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**W10D1: Inductance and Magnetic Field Energy**

Today’s Reading Assignment W10D1 Inductance & Magnetic Energy Course Notes: Sections Class 18

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**Announcements Math Review Week 10 Tuesday from 9-11 pm in 32-082**

PS 7 due Week 10 Tuesday at 9 pm in boxes outside or Next Reading Assignment W10D2 DC Circuits & Kirchhoff’s Loop Rules Course Notes: Sections Exam 3 Thursday April :30 pm –9:30 pm Class 22

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**Outline Faraday Law Problem Solving Faraday Law Demonstrations**

Mutual Inductance Self Inductance Energy in Inductors Transformers Class 22

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**Faraday’s Law of Induction**

If C is a stationary closed curve and S is a surface spanning C then The changing magnetic flux through S induces a non-electrostatic electric field whose line integral around C is non-zero Class 22

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**Problem: Calculating Induced Electric Field**

Consider a uniform magnetic field which points into the page and is confined to a circular region with radius R. Suppose the magnitude increases with time, i.e. dB/dt > 0. Find the magnitude and direction of the induced electric field in the regions (i) r < R, and (ii) r > R. (iii) Plot the magnitude of the electric field as a function r. Class 23

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**Faraday’s Law Demonstrations**

Class 23

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**Demonstration: Electric Guitar H32**

Pickups Class 21

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Electric Guitar Class 21

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Demonstration: Aluminum Plate between Pole Faces of a Magnet H Copper Pendulum Between Poles of a Magnet H13 14&show=0 13&show=0 Class 21

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**What happened to kinetic energy of pendulum?**

Eddy Current Braking What happened to kinetic energy of pendulum? Class 21

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Eddy Current Braking The magnet induces currents in the metal that dissipate the energy through Joule heating: Current is induced counter-clockwise (out from center) Force is opposing motion (creates slowing torque) Class 21

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Eddy Current Braking The magnet induces currents in the metal that dissipate the energy through Joule heating: Current is induced clockwise (out from center) Force is opposing motion (creates slowing torque) EMF proportional to angular frequency Class 21

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**Demonstration: 26-152 Levitating Magnet H28 32-082 Levitating Coil on an Aluminum Plate H15**

28&show=0 15&show=0 Class 21

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Mutual Inductance Class 22

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**Demonstration: Two Small Coils and Radio H31**

31&show=0 Class 22

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Mutual Inductance Current I2 in coil 2, induces magnetic flux F12 in coil 1. “Mutual inductance” M12: Change current in coil 2? Induce EMF in coil 1: Class 22

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**Group Problem: Mutual Inductance**

An infinite straight wire carrying current I is placed to the left of a rectangular loop of wire with width w and length l. What is the mutual inductance of the system? Class 23

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Self Inductance Class 23

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**Self Inductance Faraday’s Law**

What if is the effect of putting current into coil 1? There is “self flux”: Faraday’s Law Class 23

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**Calculating Self Inductance**

Unit: Henry Assume a current I is flowing in your device Calculate the B field due to that I Calculate the flux due to that B field Calculate the self inductance (divide out I) Class 23

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**Worked Example: Solenoid**

Calculate the self-inductance L of a solenoid (n turns per meter, length , radius R) Class 23

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Week 09, Day 2 Solenoid Inductance Class 22 22

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**Concept Question: Solenoid**

A very long solenoid consisting of N turns has radius R and length d, (d>>R). Suppose the number of turns is halved keeping all the other parameters fixed. The self inductance remains the same. doubles. is halved. is four times as large. is four times as small. None of the above. Class 23

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**Concept Q. Ans.: Solenoid**

Solution 5. The self-induction of the solenoid is equal to the total flux through the object which is the product of the number of turns time the flux through each turn. The flux through each turn is proportional to the magnitude of magnetic field which is proportional to the number of turns per unit length or hence proportional to the number of turns. Hence the self-induction of the solenoid is proportional to the square of the number of turns. If the number of turns is halved keeping all the other parameters fixed then he self inductance is four times as small. Class 23

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Group Problem: Toroid Calculate the self-inductance L of a toroid with a square cross section with inner radius a, outer radius b = a+h, (height h) and N square windings . REMEMBER Assume a current I is flowing in your device Calculate the B field due to that I Calculate the flux due to that B field Calculate the self inductance (divide out I) Class 23

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Energy in Inductors Class 23

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**Inductor Behavior L I Inductor with constant current does nothing**

Class 23

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Week 09, Day 2 Back EMF I I Class 22 28

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**Demos: 26-152 Back “emf” in Large Inductor H17 32-082 Marconi Coil H12**

17&show=0 12&show=0 Class 22

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**Marconi Coil: On the Titanic**

Week 09, Day 2 Marconi Coil: On the Titanic Another ship Same era Titanic Marconi Telegraph Class 22 30

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**Marconi Coil: Titanic Replica**

Week 09, Day 2 Marconi Coil: Titanic Replica Class 22 31

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**Big L Big dI Small dt The Point: Big EMF Huge EMF Week 09, Day 2**

Class 22 32

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**Energy To “Charge” Inductor**

1. Start with “uncharged” inductor Gradually increase current. Must do work: 3. Integrate up to find total work done: Class 23

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**Energy Stored in Inductor**

But where is energy stored? Class 23

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**Example: Solenoid Ideal solenoid, length l, radius R,**

n turns/length, current I: Volume Energy Density Class 23

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**Energy is stored in the magnetic field**

Energy Density Energy is stored in the magnetic field Magnetic Energy Density Energy is stored in the electric field Electric Energy Density Class 23

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**Worked Example: Energy Stored in Toroid**

Consider a toroid with a square cross section with inner radius a, outer radius b = a+h, (height h) and N square windings with current I. Calculate the energy stored in the magnetic field of the torus. Class 23

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**Solution: Energy Stored in Toroid**

The magnetic field in the torus is given by The stored energy is then The self-inductance is Class 23

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**Group Problem: Coaxial Cable**

Inner wire: r = a Outer wire: r = b How much energy is stored per unit length? What is inductance per unit length? HINTS: This does require an integral The EASIEST way to do (2) is to use (1) Class 23

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**Transformer Step-up transformer Flux F through each turn same:**

Ns > Np: step-up transformer Ns < Np: step-down transformer Class 22

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Demonstrations: One Turn Secondary: Nail H Many Turn Secondary: Jacob’s Ladder H Variable Turns Around a Primary Coil H9 10&show=0 11&show=0 9&show=0 Class 22

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**Concept Question: Residential Transformer**

Week 9, Day 2 Concept Question: Residential Transformer If the transformer in the can looks like the picture, how is it connected? House=Left, Line=Right Line=Left, House=Right I don’t know Class 22 42

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**Answer: Residential Transformer**

Week 9, Day 2 Answer: Residential Transformer Answer: 1. House on left, line on right The house needs a lower voltage, so we step down to the house (fewer turns on house side) Class 22 43

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**Transmission of Electric Power**

Power loss can be greatly reduced if transmitted at high voltage Class 22

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**Electrical Power Power is change in energy per unit time**

So power to move current through circuit elements: Class 14

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Power - Resistor Moving across a resistor in the direction of current decreases your potential. Resistors always dissipate power Class 14

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**Example: Transmission lines**

An average of 120 kW of electric power is sent from a power plant. The transmission lines have a total resistance of 0.40 W. Calculate the power loss if the power is sent at (a) 240 V, and (b) 24,000 V. (a) 83% loss!! (b) 0.0083% loss Class 22

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**Transmission lines We just calculated that I2R is smaller**

for bigger voltages. What about V2/R? Isn’t that bigger? Why doesn’t that matter? Class 22

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