Physics 212 Lecture 29, Slide 1 Physics 212 Lecture 29 Course Review The Topics For Today – –Electric Fields/Gauss’ Law/Potential – –Faraday’s Law – –RC/RL Circuits – –AC Circuits

Physics 212 Lecture 29 Music Who is the Artist? A)Ray Charles B)Solomon Burke C)Henry Butler D)Johnny Adams E)Otis Redding “Rediscovered” Soul singer from 60’s Absolutely Beautiful Voice Highly Recommended DVD: “One Night History of the Blues” Why? Great singer to end the course

Electric Field Physics 212 Lecture 29, Slide 3 Horizontal components cancel E from top arc points down E from bottom arc points up Top arc produces smaller horizontal components E total points down sin    Calculation:  Two curved rods each have charge +Q uniformly distributed over their length Which statement best describes the electric field due to these two rods at the midpoint between the two rods (marked by an X) A. E points upB. E points downC. E = 0 48%

Conductors, Potential Energy Physics 212 Lecture 29, Slide 4 Potential Energy is a measure of work done by E field Spherical symmetry & Gauss’ law E = 0 inside shell No work done to move q No change in potential energy ! A thin non-conducting spherical shell carries a positive charge Q spread uniformly over the surface of the shell. A positive test charge q is located inside the shell. Compare U A, the potential energy of the test charge in position A (near the shell) with U B, the potential energy of the test charge in position B (center of the shell) A. U A > U B B. U A = U B C. U A < U B 53%

Electric Potential, Gauss’ Law Physics 212 Lecture 29, Slide 5 (A) (B) (C) (E) ALWAYS START FROM DEFINITION OF POTENTIAL Spherical symmetry & Gauss’ law determines E a < r < b: A thin non-conducting spherical shell of radius a carries a uniformly distributed net surface charge 2Q. A second thin non- conducting shell of radius b carries a uniformly distributed net surface charge 3Q. The two shells are concentric. Calculate the potential difference  V = V(a) – V(b) between the inner and outer shells. 37% (D)

Electric Potential, Gauss’ Law Physics 212 Lecture 29, Slide 6 Spherical symmetry & Gauss’ law determines E r < a: (A) (B) (C) (E) A thin non-conducting spherical shell of radius a carries a uniformly distributed net surface charge 2Q. A second thin non- conducting shell of radius b carries a uniformly distributed net surface charge 3Q. The two shells are concentric. Calculate the potential difference  V = V(a) – V(0) between the inner shell and the origin. 45% (D)

Physics 212 Lecture 29, Slide 7 (A) (B) (D) (E) Charge must be induced to insure E = 0 within conducting shell Spherical symmetry & Gauss’ law determines E (C) A solid conducting sphere of radius a is centered on the origin, and carries a total charge Q 1. Concentric with this sphere is a conducting spherical shell of inner radius b and outer radius c, which carries a total charge Q 2. What is the surface charge density,  b, on the inner surface of the outer spherical shell (r = b)? 59% Gauss’ Law, Conductors

Physics 212 Lecture 29, Slide 8 r < a: E = 0 b < r < c: E = 0 Eliminate (e) Eliminate (a) and (c) Q 1 = -3  C Q 1 + Q 2 = +2  C a < r < b: E = kQ 1 /r 2 r > c: E = k(Q 1 +Q 2 )/r 2 For r > c, E must be less than the continuation of E from a to b A solid conducting sphere of radius a is centered on the origin, and carries a total charge Q 1. Concentric with this sphere is a conducting spherical shell of inner radius b and outer radius c, which carries a total charge Q 2. Which of the following graphs best represents the magnitude of the electric field at points along the positive x axis? 70% Q 1 = - 3  C Q 2 = 5  C Gauss’ Law

Electric Potential Physics 212 Lecture 29, Slide 9 ALWAYS START FROM DEFINITION OF POTENTIAL Break integral into two pieces same for conductor & insulator conductor: = 0 insulator:  0 A solid conducting sphere of radius a is centered on the origin, and carries a total charge Q 1. Concentric with this sphere is a conducting spherical shell of inner radius b and outer radius c, which carries a total charge Q 2. If the inner conducting sphere were replaced with an insulating sphere having the same charge, Q 1, distributed uniformly throughout its volume, the magnitude of the potential difference |V b – V 0 | would 53% A. increaseB. decreaseC. stay the same Q 1 = - 3  C Q 2 = 5  C

RL Circuits Physics 212 Lecture 29, Slide 10 Strategy: Back to First Principles The time constant is determined from a differential equation for the current through the inductor. The time constant is determined from a differential equation for the current through the inductor. Equation for current through inductor obtained from Kirchhoff’s Rules Equation for current through inductor obtained from Kirchhoff’s Rules I I – I 1 I1I1I1I In the circuit below, V = 6 Volts, R = 10 Ohms, L = 100 mH. The switch has been open for a long time. Then, at t = 0, the switch is closed. What is the time constant for the current through the inductor? 30% A. R/LB. R/2LC. L/R D. 2L/RE. L/2R

Faraday’s Law Physics 212 Lecture 29, Slide 11 A rectangular wire loop travels to the right with constant velocity, starting in a region of no magnetic field, moving into a region with a constant field pointing into the page (shaded rectangle below), and continuing into a region of no magnetic field. Which plot below best represents the induced current in the loop (I loop ) as it travels from the left through these three regions? Note: a positive I loop corresponds to a counter-clockwise current. Current induced only when flux is changing. Flux is changing because loop is moving. As loop enters field, current will be induced to reduce the increase in flux (Lenz’ law): counterclockwise current generates B field pointing out of the page. When loop is completely inside field, flux is constant, therefore, current is zero As loop leaves field current will be induced to produce B field pointing into the page.

Faraday’s Law Physics 212 Lecture 29, Slide 12 Two fixed conductors are connected by a resistor R = 20 Ohms. The two fixed conductors are separated by L = 2.5 m and lie horizontally. A moving conductor of mass m slides on them at a constant speed, v, producing a current of 3.75 A. A magnetic field with magnitude 5 T points out of the page. In which direction does the current flow through the moving conductor when the bar is sliding in the direction shown? A. to the rightB. to the left L H Flux is changing because area of the loop is changing. As the moving conductor slides downward the area and therefore the flux increases. Therefore a current must flow in the clockwise direction to generate a magnetic field pointing into the page

Faraday’s Law Physics 212 Lecture 29, Slide 13 Two fixed conductors are connected by a resistor R = 20 Ohms. The two fixed conductors are separated by L = 2.5 m and lie horizontally. A moving conductor of mass m slides on them at a constant speed, v, producing a current of 3.75 A. A magnetic field with magnitude 5 T points out of the page. At what speed is the conductor moving? A. 1 m/s B. 3 m/sC. 5 m/sD. 6 m/sE. 9 m/s As the moving conductor slides downward the area of the loop, and therefore the flux, increases. Faraday’s Law:

Phasers Physics 212 Lecture 29, Slide 14 65% Phasor diagram at t = 0  What is V C at t =  /(2  ) (A) (B) (C) (D) VCVC  VRVR VLVL Phasor diagram at t =  /(2  ) Voltage is equal to projection of phasor along vertical axis

Ampere’s Law Integrals Physics 212 Lecture 29, Slide 15. Two infinitely long wires carrying current run into the page as indicated. Consider a closed triangular path that runs from point 1 to point 2 to point 3 and back to point 1 as shown. Which of the following plots best shows B·dl as a function of position along the closed path? The magnetic field points in the azimuthal direction and is oriented clockwise. BB No current is contained within the loop , therefore NOT (A) The magnetic field falls off like 1/r (r is the distance from the wire). From 3 to 1, B·dl is < 0, largest magnitude near wires. From 1 to 2, B·dl is > 0

Physics 212 Lecture 29, Slide 16 62% Flux definition: Faraday’s law:

Physics 212 Lecture 29, Slide 17 65% Current is determined by time rate of change of the flux d  /dt is determined by dB/dt dB/dt (6s) = dB/dt (5 s) = -2 T/s The induced currents at t = 5s and t = 6s are equal

Faraday’s Law Physics 212 Lecture 29, Slide 18 76% Current induced because flux is changing Flux is changing beause B is changing At t = 5 seconds, B is positive, but decreasing Lenz’ law: emf induced to oppose change that brought it into being Induced current must produce positive B field Positive B field produced by counterclockwise current