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What are we doing Today? Ch. 5  Magnetism Homework Questions Check Hand in Lab 41-42 Chapter 5 Notes In class questions & homework.

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Presentation on theme: "What are we doing Today? Ch. 5  Magnetism Homework Questions Check Hand in Lab 41-42 Chapter 5 Notes In class questions & homework."— Presentation transcript:

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2 What are we doing Today? Ch. 5  Magnetism Homework Questions Check Hand in Lab 41-42 Chapter 5 Notes In class questions & homework

3 p. 174 Question 19 (b,c) textbook

4 p. 174 Question 20 (a,b)

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6 Magnetism (Chapter 5 - p.163) Magnet – An object that can attract other objects containing iron, cobalt or nickel

7 Magnetic Poles The strength of a magnet is concentrated at the ends, called north and south “poles” of the magnet. NS N E W S N Compass Bar magnet S N Iron filings

8 If you cut a magnet in half, you get 2 magnets!

9 Magnetism (Chapter 5 - p.163) Magnetic Field – Is the area of space in which the magnetic force of a magnet can act on another magnet.

10 Magnetic Fields * Magnetic field lines move away from north poles and toward south poles.

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12 Magnetic Field Lines NS We can describe magnetic field lines by imagining a tiny compass placed at nearby points. The direction of the magnetic field B at any point is the same as the direction indicated by this compass.

13 Field Lines Between Magnets NS NN Unlike poles Like poles Leave N and enter S Attraction Repulsion

14 Ferromagnetic Substance (p.167) Ferromagnetic materials are materials that have magnetic properties similar to those of iron. They can become permanently magnetized. Examples of ferromagnetic materials are nickel and cobalt.

15 ELECTROMAGNET Electromagnetism a moving charge (electricity) produces a magnetic field More coils of wire= more current = stronger magnet Bigger battery = stronger magnet A magnet with a field produced by an electric current

16 Magnetic field pattern around a straight wire. The resulting magnetic field lines form concentric circles around the wire. The Right-Hand rule can be used to predict the direction of the magnetic field

17 Magnetic field around a wire carrying current

18 The Electromagnet By the Right Hand Rule, a coil of wire with current flowing in it will create a magnetic field The strength of the magnetic field depends on The amount of current in a wire – More current means stronger magnetic field The number of turns in the coil – More turns means stronger magnetic field The material in the coil – Magnetic materials like iron and steel make the magnetic field stronger

19 The Electromagnet A simple electromagnet can be made by coiling some wire around a steel nail, and connecting a battery to it. As current rotates around the nail, a magnetic field is created with the North pole at the bottom and the South pole at the top

20 The Electromagnet Electromagnets are found in many everyday objects such as – Motors and generators – Doorbells – Speakers – Hard drives – VHS and Audio cassette tapes – Telephones

21 Solenoid (p.169) EST Solenoid: Is a cylindrical coil of live wire.

22 Magnetic Field of a Solenoid(EST) Look at p.169 in your textbook

23 Classwork P. 174  Questions 21-25

24 Exit Question(s) Draw the following circuits and which of the circuits below are connected in series, and which are connected in parallel?

25 Number 1 Answer: Parallel

26 Number 2 Answer: Series

27 Number 3 Answer: Parallel

28 Number 4 Answer: Series

29 The circuit is composed of a 6.0-V battery connected to a 10-Ω resistor. It includes an ammeter and a voltmeter connected to the resistor terminals.(look up in chapter 5 on diagrams of circuits). a)Draw the circuit diagram, including the measuring instruments.

30 Exit Questions 1) The potential difference in a circuit is determined to be 10 volts. If the resistance is 5 ohms, what must the current intensity be?

31 The potential difference in a circuit is determined to be 10 volts. If the resistance is 5 ohms, what must the current intensity be? Answer: 1.Determine what you are looking for. 2.List what you know and don’t know. V = 10 Volts R = 5 ohms I = ? 3. Use equation V = R x I  Solve for I, so I = V/R 4. I = 10 Volts / 5 ohms = 2 A Answer: The current intensity will be 2 A (Amps)

32 What To Do Now?


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