Ben Gurion University of the Negev www.bgu.ac.il/atomchip Physics 2B for Materials and Structural Engineering Lecturer: Daniel Rohrlich Teaching Assistants: Oren Rosenblatt, Shay Inbar Week 9. Inductance – Self-inductance • RL circuits • Energy in a magnetic field • mutual inductance • LC circuits • RLC circuits Source: Halliday, Resnick and Krane, 5th Edition, Chap. 36. Self-inductance We have already seen several circuit elements: battery capacitor switch resistor and now we are going to get to know the inductor Self-inductance Let’s see what an inductor does in a circuit. When current starts to flow through an inductor, it induces a magnetic flux that changes that same current. Hence the term “self” inductance. The inductance L of an inductor is defined according to the “emf” E that it produces for a given change in current: E L dI , dt which is analogous to the definition of capacitance: V q C Like C, L is defined to be positive. . Self-inductance What is the self-inductance of a long solenoid? We apply Faraday’s law, E N d B , to the solenoid, by dt calculating the magnetic flux in the solenoid: ΦB = πr2B, where N is the total number of coils in the solenoid, r is its radius and n is the number of coils per unit length. Also, the formula for B inside a solenoid is B = μ0 nI. The inductor “links” with itself N times! Self-inductance What is the self-inductance of a long solenoid? From B = μ0nI, E N E N d B dt d B dt 2 r N and ΦB = π r2B we have dB dt 2 μ 0 n r N dI dt from which we infer L = μ0 nπ r2N for a long solenoid. , Self-inductance Since the formula for inductance is E L dI dt and the “emf” E has units of volts, the unit of inductance must be volt/ (ampere/second). This unit is called the henry and denoted H: H = V·s/A . Joseph Henry Self-inductance Example 1: (a) Calculate the inductance L of a solenoid with 100 coils/cm if the volume inside the solenoid is 10–6 m3. (b) The current in the solenoid is decreasing by 0.50 A/s. What is the induced “emf”? 2 Answer: (a) Note E μ 0 n r N dI dt 2 2 μ 0 n ( r l ) dI dt where l is the length of the solenoid; thus π r2l is the volume, and we have L = μ0 n2(π r2l) = (4π × 10–7 T·m/A)(108/m2 )(10–6 m3) = 4π × 10–5 T·m2/A and we check via “E + v × B” that T·m2/A = V·s/A. , Self-inductance Example 1: (a) Calculate the inductance L of a solenoid with 100 coils/cm if the volume inside the solenoid is 10–6 m3. (b) The current in the solenoid is decreasing by 0.50 A/s. What is the induced “emf”? Answer: (b) We substitute L = 4π × 10–5 V·s/A back into the definition E = –L(dI/dt) to obtain E = 2π × 10–5 V. RL circuits Now back to the RL circuit: When the switch or circuit breaker is closed, we have (summing potentials around the circuit): Ebat IR L dI 0 , dt where Ebat is a constant, the “emf” of the battery. We E bat E bat d R can write this as I I . dt R L R RL circuits The solution is I (t ) Ebat R const. e Rt / L and from t = 0 we have E bat const . I ( 0 ) . R So if I(0) = 0 (if there is no current at t = 0, because the switch was open until then), then I (t ) Ebat R 1 e Rt / L . , RL circuits Here is a graph of I ( t ) Ebat R 1 e Rt / L : I(t) Ebat/R 0.63 Ebat/R L/R t RL circuits Example 1: A switch controls the current in an RL circuit with large L. Is sparking more likely when you initially close the switch or when you open it after t >> L/R? Answer: When you close the switch, there is no initial current, I(0) = 0, so IR = 0 and the potential on the inductor is –Ebat. The potential over the switch drops immediately to 0 and the inductor won’t immediately let current through, so sparking isn’t likely. When you open the switch, there is already a steady current Ebat/R and the inductor does not allow the current to drop to 0 immediately, so charge builds up on the switch and may cause sparking. RL circuits Example 2: Consider the setup below; assume that Switch 1 has been closed for a long time while Switch 2 has been open. At time t = 0, Switch 1 is opened and Switch 2 is closed. What is the current in the upper loop for t > 0? Switch 2 Switch 1 RL circuits Answer: At t = 0, the current has reached its maximum value Ebat/R and Kirchhoff’s equation for the circuit is IR L Thus I ( t ) I ( 0 ) e Rt / L ( Ebat / R ) e Rt / L . Switch 2 Switch 1 dI dt 0. RL circuits Example 3: We repeat the last example with two RL circuits, A and B. Which circuit has the larger L, if the resistances and batteries are the same? I(t) A B Open Switch 1, close Switch 2 t Energy in a magnetic field If we multiply each term in the circuit equation Ebat IR L dI 0 dt by I and rearrange the terms, we obtain 2 I E bat I R IL dI . dt What is IEbat? It is the rate at which charge exits the battery at the + end and enters the battery at the – end, times the difference in electric potential between the two ends. So it is the rate at which the battery delivers potential energy to the circuit. What happens to this potential energy? The last term, IL(dI/dt), must be the rate at which energy is stored in the inductor. Energy in a magnetic field What does it mean that IL(dI/dt) is the rate at which energy is stored by the inductor? As we saw in Example 2, Kirchhoff’s equation for an RL circuit without a battery is IR L dI 0, dt implying that the inductor is the source for the “emf” across the resistor. Energy in a magnetic field Hence if we integrate IL(dI/dt) with respect to time t, starting from I(0) = 0, we must obtain the total energy U stored in the inductor at time t or, equivalently, at current I(t): U 0 dI LI dt dt LI 2 dI I L . 2 0 This expression is analogous to the expression U = q2/2C for the energy stored in a capacitor. Energy in a magnetic field From U = q2/2C = C(ΔV)2/2 we derived the energy density uE in an electric field of strength E: uE U C (V ) Ad 2 0 A / d ( Ed ) 2 2 Ad 2 Ad 1 2 0E 2 . Likewise we can translate between I and B to derive the energy density uB in a magnetic field of strength B. For a long enough solenoid we have B = μ0 nI and L = μ0 nπ r2N, where N = nl, so 2 U I L 2 and u B U 2 r l B / 0 n B 2 20 . 2 2 0 n r N 2 2 2 B r l 20 Energy in a magnetic field To summarize: uE 1 uB B 2 0E 2 20 2 LC circuits At right is an LC circuit: Let’s assume that the capacitor is charged when the switch is closed at t = 0. For t > 0 the equation for the circuit is q L C dI 0, dt and since I = dq/dt, it is q C 2 L d q dt 2 2 0. current oscillates! Solving d q dt 2 q LC , we will find that the LC circuits 2 The solution of d q dt 2 q LC is q(t) = qmax sin(ωt + δ), where qmax , ω and δ are constants: qmax is the amplitude of the charge oscillations, ω is their angular frequency and δ is their phase. Substituting this solution into the differential equation, we find 2 1 LC , so 1 / LC and T = 2π/ω = 2π LC . (What about units? The units of L are the henry, H = V·s/A; the units of C are the farad, F = C/V. So LC has units s2 and 1/ LC has units s–1 = Hz.) LC circuits 2 The solution of d q dt 2 q LC is q(t) = qmax sin(ωt + δ), where qmax , ω and δ are constants: qmax is the amplitude of the charge oscillations, ω is their angular frequency and δ is their phase. Substituting this solution into the differential equation, we find 2 1 LC , so 1 / LC and T = 2π/ω = 2π LC . Note: While the charge is q(t) = qmax sin(ωt + δ), the current is I = dq/dt = ωqmax cos(ωt + δ) = Imax cos(ωt + δ). This means that the charge and current are π/2 or T/4 out of phase. LC circuits The formulas for q and I in an LC circuit are analogous to the formulas for the position x and momentum p of a body in a harmonic oscillator. Thus, we can intuitively understand LC circuits by analogy with harmonic oscillators. I=0 qmax v=0 E t=0 –qmax xmax LC circuits The formulas for q and I in an LC circuit are analogous to the formulas for the position x and momentum p of a body in a harmonic oscillator. Thus, we can intuitively understand LC circuits by analogy with harmonic oscillators. I = Imax v = vmax q=0 t = T/4 x=0 B LC circuits The formulas for q and I in an LC circuit are analogous to the formulas for the position x and momentum p of a body in a harmonic oscillator. Thus, we can intuitively understand LC circuits by analogy with harmonic oscillators. The energy of a harmonic oscillator is the sum of its kinetic and potential energies: 2 1 dx 2 2 U m m x 2 dt 2 1 . The energy in the LC circuit has the same form: 2 1 dq 2 U L q 2 dt 2C 1 In both systems, the total energy is constant. . RLC circuits Just as harmonic motion may be forced or damped, so can an electrical circuit. Adding a resistor to an LC circuit turns it into an RLC circuit, and the resistor damps the oscillations. The equation for the circuit is now q C 2 L d q dt 2 R dq 0; dt its solution is an exponential i t function q ( t ) q max e in which ω is complex. RLC circuits Substituting this solution into the equation for the circuit, we find 1 L 2 i R 0; C this is a quadratic equation and its solutions are i R 2L R LC 2 L 1 2 . For R = 0 we recover the LC result. RLC circuits Substituting this solution into the equation for the circuit, we find 1 L 2 i R 0; C this is a quadratic equation and its solutions are i R 2L R LC 2 L 1 2 . For 0 R 2 L / C , we get damped oscillations at a reduced frequency: q(t) = e–Rt/2L qmax sin(ω't + δ), where ω' = 2 2 (R / 2L) . Halliday, Resnick and Krane, 5th Edition, Chap. 36, Prob. 13: Three identical inductors (with inductance L) and two identical capacitors (with capacitance C) are connected as shown. Show the circuit has two different angular frequencies, 1/ 3 LC and 1/ LC . I1 I2 Halliday, Resnick and Krane, 5th Edition, Chap. 36, Prob. 13: Answer: The three paths from “A” to “B” must have the same potential difference. We therefore obtain q1 C L dI 1 dt L d dt ( I1 I 2 ) B I1 I2 A q2 C L dI 2 dt . Halliday, Resnick and Krane, 5th Edition, Chap. 36, Prob. 13: Answer: The three paths from “A” to “B” must have the same potential difference. We therefore obtain q1 C L dI 1 dt L d dt ( I1 I 2 ) q2 L C dI 2 . dt We can now derive equations for q1 – q2 and for q1 + q2 with different angular frequencies: q1 q 2 d d L ( I1 I 2 ) 2 L ( I1 I 2 ) , C dt dt 1/ 3 LC q1 q 2 d L ( I1 I 2 ) 0 , C dt 1/ LC , .

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