate of change of current to the induced emf: (36-1) L L di dt . B B B B This equation is similar to the defining equation for capaci- tance (V C q/C ). Like the capacitance, the inductance is always taken to be a positive quantity. Equation 36-1 shows that the SI unit of inductance is the volt second /ampere. This combination of units has been given the special name of the henry (abbreviation H), so that This unit is named after Joseph Henry (1797 – 1878), an American physicist and a contemporary of Faraday. In an electrical circuit diagram, an inductor is represented by the symbol , which resembles the shape of a solenoid. To find the relationship between the sign of L and the sign of di/dt, we use Lenz’ law. Let us decrease the current i in the solenoid of Fig. 36-1. This decrease is the change that, according to Lenz’ law, the inductance must oppose. 1 henry 1 volt second/ampere. INDUCTANCE I n Chapter 30 we studied the behavior of capacitors, which accumulate charge and thus set up an electric field in which energy is stored.
- David Halliday, Robert Resnick, Jearl Walker(Authors)
- 2023(Publication Date)
- Wiley(Publisher)
After reading this module, you should be able to . . . 30.7.1 Describe the derivation of the equation for the magnetic field energy of an inductor in an RL circuit with a constant emf source. 30.7.2 For an inductor in an RL circuit, apply the relationship between the magnetic field energy U, the inductance L, and the current i. LEARNING OBJECTIVES Energy Stored in a Magnetic Field When we pull two charged particles of opposite signs away from each other, we say that the resulting electric potential energy is stored in the electric field of the particles. We get it back from the field by letting the particles move closer together again. In the same way we say energy is stored in a magnetic field, but now we deal with current instead of electric charges. To derive a quantitative expression for that stored energy, consider again Fig. 30.6.2, which shows a source of emf ℰ connected to a resistor R and an inductor L. Equation 30.6.4, restated here for convenience, ℰ = L di __ dt + iR, (30.7.1) is the differential equation that describes the growth of current in this circuit. Recall that this equation follows immediately from the loop rule and that the loop rule in turn is an expression of the principle of conservation of energy for single-loop circuits. If we multiply each side of Eq. 30.7.1 by i, we obtain ℰi = Li di __ dt + i 2 R, (30.7.2) which has the following physical interpretation in terms of the work done by the battery and the resulting energy transfers: 1. If a differential amount of charge dq passes through the battery of emf ℰ in Fig. 30.6.2 in time dt, the battery does work on it in the amount ℰ dq. The rate at which the battery does work is (ℰ dq)/dt, or ℰi. Thus, the left side of Eq. 30.7.2 represents the rate at which the emf device delivers energy to the rest of the circuit. 2. The rightmost term in Eq. 30.7.2 represents the rate at which energy appears as thermal energy in the resistor.
eBook - PDF- William Moebs, Samuel J. Ling, Jeff Sanny(Authors)
- 2016(Publication Date)
- Openstax(Publisher)
Thus, the concepts we develop in this section are directly applicable to the exchange of energy between the electric and magnetic fields in electromagnetic waves, or light. We start with an idealized circuit of zero resistance that contains an inductor and a capacitor, an LC circuit. An LC circuit is shown in Figure 14.16. If the capacitor contains a charge q 0 before the switch is closed, then all the energy of the circuit is initially stored in the electric field of the capacitor (Figure 14.16(a)). This energy is (14.33) U C = 1 2 q 0 2 C . When the switch is closed, the capacitor begins to discharge, producing a current in the circuit. The current, in turn, creates a magnetic field in the inductor. The net effect of this process is a transfer of energy from the capacitor, with its diminishing electric field, to the inductor, with its increasing magnetic field. Figure 14.16 (a–d) The oscillation of charge storage with changing directions of current in an LC circuit. (e) The graphs show the distribution of charge and current between the capacitor and inductor. Chapter 14 | Inductance 645 In Figure 14.16(b), the capacitor is completely discharged and all the energy is stored in the magnetic field of the inductor. At this instant, the current is at its maximum value I 0 and the energy in the inductor is (14.34) U L = 1 2 LI 0 2 . Since there is no resistance in the circuit, no energy is lost through Joule heating; thus, the maximum energy stored in the capacitor is equal to the maximum energy stored at a later time in the inductor: (14.35) 1 2 q 0 2 C = 1 2 LI 0 2 . At an arbitrary time when the capacitor charge is q(t) and the current is i(t), the total energy U in the circuit is given by q 2 (t) 2C + Li 2 (t) 2 . Because there is no energy dissipation, (14.36) U = 1 2 q 2 C + 1 2 Li 2 = 1 2 q 0 2 C = 1 2 LI 0 2 . After reaching its maximum I 0 , the current i(t) continues to transport charge between the capacitor plates, thereby recharging the capacitor.
eBook - PDF- David Halliday, Robert Resnick, Jearl Walker(Authors)
- 2020(Publication Date)
- Wiley(Publisher)
(30-45) Magnetic Energy If an inductor L carries a current i, the inductor’s magnetic field stores an energy given by U B = 1 2 Li 2 (magnetic energy). (30-49) If B is the magnitude of a magnetic field at any point (in an inductor or anywhere else), the density of stored magnetic energy at that point is u B = B 2 2μ 0 (magnetic energy density). (30-55) Mutual Induction If coils 1 and 2 are near each other, a chang- ing current in either coil can induce an emf in the other. This mutual induction is described by ℰ 2 = −M di 1 dt (30-64) and ℰ 1 = −M di 2 dt , (30-65) where M (measured in henries) is the mutual inductance. 779 Problems 1 Figure 30-21 shows a uni- form magnetic field B → con- fined to a cylindrical volume of radius R. The magnitude of B → is decreasing at a constant rate of 10 mT/s. In unit-vector notation, what is the initial acceleration of an electron released at (a) point a (radial distance r = 5.0 cm), (b) point b (r = 0), and (c) point c (r = 5.0 cm)? 2 In Fig. 30-22, the inductor has 25 turns and the ideal battery has an emf of 16 V. Figure 30-23 gives the magnetic flux Φ through each turn versus the current i through the inductor. The vertical Figure 30-21 Problem 1. r r R c b a B y x Figure 30-23 Problem 2. Φ (10 –4 T • m 2 ) Φ s 0 i (A) i s Figure 30-22 Problems 2, 3, 4, and 5. S R L CHAPTER 30 INDUCTION AND INDUCTANCE 780 axis scale is set by Φ s = 4.0 × 10 –4 T· m 2 , and the horizontal axis scale is set by i s = 2.00 A. If switch S is closed at time t = 0, at what rate di/dt will the current be changing at t = 2.5τ L ? 3 In Fig. 30-22, R = 4.0 kΩ, L = 8.0 μH, and the ideal battery has ℰ = 20 V. How long after switch S is closed is the current 2.0 mA? 4 Switch S in Fig. 30-22 is closed at time t = 0, initiating the buildup of current in the 15.0 mH inductor and the 20.0 Ω resistor. At what time is the emf across the inductor equal to the potential difference across the resistor? 5 In Fig.
eBook - PDF- Edward Purcell(Author)
- 2011(Publication Date)
- Cambridge University Press(Publisher)
It is pleasant, but hardly sur- prising, to find that an exactly similar relation holds for the energy stored in an inductor. That is, we can ascribe to the magnetic field an energy density 0/81r)B 2 , and summing the energy of the whole field will give the energy Jf2LI2. To show how this works out in one case, we can go back to the toroidal coil whose inductance L we calculated in Section 7.8. We found (Eq. 58) (71) 285 286 FIGURE 7.26 Calculation of energy stored in the magnetic field of the towidai coil of Fig. 7.22. CHAPTER SEVEN The magnetic field strength B, with current I flowing, was given 2NI B=- cr (72) To calculate the volume integral of B 2 /81f we can use a volume ele- ment consisting of the cylindrical shell sketched in Fig. 7.26, with vol- ume 21frh dr. As this shell expands from r = a to r = b, it sweeps through all the space that contains magnetic field. (The field B is zero everywhere outside the torus, remember.) 1 J 1 Jb (2NI)2 N 2 hI2 (b) - B2 dv = - -- 21frh dr = -- In - 81f 81f a cr c 2 a Comparing this result with Eq. 71, we see that, indeed, J... J B2 dv = 7f LI2 81f (73) (74) The more general statement, the counterpart of our statement for the electric field in Eq. 38 of Chapter 1, is that the energy U to be associated with any magnetic field B(x. y, z) is given by: U = - B2 dv 1 1 81f Entire (75) field With B in gauss and v in cm 3 , U in Eq. 75 will be given in ergs. In Eq. 70, we may use henrys and amperes, for L and I, and then U will be given in joules. The SI equivalent of Eq. 75 for U in joules, B in teslas, and v in m 3 is U = - B2 dv 1 1 2J.lo Entire field (75') PROBLEMS 7.1 What is the maximum electromotive force induced in a coil of 4000 turns, average radius 12 cm, rotating at 30 revolutions per sec in the earth's magnetic field where the field intensity is 0.5 gauss? Ans. 0.0057 statvolt, or 1.71 volts. 7.2 A long straight wire is parallel to the y axis and passes through the point z = h on the z axis.
eBook - PDF- David Halliday, Robert Resnick, Jearl Walker(Authors)
- 2021(Publication Date)
- Wiley(Publisher)
(30.7.4) If B is the magnitude of a magnetic field at any point (in an inductor or anywhere else), the density of stored magnetic energy at that point is u B = B 2 ___ 2 μ 0 (magnetic energy density). (30.8.3) Mutual Induction If coils 1 and 2 are near each other, a changing current in either coil can induce an emf in the other. This mutual induction is described by ℰ 2 = −M di 1 ___ dt (30.9.8) and ℰ 1 = −M di 2 ___ dt , (30.9.9) where M (measured in henries) is the mutual inductance. Checkpoint 30.9.1 In Fig. 30.9.1a, consider the following three currents (in amperes and seconds) that we can set up in coil 1: (a) i a = 20.0; (b) i b = 20t; (c) i c = 10t. Rank the currents accord- ing to the magnitude of the induced emf in coil 2, greatest first. 946 CHAPTER 30 INDUCTION AND INDUCTANCE 1 If the circular conductor in Fig. 30.1 undergoes thermal expansion while it is in a uniform magnetic field, a current is induced clock- wise around it. Is the magnetic field directed into or out of the page? 2 The wire loop in Fig. 30.2a is subjected, in turn, to six uniform magnetic fields, each directed parallel to the z axis, which is directed out of the plane of the figure. Figure 30.2b gives the z components B z of the fields versus time t. (Plots 1 and 3 are parallel; so are plots 4 and 6. Plots 2 and 5 are parallel to the time axis.) Rank the six plots according to the emf induced in the loop, greatest clockwise emf first, greatest counterclock- wise emf last. Questions Figure 30.1 Question 1. Figure 30.2 Question 2. x y 2 1 3 4 5 6 t B z ( a) ( b ) 8 The switch in the circuit of Fig. 30.6.1 has been closed on a for a very long time when it is then thrown to b. The result- ing current through the inductor is indicated in Fig. 30.8 for four sets of values for the resistance R and inductance L: (1) R 0 and L 0 , (2) 2R 0 and L 0 , (3) R 0 and 2L 0 , (4) 2R 0 and 2L 0 .
eBook - PDFCircuit Analysis with PSpice
A Simplified Approach
- Nassir H. Sabah(Author)
- 2017(Publication Date)
- CRC Press(Publisher)
Variation of i and v when v SRC decreases (a) or increases (b) in Figure 7.18. + – v = 0 (a) (b) + – v t i i Slope = V SD /L L I SD I SD L + – v = 0 – + V SD FIGURE 7.20 (a) Inductor as a short circuit when a steady current flows. (b) Inductor current increasing with time due to a steady applied voltage. Capacitors, Inductors, and Duality 179 7.2.6 Stored Energy The energy stored in the magnetic field of an inductor is equal to the work done in establishing the flux in the inductor against the induced voltage, which opposes the increase in inductor current during the establish- ment of the flux. Consider that the assigned direction of i through an inductor is that of a voltage drop v across L, as in Figure 7.16a. The instantaneous power input to the inductor is p = vi. Assuming that the voltage is applied at t = 0 with i = 0 and λ = 0 for t < 0, the energy supplied over the interval from t = 0 to t is w t pdt vidt t t ( ) = = ò ò 0 0 (7.34) Substituting dλ = vdt from Faraday’s law and dλ = Ldi from Equation 7.18, Equation 7.34 becomes w t id L idi Li L i t t ( ) = = = = = ò ò l l l 0 2 2 0 1 2 1 2 1 2 (7.35) It should be kept in mind that electric energy in an electric circuit is represented by energy stored in capaci- tors. It is a function of voltages across these capacitors and represents potential energy of current carriers with respect to an arbitrary zero reference. On the other hand, magnetic energy in an electric circuit is represented by Energy Stored in Inductors. It is a function of currents through these inductors and represents kinetic energy of current carriers. Primal Exercise 7.7 Determine the inductance of a coil of 500 turns on a toroidal core, as in Figure 7.15, having a mean diam- eter of 10 cm, a cross-sectional area of 0.8 cm 2 , and μ r = 2000. Ans. 1.6 H. Example 7.3: Inductor Response to Trapezoidal Current The current shown in Figure 7.21 is applied to a 1 μH inductor of zero initial flux.
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