0 thru 2 4 r Biot-Savart Ampere Inductance Flux B - PDF document

Inductance Important fact: Magnetic Flux Φ B is proportional to the current making the Φ B All our equations for B-fields show that B α I µ × I d l v r ˆ ⋅ = µ B d l I = v v d B v 0 ∫ 0 thru π 2 4 r Biot-Savart Ampere Inductance Φ ∝ ∝ Flux B I B Φ B ∝ Flux I Assumes leaving everything else the same. If we double the current I, we will double the magnetic flux through any surface. i 1

Self-inductance Self-Inductance (L) of a coil of wire Φ B ≡ LI Φ This equation defines self-inductance. ≡ L B Note that since Φ B α I, L must be I independent of the current I. L has units [L] = [Tesla meter 2 ]/[Amperes] New unit for inductance = [Henry]. Inductors (coil of wire) An inductor is just a coil of wire. The Magnetic Flux created by the coil, through the coil itself: Φ = ⋅ B v d A v B ∫ This is quite hard to calculate for a single loop. Earlier we calculated B at the center, but it varies over the area. 2

Inductors Consider a simpler case of a solenoid N = µ = µ | B inside | nI I v 0 0 L Recall that the B-field inside a solenoid is uniform! N = number of loops L = length of solenoid * Be careful with symbol L ! Inductors N = µ = µ | B inside | nI I v 0 0 L Φ = ⋅ = = µ N B v d A v N BA N ( nI ) A B ∫ 0 Φ = = µ = µ 2 L B NnA An L 0 0 I Self-Inductance Length 3

Clicker Question Inductor 1 consists of a single loop of wire. Inductor 2 is identical to 1 except it has two loops on top of each other. How do the self-inductances of the two loops compare? A) L 2 = 2 L 1 B) L 2 > 2L 1 C) L 2 < 2L 1 I Coil 2 Coil 1 I Answer: L2 > 2L1 , in fact L2 = 4 L1. The self inductance L increases by 4. L = Φ /i. If we keep i fixed, but double the number N of turns, Φ increases by 4 . Total flux. N doubles, but Φ 1 = BA also doubles because when we double the number of turns, then B is doubled. Clicker Question Two long solenoids, each of inductance L, are connected together to form a single very long solenoid of inductance L total . What is L total ? A) 2L + = B) 4L C) 8L L D) none of these/don't know L L total = ? Answer: 2L. The inductance of a solenoid is L = µ on2A l , where n is the number of turns per length n = N/l and l is the length. In this case, we did not change n, but l (length) doubled, so L doubles. 4

Self-induced EMF What does this inductance tell us? Φ = L B I Φ B = LI Φ d dI L is independent of time. B = L Depends only on geometry of inductor dt dt (like capacitance). Self-induced EMF Φ Φ d dI d B = ε = − L B Recall Faraday’s Law dt dt dt dI − ε = L dt Changing the current in an inductor creates dI an EMF which opposes the change in the ε = − L current. dt Sometimes called “back EMF” 5

Self-induced EMF dI ε = − L dt It is difficult (requires big external Voltage) to change quickly the current in an inductor. The current in an inductor cannot change instantly. If it did (or tried to), there would be an infinite back EMF. This infinite back EMF would be fighting the change! Magnetic Field energy Recall that for a capacitor C, there is stored potential energy in the electric field. 1 CV U = 2 2 The energy is stored in the electric field and the density is: U 1 = = ε 2 u E | E v | 0 Volume 2 6

Magnetic Field energy The power supplied to an inductor is di = = P Vi Li dt The energy dU supplied to an inductor in an interval dt is: = = dU Pdt Lidi The total energy U supplied while the current increases from 0 to a final value is I 1 LI = ∫ = 2 U Lidi 2 0 Magnetic energy density For an inductor L, with current I, there is stored energy in the magnetic field. 1 LI U = 2 2 µ 2 1 N A For a coil this is = 2 U 0 I π 2 2 r 2 2 U U 1 N I The energy density is = = = µ u B 0 π π 2 Volume 2 rA 2 ( 2 r ) µ NI But we know that = B 0 π 2 r 2 B → u B = 2 µ 0 7

Clicker Question The same current i is flowing through solenoid 1 and solenoid 2. Solenoid 2 is twice as long and has twice as many turns as solenoid 1, and has twice the diameter. (Hint) for a solenoid B = µ o n i ) What is the ratio of the magnetic energy contained in U solenoid 2 to that in solenoid 1, that is, what is 2 U 1 I A)2 B) 4 1 C) 8 D) 16 2 I E) None of these. Answer: There is 8 times as much magnetic field energy in the large solenoid as in the small solenoid. The B-field is the same in both solenoids (same n = turns/length so same B = µ o n I) so both solenoids contain the same energy per volume u = U/vol = B2/(2 µ o). The larger solenoid has 8 times the volume (2X the length, 4X the cross-sectional area) so it has 8 the energy. Magnetic Field energy It takes work to get current flowing through an inductor. You must work against the back EMF which opposes any change in the current. That work = potential energy stored = U = ½ LI 2 And is thus stored in the inductor’s magnetic field. 8

Inductors as circuit elements What do these inductors do in circuits? Just recall that the EMF or Voltage across an inductor is: dI ε = − L dt So, when we add them to circuits, we can apply the usual Kirchhoff’s Voltage Law and include the inductors. Inductors in a circuit Consider a circuit with a battery, resistor and inductor (RL circuit) switch R a Battery=V b L Switch is in position (a) for a long time. In steady state (after a long time), the current will no longer be changing and thus the inductor looks like a regular wire! 9

RL circuit switch R a Battery=V b L If after a long time the inductor acts like a wire: V I = ∆ = − = V V IR 0 R RL circuit switch R a Battery=V b L At t=0, move the switch to (b). Normally one might expect there to immediately be zero current. However, the inductor has stored energy! 10

LR circuit switch R a Battery=V b L dI ∆ = − − = V IR L 0 dt di ⎛ R ⎞ We need to solve this differential = − i ⎜ ⎟ dt L equation for i(t). ⎝ ⎠ RL circuit dI R ⎛ ⎞ = − I ⎜ ⎟ dt L ⎝ ⎠ V ⎛ R ⎞ − t ⎜ ⎟ I = = L I ( t ) I e ⎝ ⎠ where 0 R 0 ⎛ L ⎞ − t / ⎜ ⎟ Current exponentially decays with = R I ( t ) I e ⎝ ⎠ Time Constant = τ = L/R (units of seconds). 0 11

Clicker Question Rank in order, from largest to smallest, the time constants in the three circuits. A. τ 1 > τ 2 > τ 3 B. τ 2 > τ 1 > τ 3 C. τ 2 > τ 3 > τ 1 D. τ 3 > τ 1 > τ 2 E. τ 3 > τ 2 > τ 1 L ⎛ ⎞ − t / ⎜ ⎟ = i ( t ) i e R ⎝ ⎠ 0 Clicker Question The switch in the circuit below is closed at t=0. R = 20 Ω What is the initial rate of change of current di/dt in the inductor, V = 10V immediately after the switch is L = 10H closed ? (Hint) what is the initial voltage across the inductor?) A) 0 A/s B) 0.5A/s C)1A/s D) 10A/s E) None of these. Answer: 1A/s. By the Loop Law, V(battery voltage) = iR + Ldi/dt. Immediately after the switch is closed , the current is zero (because the current thru the inductor cannot change instantly), so iR is zero, so V = L di/dt. So di/dt = V/L = 10V/10H = 1 A/s. 12

Clicker Question An LR circuit is shown below. Initially the switch is open. At time t=0, the switch is closed. What is the current thru the inductor L immediately after the switch is closed (time = 0+)? R 1 = 10 Ω A) Zero B) 1 A V = 10V L = 10H R 2 = 10 Ω C) 0.5A D) None of these. Clicker Question An LR circuit is shown below. Initially the switch is open. At time t=0, the switch is closed. R 1 = 10 Ω V = 10V L = 10H R 2 = 10 Ω After a long time, what is the current from the battery? A) 0A B) 0.5A C) 1.0A D) 2.0A E) None of these. 13

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