Handout 10: Inductance. SelfInductance and inductors


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1 1 Handout 10: Inductance SelfInductance and inductors In Fig. 1, electric current is present in an isolate circuit, setting up magnetic field that causes a magnetic flux through the circuit itself. This flux changes when the current changes. Thus, any circuit with varying current will have an emf induced in it by variation of its own magnetic field. Such emf is called a selfinduced emf. By Lenz s law, the selfinduced emf always opposes the change in the current that caused the emf. From Fig. 1, the current I in the circuit causes the total magnetic flux Φ through the coil. We define selfinductance L of the circuit as Figure 1: Current in the coil causes magnetic field and hence magnetic flux through the coil itself L = Φ I. The unit of the inductance is henry (H). The selfinductance is also called simply the inductance. From Faraday s law, we have an expression for the selfinduced emf Figure 2: A solenoid E = dφ dt = d LI dt = L di dt. A device that is designed to have a particular inductance is called an inductor. The usual symbol for an inductor is... Solenoid Consider a solenoid (Fig. 2) of lengths l containing N turns. The crosssection area of the solenoid is A. When the current I flows in it, from Ampere s law, the magnetic field inside the solenoid is given by B = μ 0 NI l. The total magnetic flux is Φ = B NA = μ 0 N 2 IA l. Since inductance L = Φ I, we have an expression of inductance of the solenoid L = μ 0N 2 A. l Note that the inductance is a constant depending on the dimensions and the number of turns of the solenoid.
2 Example 1 A cylindrical solenoid of length 0.50 m contains 150 turns. The radius of the solenoid is 2.0 cm. Calculate the inductance. 2 Example 2 An inductor has inductance 0.26 H and carries a current what is increasing at uniform rate As 1. Determine the selfinduced emf. Example 3 An aircore toroid with crosssection area A and mean radius R is closely wound with N turns. a) Show that the inductance is given by L = μ 0N 2 A 2πR. b) Suppose N = 200 turns, A = 5.0 cm 2 and R = 0.1 m, calculate L. Energy stored in inductor Because of the selfinduced emf E, the voltage across the inductor is V = L di dt. The energy is transferred to the inductor at rate P = IV = LI di dt. The energy supplied, U, is equal to the integral Pdt. While the current increases from zero to any I, the energy store is given by I U = LI di = LI2. It must be noted that resistors and inductors behave differently. When the current (either increasing
3 or decreasing) passes through a resistor, energy is dissipated as heat. By contrast, when the current in the inductor is increasing, the energy is stored in the inductor. When the current is decreasing, the energy is released from the inductor. When the current is constant, there is no energy flow in or out. Solenoid Selfinductance of a solenoid is given by L = μ 0 N 2 A l. Therefore, the energy stored 3 U = 1 2 LI2 = 1 2 μ 0 N 2 A l I 2 = 1 2μ 0 μ 0 NI l 2 Al. Since μ 0 NI l = B inside the solenoid and Al is the volume of the solenoid. The energy density (energy per unit volume), is given by u = U Al = B 2. 2μ 0 This is a general expression for energy density stored in magnetic field. It should be compared with expression for energy density stored in electric field u = 1 2 ε 0 E 2 (see handout 4). Example 4 A electricpower industry wants to store energy in a large inductor. What inductance would be needed to store 1.00 kw h 1 of energy in a coil carrying a 200A current? Example 5 On average, electromagnetic waves contain equal amount of magnetic energy density and electric energy density. Determine the ratio of electric field E to to magnetic field B. Comment on this result.
4 4 RL circuit Circuit in Fig. 3 consists of a battery with emf E, a resistor with resistance R and an inductor with inductance L. Initially, the switch S 1 and S 2 are open. When S 1 is closed, the current I flows from the battery and it is increasing from zero. At any instant, the voltage across the resistor is V R = IR and the selfinduced emf of the inductor is E L = L di dt. From Kirchhoff s law, E = V, Figure 3: RL circuit E L di dt = IR L di R dt = I E R. We can solve the above equation as follows: I di 0 I E R = t R L dt 0 ln ln I E R I E R E R 0 I = R L t = R L t. Figure 4: The maximum current E R. After one time constant τ, the current is 1 1 e 63% of the final value The above equation can be rearranged, yielding current I as a function of time t: I = E R 1 e R L t. The current increases with time and it reaches the maximum value E R as t. This can be seen from the graph in Fig. 4. The time constant for an RL circuit is defined as τ = R L. When t = τ, the current is 1 1 e 63% of the maximum value. Figure 5: The decay of current in an RL circuit. When the switch S 1 in Fig. 3 has been closed for a while and that the current reaches the maximum value I = E R. The switch S 1 is open and S 2 is closed at the same time. From Kirchhoff s law, L di dt = IR. The initial current is I = E R. One can solve the above equation to obtain
5 5 I = E R e R L t. This equation is plotted in Fig. 5. When t = τ, the current is 1 e 37% of the initial value. LC circuit An LC circuit, consisting of an inductor and a capacitor, is an example of oscillatory circuit. The charge can current oscillates in time, in contrast to the RL circuit where the current and charge vary exponentially with time. Consider an LC circuit in Fig. 6. Initially, the capacitor has charge q 0 and the switch is open. When the switch is closed, the charge flows from the capacitor to the inductor. The potential difference across the capacitor is q C. The increasing current in the inductor causes a selfinduced emf E L = L di dt. The current can be written as I = dq dt. From Kirchhoff s law, Figure 6: LC circuit L di dt + q c = 0 d 2 q dt 2 + q LC = 0. The above equation looks like a SHM equation for electric charge q with angular frequency ω = 1 LC. The solution, subject to the initial condition q 0 = q 0, is given by Figure 7: Oscillating charge of LC circuit with period T = 2π ω = 2π LC q = q 0 cos ωt. The amplitude of charge oscillation is q 0. The variation of charge is illustrated in Fig. 7. The current varies as I = dq dt = ωq 0 sin ωt. The amplitude of the current is ωq 0 and the phase of the current differs from that of charge by π 2. Figure 8 shows the graphs of energy against time. The energy in the capacitor is given by Figure 8: Variation of energies stored in capacitor, inductor and total energy over time. U C = q 2 2 2C = q 0 2C cos2 ωt, and the energy in the inductor by
6 6 U L = 1 2 LI2 = 1 2 Lω2 q 0 2 sin 2 ωt = q 0 2 2C sin2 ωt. The total energy U C + U L = q 2 0 2C is constant over time. Therefore, there is no energy lost. RLC circuit in series In the LC circuit, the total energy is constant and the charge in the capacitor oscillates with constant amplitude. Introducing a resistor into the circuit will cause energy lost and, hence, the amplitude of charge oscillation decreases with time. This is analogous to damped harmonic oscillator in mechanics. Consider an RLC circuit shown in Fig. 9. Intially, the charge q 0 is stored in the capacitor. The capacitor discharges. The voltage across the capacitor is q C. The induced emf across the inductor is E L = L di dt. The voltage across the resistor is IR. From Kirchhoff s law, Figure 9: RLC circuit L di dt q c = IR. By writing I = dq dt, Kirchhoff s law can be expressed in the form d 2 q dt 2 + R dq L dt + 1 q = 0. LC This secondorder differential equation is analogous to damped harmonic motion where the mass is replaced by the charge q. Figure 10: Graph of charge against time for an underdamped RLC circuit The solution is different for the underdamped (small R) and overdamped (large R) cases. Here we are interested in a specific case when R 2 < 4L C. The solution takes the form q = q 0 e (R 2L)t cos ωt, ω = 1 LC R2 4L 2. The plot of charge against time is shown in Fig. 10. The amplitude of the oscillation decreases exponentially.
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