1. Physics 2 for Electrical Engineering Ben Gurion University of the Negev www.bgu.ac.il/atomchipwww.bgu.ac.il/atomchip, www.bgu.ac.il/nanocenterwww.bgu.ac.il/nanocenter Lecturers: Daniel Rohrlich, Ron Folman Teaching Assistants: Ben Yellin, Yoav Etzioni Grader: Gady Afek Week 9. Inductance – Self-inductance RL circuits Energy in a magnetic field mutual inductance LC circuits RLC circuits Sources: Halliday, Resnick and Krane, 5 th Edition, Chap. 36.

2. Self-inductance We have already seen several circuit elements: battery capacitor switch resistor and now we are going to get to know the inductor

3. Self-inductance Inductors come in various sizes and colors: Get yours today!

4. Self-inductance Let’s see what an inductor does in a circuit.

5. 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: which is analogous to the definition of capacitance: Like C, L is defined to be positive.

6. Self-inductance What is the self-inductance of a long solenoid?

7. Self-inductance What is the self-inductance of a long solenoid? We apply Faraday ’ s law, to the solenoid, by calculating the magnetic flux in the solenoid: Φ B = πr 2 B, 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 = μ 0 nI.

8. Self-inductance What is the self-inductance of a long solenoid? We apply Faraday ’ s law, to the solenoid, by calculating the magnetic flux in the solenoid: Φ B = πr 2 B, 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 = μ 0 nI. But what is N doing in the formula for E ?

9. Self-inductance What is the self-inductance of a long solenoid? We apply Faraday ’ s law, to the solenoid, by calculating the magnetic flux in the solenoid: Φ B = πr 2 B, 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 = μ 0 nI. The inductor “links” with itself N times!

10. Self-inductance What is the self-inductance of a long solenoid? From B = μ 0 nI, and Φ B = π r 2 B we have from which we infer L = μ 0 nπ r 2 N for a long solenoid.

11. Self-inductance Since the formula for inductance is 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

12. 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 m 3. (b) The current in the solenoid is decreasing by 0.50 A/s. What is the induced “emf”?

13. 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 m 3. (b) The current in the solenoid is decreasing by 0.50 A/s. What is the induced “emf”? Answer: (a) Note where l is the length of the solenoid; thus π r 2 l is the volume, and we have L = μ 0 n 2 (π r 2 l) = (4π × 10 –7 T·m/A)(10 8 /m 2 )(10 –6 m 3 ) = 4π × 10 –5 T·m 2 /A and we check via “E + v × B” that T·m 2 /A = V · s/A.

14. 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 m 3. (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.

15. Self-inductance Example 2: We fill the solenoid with (a) a diamagnetic or (b) a paramagnetic or (c) a ferromagnetic material. What happens to L?

16. Self-inductance Example 2: We fill the solenoid with (a) a diamagnetic or (b) a paramagnetic or (c) a ferromagnetic material. What happens to L? Walk-through metal detectors: Get yours today!

17. RL circuits Now back to the RL circuit: When the switch or circuit breaker is closed, we have (summing potentials around the circuit): where E bat is a constant, the “emf” of the battery. We can write this as

18. RL circuits The solution is and from t = 0 we have So if I(0) = 0 (if there is no current at t = 0, because the switch was open until then), then

19. RL circuits Here is a graph of I(t)I(t) t E bat /R L/RL/R 0.63 E bat /R

20. RL circuits And here is a graph of dI/dt E bat /L t

21. 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?

22. 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 – E bat. 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 E bat /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.

23. 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

24. RL circuits Answer: At t = 0, the current has reached its maximum value E bat /R and Kirchhoff’s equation for the circuit is Thus Switch 2 Switch 1

25. 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)I(t) t A B Open Switch 1, close Switch 2

26. RL circuits Answer: The current depends on an exponential in t/τ where the time constant τ of the circuit is τ = L/R. Since B has a larger τ (slower exponential decay), its L is larger. I(t)I(t) t A B Open Switch 1, close Switch 2

27. Energy in a magnetic field If we multiply each term in the circuit equation by I and rearrange the terms, we obtain What is I E bat ? 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?

28. Energy in a magnetic field If we multiply each term in the circuit equation by I and rearrange the terms, we obtain What is I E bat ? 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? We recognize I 2 R as the rate of energy (heat) dissipation in the resistor.

29. Energy in a magnetic field If we multiply each term in the circuit equation by I and rearrange the terms, we obtain What is I E bat ? 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.

30. 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 implying that the inductor is the source for the “emf” across the resistor.

31. 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): This expression is analogous to the expression U = q 2 /2C for the energy stored in a capacitor.

32. Energy in a magnetic field From U = q 2 /2C = C(ΔV) 2 /2 we derived the energy density u E in an electric field of strength E:

33. Energy in a magnetic field From U = q 2 /2C = C(ΔV) 2 /2 we derived the energy density u E in an electric field of strength E: Likewise we can translate between I and B to derive the energy density u B in a magnetic field of strength B. For a long enough solenoid we have B = μ 0 nI and L = μ 0 nπ r 2 N, where N = nl, so and

34. Energy in a magnetic field To summarize:

35. Source Mutual inductance If two electrical coils are close together, a changing flux due to a changing current in one coil may induce a current in the other coil, and vice versa. This is the mutual induction of two coils and depends on their mutual inductance M. B Coil 2 Coil 1

36. Mutual inductance Let Φ 12 denote the magnetic flux through one turn of Coil 2 due to Coil 1. We will assume that Φ 12 is proportional to I 1, since Ampère’s law implies that B is proportional to I. We write N 2 Φ 12 = MI 1 where N 2 is the number of turns in Coil 2. It can be shown that the inductance M is mutual: N 2 Φ 12 = MI 1 and also N 1 Φ 21 = MI 2 with the same M.

37. Mutual inductance Given that N 2 Φ 12 = MI 1 and N 1 Φ 21 = MI 2, we calculate the induced “emf” in each coil as and now the formal similarity of mutual inductance and self- inductance is apparent.

38. Mutual inductance Given that N 2 Φ 12 = MI 1 and N 1 Φ 21 = MI 2, we calculate the induced “emf” in each coil as and now the formal similarity of mutual inductance and self- inductance is apparent. Can there be mutual conductance with self-conductance? Can there be self-inductance without mutual inductance?

39. Mutual inductance Example 1: Calculate the mutual inductance M of two coils, if Coil 1 is a long solenoid of radius r, length l and N 1 turns, and Coil 2 (with N 2 turns) encloses Coil 1.

40. Mutual inductance Example 1: Calculate the mutual inductance M of two coils, if Coil 1 is a long solenoid of radius r, length l and N 1 turns, and Coil 2 (with N 2 turns) encloses Coil 1. Answer: If Coil 2 encloses Coil 1, then all the magnetic flux in Coil 1 links all the turns of Coil 2, and so Φ 12 is just the flux πr 2 B of the solenoid. Thus N 2 Φ 12 = N 2 πr 2 B = N 2 (πr 2 μ 0 N 1 /l) I 1, so for this case M = πr 2 μ 0 N 1 N 2 /l.

41. Mutual inductance Example 2: The battery inside this electric toothbrush, in its handle, is charged by mutual induction. The base has a coil that just fits into the handle. Given that N base = 800, N handle = 1500, that the cross-sectional area of the coils is 1.0 × 10 –4 m 2, and that l = 0.02 m, what is the mutual inductance of the two solenoids?

42. Mutual inductance Example 2: The battery inside this electric toothbrush, in its handle, is charged by mutual induction. The base has a coil that just fits into the handle. Given that N base = 800, N handle = 1500, that the cross-sectional area of the coils is 1.0 × 10 –4 m 2, and that l = 0.02 m, what is the mutual inductance of the two solenoids? Answer: We already calculated M = πr 2 μ 0 N 1 N 2 /l for a solenoid; we now have πr 2 = 1.0 × 10 –4 m 2, N 1 N 2 = 1.2 × 10 6 and l = 0.02 m, so M = 7.5 mH. l N base N handle Coil 1 (base) Coil 2 (handle)

43. 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 and since I = dq/dt, it is Solving we will find that the current oscillates!

44. LC circuits The solution of is q(t) = q max sin(ωt + δ), where q max, ω and δ are constants: q max 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 so and T = 2π/ω = 2π (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 s 2 and has units s –1 = Hz.)

45. LC circuits The solution of is q(t) = q max sin(ωt + δ), where q max, ω and δ are constants: q max 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 so and T = 2π/ω = 2π Note: While the charge is q(t) = q max sin(ωt + δ), the current is I = dq/dt = ωq max cos(ωt + δ) = I max cos(ωt + δ). This means that the charge and current are π/2 or T/4 out of phase.

46. 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. v = 0 x max I = 0 q max –q max t = 0 E

47. 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. x = 0 q = 0 I = I max t = T/4 v = v max B

48. 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. v = 0 x max t = T/2 I = 0 –q max q max E

49. 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. x = 0 q = 0 I = I max t = 3T/4 v = v max B

50. 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. v = 0 x max I = 0 q max –q max t = T E

51. 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: The energy in the LC circuit has the same form: In both systems, the total energy is constant.

52. 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 its solution is an exponential function in which ω is complex.

53. RLC circuits Substituting this solution into the equation for the circuit, we find this is a quadratic equation and its solutions are For R = 0 we recover the LC result.

54. RLC circuits Substituting this solution into the equation for the circuit, we find this is a quadratic equation and its solutions are For we get damped oscillations at a reduced frequency: q(t) = e –Rt/2L q max sin(ω't + δ), where ω' =

55. RLC circuits Substituting this solution into the equation for the circuit, we find this is a quadratic equation and its solutions are For we get damped oscillations at a reduced frequency: q(t) = e –Rt/2L q max sin(ω't + δ), where ω' =

56. Halliday, Resnick and Krane, 5 th 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, and I1I1 I2I2

57. Halliday, Resnick and Krane, 5 th Edition, Chap. 36, Prob. 13: Answer: The three paths from “A” to “B” must have the same potential difference. We therefore obtain A B I1I1 I2I2

58. Halliday, Resnick and Krane, 5 th Edition, Chap. 36, Prob. 13: Answer: The three paths from “A” to “B” must have the same potential difference. We therefore obtain We can now derive equations for q 1 – q 2 and for q 1 + q 2 with different angular frequencies: