What are we doing Today?Ch. 5 Magnetism

• Homework Questions Check • Hand in Lab 41-42• Chapter 5 Notes• In class questions & homework

p. 174 Question 19 (b,c)textbook

p. 174 Question 20 (a,b)

Magnetism (Chapter 5 - p.163)

Magnet – An object that can attract other objects containing iron, cobalt or nickel

Magnetic PolesThe The strengthstrength of a magnet is of a magnet is concentrated at the ends, concentrated at the ends, called north and south called north and south ““polespoles” of the magnet.” of the magnet.

NNSS

N

E

W

SNN

CompassCompassBar magnetBar magnet

S

N

Iron filings

If you cut a magnet in If you cut a magnet in half,half,

you get 2 magnets!you get 2 magnets!

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.

Magnetic Fields

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

Magnetic Field Lines

N S

We can describe We can describe magnetic field linesmagnetic field lines by imagining a tiny by imagining a tiny compass placed at compass placed at nearby points.nearby points.

The The directiondirection of the of the magnetic field magnetic field BB at at any point is the same any point is the same as the direction as the direction indicated by this indicated by this compass. compass.

Field Lines Between Magnets

N S

N N

Unlike poles

Like poles

Leave N and enter S

Attraction

Repulsion

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.

ELECTROMAGNET

Electromagnetisma 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

• 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

Magnetic field around a wire carrying current

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 onThe amount of current in a wire – More current

means stronger magnetic fieldThe number of turns in the coil – More turns

means stronger magnetic fieldThe material in the coil – Magnetic materials like

iron and steel make the magnetic field stronger

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

The Electromagnet

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

Solenoid (p.169) EST

• Solenoid: Is a cylindrical coil of live wire.

Magnetic Field of a Solenoid(EST)

Look at p.169 in your textbook

Classwork

• P. 174 Questions 21-25

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

Number 1

Answer: Parallel

Number 2

Answer: Series

Number 3

Answer: Parallel

Number 4

Answer: Series

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.

Exit Questions1) The potential difference in a circuit is determined to be 10 volts. If the resistance is 5 ohms, what must the current intensity be?

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/R4. I = 10 Volts / 5 ohms = 2 AAnswer: The current intensity will be 2 A (Amps)

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