Chapter 30 Inductance

Transcription

1 Chapter 30 Inductance In this chapter we investigate the properties of an inductor in a circuit. There are two kinds of inductance mutual inductance and self-inductance. An inductor is formed by taken a length of copper wire and wrapping it around a cylinder to form a coil. If a changing current is applied to the coil it induces an emf in adjacent coils (mutual inductance) or itself (self-inductance). A second property of inductors is that it stores energy in its solenoidal magnetic field. Similar to a fully-charged capacitor where the energy is stored in the electric field, an inductor, supplied with a steady current, also stores energy in the form of a magnetic field. 1 Mutual inductance Consider two neighboring coils of wire as shown in the figure. A current flowing in coil 1 produces a magnetic field B and hence a magnetic flux through coil 2. If the current in coil 1 changes, the flux through coil 2 changes as well; and according to Faraday s law, this induces an emf in coil 2. As a result, a change in the current in one circuit can induce a current in a second circuit. Figure 1: The current i 1 in coil 1 gives rise to a magnetic flux through coil 2. 1

2 E 2 = N 2 dφ B2 dt We would like to write an equation that expresses the relationship between the flux in the 2nd coil in terms of the current i 1 in the first coil. N 2 Φ B2 = M 21 i 1 where Φ B2 is the flux for a single turn of coil 2, and M 21 is the mutual inductance of the two coils. Using this equation, we have a working definition for the mutual inductance Unit of Inductance M 21 = N 2 Φ B2 i 1 (Mutual Inductance) (1) 1 H = 1 W b/a = 1 V s/a The emf produced in the 2nd coil E 2 is N 2 dφ B2 /dt, so, we can write the following: E 2 = M 21 di 1 dt (2) 1.1 Calculating Mutual Inductance B 1 = µ o n 1 i 1 = µ o N 1 i 1 l The flux through a cross section of the solenoid equals B 1 A. This also equals the flux Φ B2 through each turn of the outer coil, independent of its cross-section area. M = N 2 Φ B2 i 1 = N 2 B 1 A = N 2 µ o N 1 i 1 A = µ o A N 1 N 2 i 1 i 1 l l 2

3 Figure 2: A long solenoid with cross-sectional area A and N 1 turns is surrounded at its center by a coil with N 2 turns. In example 30.1 in the book, M = 25 µh. Ex. 4 A solenoidal coil with 25 turns of wire is wound tightly around another coil with 300 turns. The inner solenoid is 25.0 cm long and has a diameter of 2.00 cm. At a certain time, the current in the inner solenoid is A and is increasing at a rate of 1750 A/s. (a) For this time, calculate the average magnetic flux through each turn of the inner solenoid. (b) For this time, calculate the mutual inductance of the two solenoids. (c) the emf induced in the outer solenoid due to the changing current in the inner solenoid. 2 Self-Inductance and Inductors Self-induced emfs can occur in any circuit, since there is always some magnetic flux through the closed loop of a current-carrying circuit. However, the effect is enhanced if the circuit includes a coil with N turns of wire. As a result of the current i, there is an average magnetic flux Φ B through each turn of the coil. Similar to the mutual inductance defined earlier, we can define the self-inductance as: L = NΦ B i (Self-Inductance) (3) From Faraday s law for a coil with N turns, the self-induced emf is E = N dφ B /dt, so it follows that: E = L di dt (4) 3

4 Figure 3: The current i in the circuit causes a magnetic field B in the coil and hence a flux through the coil. Ex. 8 A toroidal solenoid has 500 turns, cross-sectional area 6.25 cm 2, and mean radius 4.00 cm. (a) Calculate the coil s self-inductance, and (b) if the current decreases uniformly from 5.00 A to 2.00 A in 3.00 ms, calculate the self inductance emf in the coil. (c) The current is directed from terminal a of the coil to terminal b. Is the direction of the induced emf from a to b or from b to a? 2.1 Inductors as Circuit Elements According to Faraday s Law E = E d l = dφb /dt However, E n 0 only in the inductor, so, E n d l = L di dt E = b a E n d l When applying this equation, reference Fig. 4 below. V ab = b a E dl = L di dt 4

5 Figure 4: A circuit containing an emf source and an inductor. The emf source is variable, so the current i and its rate of change di/dt can be varied. Figure 5: The potential difference across a resistor depends on the current, whereas the potential difference across an inductor (b), (c), (d) depends on the rate of change of the current. 5

6 2.2 Calculating Self-Inductance for a toroidal solenoid Figure 6: Determining the self-inductance of a closely wound toroidal solenoid. Only a few turns of the winding are shown. Part of the toroid is cut away to show the cross-sectional area A and radius r. Φ B = BA = µ 0NiA 2πr The self-inductance of a toroidal solenoid would be: L = NΦ B i 3 Magnetic-Field Energy = µ on 2 A 2πr How much energy can be stored in an inductor? First, we look at the instantaneous power going into an inductor this simultaneously causes a back emf and the build up of the magnetic field. P = V ab i = L i di dt The energy du supplied to the inductor during an infinitesimal time interval dt is du = P dt, so, du = L i di 6

7 Figure 7: A resistor is a device in which energy is irrecoverably dissipated. By contrast, energy stored in a current-carrying inductor can be recovered when the current decreases to zero. Energy stored in an Inductor 3.1 Magnetic Energy Density I U = L i di = L I2 Let s derive an equation for the magnetic energy density stored in a toroidal solenoid. Recall that the magnetic field in a toroidal solenoid (Ex ) is: B = µ oni 2πr Figure 8: Calculating the magnetic field in a toroidal solenoid using Ampere s Law. 7

8 Using our definition of stored energy in an inductor, we can write the total stored energy as: U = 1 2 L I2 = 1 2 µ o N 2 A 2πr To obtain the energy density, we divide by the volume, 2πr area Pappus theorem. u = U 2πrA = 1 2 µ N 2 I 2 o (2πr) 2 However, B = µ o NI/2πr, so, we can write that: u = B2 2µ o (Magnetic Energy Density in a Solenoid) I 2 4 The R-L Circuit Referring to the figure below, we see two switches. The first switch (S 1 ) is closed to put energy into the inductor, L. Afterwards, the switch S 1 is opened and the switch S 2 is closed. This causes the inductor to release its energy. Figure 9: An R-L circuit. 8

9 Current Growth in an R-L Circuit Here is the equivalent circuit for putting energy into the inductor. We apply Krichhoff s rules to determine the current as a function of time i(t). E ir L di dt = 0 Solving this differential equation, we find the following: i(t) = E R (1 e (R/L)t ) and this is shown in the figure below. Figure 10: Graph of i versus t for growth of current in an R-L circuit with an emf in series. The final current is I = E/R; after one time constant τ, the current is 1 1/e of this value. We ve seen this function relationship before in the RC circuit where the time constant τ could be written as τ = RC. In the case of our RL circuit, we can write 9

10 the time constant as τ = L/R, and write: i(t) = E R (1 e t/τ ) where τ = L R 10

11 Current Decay in an R-L Circuit Let s take a look at an inductor that is fully energized, having a potential energy U = 1 2 LI2. Once again, we can apply Kirchhoff s rule and find I(t). I(t) = I o e (R/L)t (Decaying current in an inductor) The decaying current is shown in the figure below. Figure 11: Graph of i versus t for decay of current in an R-L circuit. After one time constant τ, the current is 1/e of its initial value. If we examine the instantaneous power of this decaying circuit, we find the following: where di/dt < 0. 0 = i 2 R + Li di dt 11

12 5 The L-C Circuit The electrical oscillations in an LC circuit are very similar to the mechanical oscillations presented in Sections 14.2 and With the mechanical oscillations of the mass and spring system, the kinetic energy and potential energy are transferred back-and-forth between the mass and the spring. In our electrical counterpart, we have the potential energies transferred back-and-forth between the capacitor and the inductor. Figure 12: In an oscillating L-C circuit, the charge on the capacitor and the current through the inductor both vary sinusoidally with time. Energy is transferred between magnetic energy in the inductor (U B ) and electrical energy in the capacitor (U E ). As in simple harmonic motion, the total energy E remains constant. 5.1 Electrical Oscillations in an L-C Circuit Applying Kirchhoff s loop rule to the L-C circuit, we can determine the charge on the capacitor q(t). L di dt q C = 0 d 2 q dt LC q = 0 L-C circuit) 12

13 Figure 13: In an oscillating L-C circuit, the charge on the capacitor and the current through the inductor both vary sinusoidally with time. Energy is transferred between magnetic energy in the inductor (U B ) and electrical energy in the capacitor (U E ). As in simple harmonic motion, the total energy E remains constant. The general solution to this differential equation is: q(t) = Q cos(ωt + φ) (for an L-C circuit) where Q = EC, the initial charge on the capacitor when there s no magnetic field in the inductor. The angular frequency is determined in the same manner as Chapter 14 where: ω = 1 LC Meanwhile, the current in the L-C circuit can be determined from the definition of the current, i(t) = dq/dt. i(t) = ωq sin(ωt + φ) (for an L-C circuit) 13

14 5.2 Energy Conservation in an L-C Circuit U M + U E = constant 6 The L-R-C Series Circuit 1 2 Li2 + q2 2C = Q2 2C Figure 14: In this figure, the switch is initially in the up position to charge the capacitor, and then moved to the down position (at time t = 0) to start the damped oscillation in the L-R-C circuit. Once again, applying Kirchhoff s loop rule, we find the following differential equation: ir L di dt q C = 0 14

15 As before, we substitute i = dq/dt into the equation and solve the following 2nd order differential equation for q(t). d 2 q dt 2 The solution to this equation is: + R L q(t) = Ae (R/2L)t cos dq dt + 1 LC q = 0 ( ) 1 LC R2 4L t + φ 2 The angular frequency of the underdamped oscillations in an L-R-C series circuit is: ω = 1 LC R2 4L 2 15

16 Figure 15: Graphs showing the capacitor charge as a function of time in an L-R-C series circuit with initial charge Q. 16