2. Physics 1502: Lecture 20 Today’s Agenda Announcements: –Chap.27 & 28 Homework 06: FridayHomework 06: Friday Induction

3. Faraday's Law dS B B v B N S v B S N

4. A Loop Moving Through a Magnetic Field  (t) = ?  (t) = ? F(t) = ?

5. Schematic Diagram of an AC Generator  d dt B      d (cos(  t)) dt   sin(  t))

6. Schematic Diagram of an DC Generator

8. (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.

9. Demo E-M Cannon v Connect solenoid to a source of alternating voltage. ~ side view   F F  B B B top view 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.

10. 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 F 1 be the force on Ring 1; F 2 be the force on Ring 2. (a) F 2 < F 1 (b) F 2 = F 1 (c) F 2 > F 1

11. 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 a 1 be the acceleration of ring 1 and a 2 be the acceleration of Ring 2. (a) a 2 < a 1 (b) a 2 = a 1 (c) a 2 > a 1 Ring 1 Ring 2

12. 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

13. 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

14. Induction Self-Inductance, RL Circuits L/R V t 0 L  X X X X X X X X X

15. Recap from the last Chapter: 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 Faraday's Law of Induction v B N S v B S N B dI/dt Can time varying current in a conductor induce EMF in in that same conductor ?

16. Self-Inductance X X X X X X  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. X X X X X X Consider the loop at the right.  magnetic field produced in the area enclosed by the loop.  flux through loop changes switch closed  current starts to flow in the loop.

17. Self-Inductance I 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.

18. 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)

20. 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) Archetypal inductor is a long solenoid, just as a pair of parallel plates is the archetypal capacitor. r << l SI UNITS for L : Henry

21. Calculation l r N turns Long Solenoid: N turns total, radius r, Length l 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)

22. RL Circuits At t=0, the switch is closed and the current I starts to flow. Loop rule: Note that this eqn is identical in form to that for the RC circuit with the following substitutions:    I a b L I RC: RC  RL:

23. Lecture 20, ACT 5 At t=0 the switch is thrown from position b to position a in the circuit shown: –What is the value of the current I  a long time after the switch is thrown? (a) I  = 0 (b) I  =  / 2R (c) I  = 2  / R (a) I  = 0 (b) I  =  / 2R (c) I  = 2  / R 1A What is the value of the current I  immediately after the switch is thrown? 1B

24. Lecture 20, ACT 5 At t=0 the switch is thrown from position b to position a in the circuit shown: –What is the value of the current I  a long time after the switch is thrown? (a) I  = 0 (b) I  =  / 2R (c) I  = 2  / R (a) I  = 0 (b) I  =  / 2R (c) I  = 2  / R 1A What is the value of the current I  immediately after the switch is thrown? 1B

25. 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: We therefore write: The voltage drop across the inductor is given by: a b L II 

26. RL Circuit (  on) L/R t I 2L/R 0  R VLVL 0 t  Current Max =  R 63% Max at t=L/R Voltage on L Max =  /R 37% Max at t=L/R

27. 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 b L II

28. RL Circuit (  off) 0 -- VLVL t L/R t 2L/R I 0  R Current Max =  R 37% Max at t=L/R Voltage on L Max = -  37% Max at t=L/R

29.  on  off t 0 -- I t 0  R L/R t 2L/R 0  R I 0 t  L/R 2L/R VLVL VLVL