Physics 121: Electricity & Magnetism – Lecture 12 Induction II & E-M Oscillations

Physics 121: Electricity & Magnetism – Lecture 12 Induction II & E-M Oscillations Dale E Gary Wenda Cao NJIT Physics Department Induction Review     Faraday’s Law: A changing magnetic flux through a coil of wire induces an EMF in the wire, proportional to the number of turns, N Lenz’s Law: The direction of the current driven by the EMF is such that it creates a magnetic field to oppose the flux change   Induction and energy transfer: P = F ⋅ v = Fv The forces on the loop oppose the motion of the loop, and P = iε the power required to move the loop provides the electrical power in the loop   dΦ B ε = E ⋅ d s = − N ∫ A changing magnetic field dt creates and electric field ε = − N dΦ B November 7, dt Induction and Inductance     When we try to run a current through a coil of wire, the changing current induces a “back-EMF” that opposes the current That is because the changing current creates a changing magnetic field, and the increasing magnetic flux through the coils of wire induce an opposing EMF We seek a description of this that depends only on the geometry of Inductance units: henry (H), H = T-m /A the coils (i.e., independent of the current through the coil) We call this the inductance (c.f L = NΦ B Inductance i capacitance) It describes the proportionality between the q current through a coil and the C= magnetic flux induced in it V November 7, Inductance of a Solenoid  Consider a solenoid Recall that the magnetic field inside a solenoid is B = µ 0in Φ B = ∫ B ⋅ dA = µ 0inA Number The magnetic the nsolenoid is then of turnsflux per through unit length = N/l   L= The inductance of the solenoid is then: NΦ B Nµ 0inA = = nlµ nA = µ n 2lA i i Note that this only on the geometry Since N = nl, this ε Acan µ Ndepends A also be written C= L= Compare with capacitance of a capacitor l l Can also write à0= ì107 H/m = 1.257 àH/m Compare with ε0 = 8.85 pF/m November 7, Self-Induction You should be comfortable with the notion that a changing current in one loop induces an EMF in other loop  You should also be able to appreciate that if the two loops are part of the same coil, the induction still occurs—a changing current in one loop of a coil induces a back-EMF in another loop of the same coil  In fact, a changing current in a single loop induces a back-EMF in itself This iL = NΦ B is called self-induction NΦ B  Since for any inductorL = dΦ B then di i L =N dt dt   But Faraday’s Law saysε L dΦ B di The self-induced EMF is = −N = −L opposite to the direction of dt dt change of current November 7, Induced EMF in an Inductor Which statement describes the current through the inductor below, if the induced EMF is as shown? εL → A B C D E Constant and rightward Constant and leftward Increasing and rightward Decreasing and leftward Increasing and leftward November 7,  Inductors in Circuits—The RL Circuit Inductors, or coils, are common in electrical circuits  They are made by wrapping insulated wire around a core, and their main use is in resonant circuits, or filter circuits  Consider the RL circuit, where a battery with EMF ε drives a current around the loop, producing a back EMF εL in the inductor ε − iR − L di = dt  Kirchoff’s loop rule gives  ε differential equation Solving this i = (1 − e − Rt / L ) Rise of current for i gives R November 7, RL Circuits ε  When t is large:i =  Inductor acts like an When t is small (zero), i = 0.open circuit  R Inductor acts like a wire ε (1 − e − Rt / L ) R The current starts from zero and =ε /R increases up to a maximumi of with a time constant given by τL = L R Compare: τ C = RC Inductor time constant Capacitor time constant The voltage across the resistorVRis= iR = ε (1 − e  The voltage across the inductor is VL = ε − VR = ε − ε (1 − e − Rt / L ) = ε e − Rt / L  i= − Rt / L ) November 7, Inductive Time Constant The three loops below have identical inductors, resistors, and batteries Rank them in terms of inductive time constant, L/R, greatest first A I, then II & III (tie) II, I, III III & II (tie), then I III, II, I II, III, I B C D E I II III November 7, RL Circuits What happens when the switch is thrown from a to b?  Kirchoff’s Loop Rule was: ε − iR − L di = dt  Now it is: di iR + L = dt  The decay of the current, then, is ε given by i = e − Rt / L Decay of current R  Voltage across resistor: VR = iR = ε e − Rt / L  VR (V)  Voltage across inductor: di ε d e − Rt / L = −ε e − Rt / L VL = L = L dt R dt November 7, What is Happening?      When the battery is removed, and the RL series circuit is shorted, the current keeps flowing in the same direction it was for awhile How can this be? In the case of an RC circuit, we would see the current reverse as the stored charge flowed off the capacitor But in the case of an RL circuit the opposite happens—charge continues to flow in the same direction What is happening is that the current tries to drop suddenly, but this induces an EMF to oppose the change, causing the current to keep flowing for awhile Another way of thinking about it is that the magnetic field that was stored in the inductor is “collapsing.” There is energy stored in the magnetic field, and when the source of current is removed, the energy flows from the magnetic field back into the circuit November 7, Make Before Break Switches        The switch in a circuit like the one at right has to be a special kind, called a “make before break” switch The switch has to make the connection to b before breaking the connection with a If the circuit is allowed to be in the state like this… even momentarily, midway between a and b, then a big problem results Recall that for a capacitor, when we disconnect the circuit the charge will merrily stay on the capacitor indefinitely Not so on an inductor The inductor needs current, i.e flowing charge It CANNOT go immediately to zero The collapsing magnetic field in the inductor will force the current to flow, even when it has no where to go The current will flow in this case by jumping the air gap Link to video You have probably seen this when unplugging something with a motor—a spark that jumps from the plug to the socket November 7, Example Circuit  This circuit has three identical resistors R = Ω, and two identical inductors L = 2.0 mH The battery has EMF ε = 18 V (a) What is the current i through the battery just after the switch is closed? ε =2A i = (acts like open wire) R (b) What is the current i through the battery a long time after the switch is closed? 3ε (acts like straight wire) i= =6A R (c) What is the behavior of the current between these times? Use Kirchoff’s Loop Rule on each loop to find out November 7, Current Through Battery The three loops below have identical inductors, resistors, and batteries Rank them in terms of current through the battery just after the switch is closed, greatest first A I, II, III II, I, III III, I, II III, II, I II, III, I B C D E I II III November 7, Current Through Battery The three loops below have identical inductors, resistors, and batteries Rank them in terms of current through the battery a long time after the switch is closed, greatest first A I, II, III II, I, III III, I, II III, II, I II, III, I B C D E I II III November 7,  Energy Stored in Magnetic Field di By Kirchoff’s Loop Rule, we have ε = iR + L dt We can find the power in the circuit by multiplying by i εi = i R + Li di dt power stored in magnetic field power power provided by dissipated in battery resistor dU B di  Power is rate that work is done, i.e P= = Li dt dt  So dU B = Li di , or after integration U B = Li Energy in magnetic field  q2 Recall for electrical energy in a capacitor: UE = = CV 2C November 7, The LC Circuit      What happens when we make a circuit from both an inductor and capacitor? If we first charge the capacitor, and then disconnect the battery, what will happen to the charge? Recall that initially the inductor acts like an open circuit, so charge does not flow immediately However, over longer times the inductor acts like a simple, straight wire, so charge will eventually flow off from the capacitor As the charge begins to flow, it develops a magnetic field in the inductor November 7, Electromagnetic Oscillations q2 1 U E U= = =Li 2CV B2C 2 November 7, Oscillations Forever? What you think (physically) will happen to the oscillations over a long time? A They will stop after one complete cycle They will continue forever They will continue for awhile, and then suddenly stop They will continue for awhile, but eventually die away There is not enough information to tell what will happen B C D E November 7, Ideal vs Non-Ideal In an ideal situation (no resistance in circuit), these oscillations will go on forever  In fact, no circuit is ideal, and all have at least a little bit of resistance  In that case, the oscillations get U = mgh smaller with time They are said to be “damped oscillations.” K = 12 mv ω = g l  This is just like the situation with a pendulum, which is another kind of oscillator  There, the energy oscillation is between potential energy and kinetic energy  Damped Oscillations Spring AnimationU = 12 kx K = 12 mv ω = November 7, k m   Derivation of Oscillation Frequency We have shown qualitatively that LC circuits act like an oscillator We can discover the frequency of oscillation by looking at the equations governing the total energy q2 U = UE +UB =      + Li 2C Since the total energy is constant, the time derivative should be zero: dU q dq di = + Li = dt C dt dt dq di d q d 2q q i= = L + =0 dt and dt dt , so making these substitutions: dt C But This is a second-order, differential equation, q = Q cos(ωt + φhomogeneous ) whose solution is i.e the charge varies according to a cosine wave with amplitude dq d q Q and frequency ω Check by taking = −Qω sin(ωt + φ ) = −Qω cos(ωt + φ ) dt dt two time derivatives of charge: Plugd 2into 1 q qoriginal equation: Q L dt + C = − LQω cos(ωt + φ ) + C cos(ωt + φ ) = − Lω + C =0 ω= November 7, LC Example a) What is the expression for the voltage change across the capacitor in the circuit below, as a function of time, if L = 30 mH, and C = 100 µF, and the capacitor is fully charged with 0.001 C at time t=0? First, the angular frequency of oscillation is 1 ω= = = 577.4 rad/s −2 −4 LC (3 × 10 H)(10 F) Because the voltage across the capacitor is proportional to the charge, it has the same expression as the charge: q Q cos(ωt + φ ) = C C At time t = 0, q = Q, so φ = Therefore, the full expression for the voltage 10 −3 C across the capacitor is VC = −6 cos(577t ) = 1000 cos(577t ) volts 10 F VC = November 7, Example, cont’d b) What is the expression for the current in the circuit? The current isi = dq = −Qω sin(ωt ) dt i = −(10 −3 C)(577 rad/s) sin(577t ) = −0.577 sin(577t ) amps c) How long until the capacitor charge is reversed? 2π T =is = That happens after ½ period, where the period f ω T π = = 5.44 ms ω November 7, Summary NΦ B Inductance (units, henry H) is given Lby = i  Inductance of a solenoid is: µ N A L= (depends only on geometry) l  EMF, in terms of inductance, is: ε L = − N dΦ B = − L di dt dt   RL circuits Rise of current Decay of current Inductor time constant L ε ε τL = i = e − Rt / L i = (1 − e − Rt / L ) R R R  Energy in inductor:  Energy in magnetic field Li 2 q LC circuits: total electric + magnetic energy is conserved U = UE +UB = + Li 2C UB = Charge equation Current equation Oscillation frequency q = Q cos(ωt + φ ) i = −Qω sin(ωt + φ ) ω= LC November 7,