Transcript Lecture 20
Physics 1502: Lecture 20 Today’s Agenda • Announcements: – Chap.27 & 28 • Homework 06: Friday • Induction Faraday's Law dS B N S v B S N v B B A Loop Moving Through a Magnetic Field Schematic Diagram of an AC Generator dF B d (cos( wt)) = - NAB e= -N dt dt = - NAB w sin( wt)) Schematic Diagram of an DC Generator (a) As the conducting plate enters the field (position 1), the eddy currents are counterclockwise. As the plate leaves the field (position 2), the currents are clockwise. In either case, the force on the plate is opposite to the velocity, and eventually the plate comes to rest. (b) When slots are cut in the conducting plate, the eddy currents are reduced and the plate swings more freely through the magnetic field. Demo E-M Cannon • Connect solenoid to a source of alternating voltage. • The flux through the area ^ to axis of solenoid therefore changes in time. • A conducting ring placed on top of the solenoid will have a current induced in it opposing this change. • There will then be a force on the ring since it contains a current which is circulating in the presence of a magnetic field. v ~ side view F B B F B top view Lecture 20, ACT 1 • Suppose two aluminum rings are used in the demo; Ring 2 is identical to Ring 1 except that it has a small slit as shown. Let F1 be the force on Ring 1; F2 be the force on Ring 2. Ring 1 Ring 2 (a) F2 < F1 (b) F2 = F1 (c) F2 > F1 Lecture 20, ACT 2 • Suppose one copper and one aluminum rings are used in the demo; the resistance of the two rings is similar but the aluminum ring has less mass. Let a1 be the acceleration of ring 1 and a2 be the acceleration of Ring 2. (a) a2 < a1 (b) a2 = a1 (c) a2 > a1 Ring 1 Ring 2 Lecture 20, ACT 3 • Suppose you take the aluminum ring, shoot it off the cannon, and try to nail your annoying neighbor. Unfortunately, you just miss. You think, maybe I can hit him (her) if I change the temperature of the ring. In order to hit your neighbor, do you want to heat the ring, cool the ring, or is it just hopeless? (a) heat (b) cool (c) hopeless Hot Ring Cool Ring Lecture 20, ACT 4 • Suppose the alternating magnetic field is kept at a level where the ring just levitates, but doesn’t jump off. If I keep the ring suspended for about 5 minutes, is it safe to pick it up? (a) No (b) Yeah, I’ll do it ~ side view Induction Self-Inductance, RL Circuits I a I R XXX XXXX XX b L e L/R e1 1 VL f( x ) 0.5 0.0183156 0 0 1 2 3 4 t Recap from the last Chapter: Faraday's Law of Induction B N S v B S N v Can time varying current in a conductor induce EMF in in that same conductor ? • Time dependent flux is generated by change in magnetic field strength due motion of the magnet • Note: changing magnetic field can also be produced by time varying current in a nearby loop dI/dt B Self-Inductance • Consider the loop at the right. • switch closed current starts to flow in the loop. X XX X X XX XX X X XX X • \ magnetic field produced in the area enclosed by the loop. • \ flux through loop changes • \ emf induced in loop opposing initial emf • Self-Induction: the act of a changing current through a loop inducing an opposing current in that same loop. Self-Inductance • The magnetic field produced by the current in the loop shown is proportional to that current. • The flux, therefore, is also proportional to the current. • We define this constant of proportionality between flux and current to be the inductance L. • We can also define the inductance L, using Faraday's Law, in terms of the emf induced by a changing current. I Self-Inductance • The inductance of an inductor ( a set of coils in some geometry, e.g., solenoid, toroid) then, like a capacitor, can be calculated from its geometry alone if the device is constructed from conductors and air. • If extra material (e.g. iron core) is added, then we need to add some knowledge of materials as we did for capacitors (dielectrics) and resistors (resistivity) Self-Inductance • The inductance of an inductor ( a set of coils in some geometry ..eg solenoid, toroid) then, like a capacitor, can be calculated from its geometry alone if the device is constructed from conductors and air. • If extra material (eg iron core) is added, then we need to add some knowledge of materials as we did for capacitors (dielectrics) and resistors (resistivity) SI UNITS for L : Henry • Archetypal inductor is a long solenoid, just as a pair of parallel plates is the archetypal capacitor. l r N turns r << l A ++++ d ----- Calculation l • Long Solenoid: N turns total, radius r, Length l r N turns For a single turn, The total flux through solenoid is given by: Inductance of solenoid can then be calculated as: This (as for R and C) depends only on geometry (material) RL Circuits • At t=0, the switch is closed and the current I starts to flow. a I I R b • Loop rule: e Note that this eqn is identical in form to that for the RC circuit with the following substitutions: RCRL: RC: \ L Lecture 20, ACT 5 • At t=0 the switch is thrown from position b to position a in the circuit shown: 1A – What is the value of the current I a long time after the switch is thrown? I a I R b L e R (a) I = 0 (b) I = e / 2R (c) I = 2e / R 1B • What is the value of the current I0 immediately after the switch is thrown? (a) I0 = 0 (b) I0 = e / 2R (c) I0 = 2e / R Lecture 20, ACT 5 • At t=0 the switch is thrown from position b to position a in the circuit shown: 1A – What is the value of the current I a long time after the switch is thrown? I a I R b L e R (a) I = 0 (b) I = e / 2R (c) I = 2e / R 1B • What is the value of the current I0 immediately after the switch is thrown? (a) I0 = 0 (b) I0 = e / 2R (c) I0 = 2e / R RL Circuits • To find the current I as a fct of time t, we need to choose an exponential solution which satisfies the boundary condition: a I I R b e L RL = R • We therefore write: • The voltage drop across the inductor is given by: L RL Circuit (e on) e/1 R Max = e/R Q Current L/R 2L/R 1 2 f( x ) I 0.5 63% Max at t=L/R 00 0 Voltage on L Max = e/R 1 t 3 4 t 3 4 x t/RC e1 f( xV ) 0.5 L 37% Max at t=L/R 0.0183156 0 0 1 2 RL Circuits • After the switch has been in position a for a long time, redefined to be t=0, it is moved to position b. • Loop rule: • The appropriate initial condition is: • The solution then must have the form: a I I R b e L RL Circuit (e off) e/R 1 1 Current L/R 2L/R 1 2 f( x ) 0.5 I Max = e/R 37% Max at t=L/R 0.0183156 0 0 01 0 Q Voltage on L Max = -e t 3 x 4 4 f( x ) V 0.5L 37% Max at t=L/R -e 0 0 1 2 t 3 4 e on L/R e/1 R e off 2L/R e/R 1 1 I f( x ) 0.5 00 1 2L/R 1 2 f( x ) 0.5 I 0 1 e1 2 t 0.0183156 3 4 01 0 Q 0 0 1 2 0 0 x t/RC f( xV ) 0.5 L 83156 L/R t 3 4 t 3 x 4 4 f( x )V 0.5 L -e0 0 1 2 t 3 4