Welcome Back to Physics 1308 Induction and Inductance Heinrich Friedrich Emil Lenz 12 February 1804 10 February 1865
Announcements Assignments for Thursday, November 8th: - Reading: Chapter 33.1 - Watch Videos: - https://youtu.be/4qwglaws_y8 Lecture 21 - The Nature of Light Homework 10 Assigned - due before class on Tuesday, November 13th. Midterm Exam 3 will be in class on Thursday, November 15th. It will cover material corresponding to that covered through chapter 29 in your textbook, and all associated lecture, homework and in class material.
Review Question 1 Consider the situation shown. A triangular, aluminum loop is slowly moving to the right. Eventually, it will enter and pass through the uniform magnetic field region represented by the tails of arrows directed away from you. Initially, there is no current in the loop. When the loop is exiting the magnetic field, what will be the direction of any induced current present in the loop? A) clockwise B) counterclockwise C) no current is induced The induced flux will want to oppose the change hence creating a magnetic pointing out of the page. According to the right hand rule, the flux should point out of the page.
Review Question 2 Consider the situation shown. A triangular, aluminum loop is slowly moving to the right. Eventually, it will enter and pass through the uniform magnetic field region represented by the tails of arrows directed away from you. Initially, there is no current in the loop. When the loop is exiting the magnetic field, what will be the direction of any induced current present in the loop? A) clockwise B) counterclockwise C) no current is induced The induced flux will want to oppose the change hence maintaining the magnetic pointing into the page. According to the right hand rule, the flux should point into the page.
Key Concepts RC Circuits: If a constant emf E is introduced into a single-loop circuit containing a resistance R and an inductance L, the current rises to an equilibrium value of E/R according to When the source of constant emf is removed and replaced by a conductor, the current decays from a value i0 according to
Question 1 When the switch is closed the current through the circuit exponentially approaches a value I = E/R. If we repeat this experiment with an inductor having twice as the number of turns per unit length, the time it takes for the current to reach a value of I/2 A) increases B) decreases C) stays the same The inductance of an inductor is defined as L = N i So an inductor with more turns will have a larger inductance. We also know that current and the time constant are related mathematically. i = E R (1 e t/ L ), L = L R From here you can see as L increases, the time constant increases. Thus, it takes longer to reach a value of I/2.
Question 2 A circuit contains a battery, a switch, an inductor, and a resistor connected in series. Initially, the switch is open. In which one of the following intervals does the energy stored in the inductor have the largest value? A)before the switch is closed B) immediately after the switch is closed when the current in the circuit is increasing C) a long time after the switch is closed The largest value for energy will be when the U = 1 2 Li2 most current runs through the inductor when it acts like an ordinary connecting wire.
Instructor Problem: Solenoid w/ Inductance A solenoid having an inductance of 6.30 µh is connected in series with a 1.2 kω resistor. A) If a 14.0 V battery is connected across the pair, how long will it take for the current through the resistor to reach 80.0% of it final value? B) What is the current through the resistor at time t = 1.0τL?
Given: L =6.30 µh E = 14 V R =1.2 k Part A: If the battery is connected at t = 0, at a later time the current is given by the equation i = E R (1 e t/ L ) where τl = L/R. We want to find the point when i reaches 80.0% of its final value: i =.800 E R Putting this together gives.800 E R = E R (1 e t/ L ).800 = 1 e t/ L e t/ L =0.200 (t/ L )=ln(0.200) t = 6.30 10 6 H 1.2 10 3 ln (0.200) t = (ln (0.200)) t = L R ln (0.200) t =8.45 10 9 s
Part B: At t = 1.0τL the current in the circuit is i = E R (1 e t/ L ) = 14 V 1.2 10 3 (1 e 1.0 ) i =7.37 10 3 A Note: at t = 0, the current in the circuit is zero. However, after a very long time, the inductor acts like an ordinary connecting wire, so the current is i = E R = 14 V 1.20 10 3 =0.0117 A
Student Problem: Series RL Circuit A battery is connected to a series RL circuit at time t = 0. At what multipole of τl will the current be 0.100% less than its equilibrium value?
The inductor in the RL circuit initially acts to oppose changes in current through it. So, starting at t = 0 (the moment the switch closes) the current increases according to the expression i = E R (1 e t/ L ) where τl = L/R. We are asked for the current to be 0.100% less than its equilibrium rate. That translates to a value of i =0.9990 E R Putting this together, we have 0.9990 E R = E R (1 e t/ L ) t/ L =ln(0.0010) 0.9990 = 1 e t/ L t/ L =6.91 0.0010 = e t/ L
Student Problem: RL Circuit The figure below shows three circuits with identical batteries, indicators and resistors. Rank the circuits according to the current through the battery just after the switch is closed. 2 > 3 > 1 (zero)
Student Problem: RL Circuit The figure below shows three circuits with identical batteries, indicators and resistors. Rank the circuits according to the current through the battery a long time after the switch is closed. 2 > 3 > 1
The End for Today!