Intro Lesson

Magnetism Chapter 19

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Chapter 19

Section 1 Magnets and Magnetic Fields

Magnets •  Magnets attract iron-containing objects. •  Magnets have two distinct poles called the north pole and the south pole. These names are derived from a magnet’s behavior on Earth. •  Like poles of magnets repel each other; unlike poles attract each other.

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Properties of Magnets   Magnets have two opposite poles. — north — south   Magnets exert forces on each other.   The forces depend on the alignment of the poles.

Magnets and Magnetic Fields Magnets have two ends – poles – called north and south. Like poles repel; unlike poles attract.

Magnets and Magnetic Fields However, if you cut a magnet in half, you don’t get a north pole and a south pole – you get two smaller magnets.

Chapter 19 Magnetic Fields •  A magnetic field is a region in which a magnetic force can be detected. •  Magnetic field lines can be drawn with the aid of a compass.

Magnets and Magnetic Fields Magnetic fields can be visualized using magnetic field lines, which are always closed loops.

Chapter 19

Section 1 Magnets and Magnetic Fields

The Direction of a Magnetic Field Field lines are conventionally drawn FROM the north pole TO the south pole.

The direction of a magnetic field at any location is the direction that the north pole of a compass at that location would point. (The arrow head represents the north pole of a magnet and points to the nearest south pole). Chapter menu

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Properties of magnets   Materials that make good permanent magnets are called hard magnets.   Steel, which contains iron and carbon, is a common and inexpensive material used to create hard magnets.   Materials that lose their magnetism quickly are called soft magnets.

The Magnetic Field of the Earth How do we use Earth’s magnetic field to tell direction?

The Magnetic Field of the Earth   As early as 500 B.C. people discovered that some naturally occurring materials— such as lodestone and magnetite—have magnetic properties.   By 1200, explorers from Italy were using a compass to guide ocean voyages beyond the sight of land.

The Magnetic Field of the Earth   The Earth’s magnetic poles are defined by the planet’s magnetic field.   That means the south magnetic pole of the planet is near the north geographic pole.   When you use a compass, the north-pointing end of the needle points toward a spot near (but not exactly at) the Earth’s geographic north pole.

Chapter 19

Section 1 Magnets and Magnetic Fields

Earth’s Magnetic Field

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Chapter 19

Section 1 Magnets and Magnetic Fields

Magnetic Fields, continued •  Earth’s magnetic field is similar to that of a bar magnet. •  The magnetic south pole is near the Geographic North Pole. The magnetic north pole is near the Geographic South Pole. •  Magnetic declination is a measure of the difference between true north and north indicated by a compass. Chapter menu

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The magnetic field produced by certain arrangements of bar magnets are represented in the diagrams shown below

MAGNETS AND MAGNETIC FIELDS   The magnetic field lines drawn to represent the magnetic field

produced by certain arrangements of bar magnets are represented in the diagrams shown below

Notice the similarity between the lines of this slide and the previous slide particle distribution

The Why & How of Magnets   The sources of nearly all magnetic effects in matter are the electrons in atoms.   There are two ways in which electrons create magnetism: 1.  Electrons’ motion around the nucleus makes the entire atom a small magnet. 2.  Electrons themselves act as though they were magnets.

The Why & How of Magnets

  All atoms have electrons, so you might think that all materials should be magnetic, but there is great variability in the magnetic properties of materials.   The electrons in some atoms align to cancel out one another’s magnetic influence.   While all materials show some kind of magnetic effect, the magnetism in most materials is too weak to detect without highly sensitive instruments.

Iron, Cobalt, Nickel

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Magnesium

Note: Magnetism is a property not just of the chemical make-up of a material, but of its crystalline structure and microscopic organization.

The Why & How of Magnets   In diamagnetic materials, the electrons are oriented so their individual magnetic fields cancel each other out within an atom.   Individual atoms in paramagnetic materials are magnetic but the atoms themselves are randomly arranged so the overall magnetism of a sample is zero.

But when paramagnetic materials are placed in a magnetic field, the atoms align so that the material is weakly magnetic.

The Why & How of Magnets   A small group of metals have very strong magnetic properties, including iron, nickel, and cobalt.   These metals are the best known examples of ferromagnetic materials.   Atoms with similar magnetic orientations line up with neighboring atoms in groups called magnetic domains.

Induced Magnetism "   Metals that are ferromagnetic have atoms

that behave like mini magnets, but on a grand scale they cancel out. When placed in external magnetic field, these can be lined up and the metal then becomes magnetized.

Magnetizing Iron Magnetic domains in iron nails are induced to align by proximity of the strong magnet Each nail becomes itself a magnet, which in turn magnetizes the nail below it, forming a chain. When the strong magnet is removed, most of the domains unalign and the nails lose most of their magnetization.

Demagnetizing Iron Magnetic domains can be scrambled by heating the iron, striking it with great force, or other disruptions of alignment. Magnetized Test tube of un-magnetized iron filings

Demagnetized

SHAKE

S

S

N

Magnetizer

N

Demagnetizing Iron Iron nail is attracted to the large magnet due to alignment of domains in the nail. Heat the nail to a high temperature and the domains become randomized so the nail is no longer attracted to the magnet.

The Why & How of Magnets   Magnetic domains in a ferromagnetic material will always orient themselves to attract a permanent magnet. —  If a north pole approaches, domains grow that have south poles facing out. —  If a south pole approaches, domains grow that have north poles facing out.

Chapter 19

Section 1 Magnets and Magnetic Fields

Magnetic Domains •  Magnetic Domain A region composed of a group of atoms whose magnetic fields are aligned in the same direction is called a magnetic domain. •  Some materials can be made into permanent magnets. –  Soft magnetic materials (for example iron) are easily magnetized but tend to lose their magnetism easily. –  Hard magnetic materials (for example nickel) tend to retain their magnetism. Chapter menu

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Names of some magnetic materials Examples of ferromagnetic materials: iron, cobalt, nickel Examples of paramagnetic materials: aluminum, manganese, chromium, oxygen, sodium, lithium Examples of diamagnetic materials: water, pyrolitic graphite, copper, gold, bismuth

Diamagnetism •  Diamagnetism is the property of an object or material which causes it to create a magnetic field in opposition to an externally applied magnetic field.

Diamagnetism •  Diamagnetism is the property of an object or material which causes it to create a magnetic field in opposition to an externally applied magnetic field. •  Diamagnetism is believed to be due to quantum mechanics (and is understood in terms of Landau levels) and occurs because the external field alters the orbital velocity of electrons around their nuclei, thus changing the magnetic dipole moment. According to Lenz’s law, the field of these electrons will oppose the magnetic field changes provided by the applied field.

Magnetism: Lesson 2

On July 21, 1820, Hans Oersted, professor of physics at Copenhagen, delivered a lecture to students. By chance, a wire leading to a voltaic pile was nearly parallel to & above a compass that happened to be on the table. When the circuit was closed, the needle swung around almost perpendicular to the current carrying wire, as if gripped by a powerful magnet.

Andre Marie Ampere heard the news & immediately performed some experiments of his own. Within two weeks, Ampere showed that a current-carrying wire always induces a magnetic force that acts at right angles to the current.

Chapter 19

Section 2 Magnetism from Electricity

Magnetic Field of a Current-Carrying Wire

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Chapter 19

Section 2 Magnetism from Electricity

Magnetic Field of a Current-Carrying Wire

Negative Current

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Direction of B

42

Magnetic Field Produced by a Current-Carrying Wire

Chapter 19

Section 2 Magnetism from Electricity

The Right-Hand Rule •  Right Hand Rule

–  Grasp the wire in your right hand

–  Point your thumb in the direction of the current

–  Your fingers will curl in the direction of the field

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Chapter 19

Section 2 Magnetism from Electricity

Magnetic Field of a Current-Carrying Wire

Negative Current

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Chapter 19

Section 2 Magnetism from Electricity

The Right-Hand Rule •  Right Hand Rule

–  Grasp the wire in your right hand

–  Point your thumb in the direction of the current

–  Your fingers will curl in the direction of the field

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Chapter 19

Section 2 Magnetism from Electricity

Magnetic Field of a Current Loop •  Solenoids produce a strong magnetic field by combining several loops. •  A solenoid is a long, helically wound coil of insulated wire.

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"  So we have learned that moving

charges (currents) exert a force on a compass needle (a magnet).

~moving charges exert force on magnets~

"  It follows, from Newton’s Third Law, that magnets (or a magnetic field) ought to exert forces on currents (moving charged particles).

~magnetic fields exert force on moving charges~

Chapter 19

Section 3 Magnetic Force

Charged Particles in a Magnetic Field •  A charge moving through a magnetic field experiences a force proportional to the charge, velocity, and the magnetic field.

F = qvB or

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Defining Magnetic Field, B Fe E= q "  Recall that the electric field, E, is defined as electric force per unit charge exerted on a moving charged particle. € "  Similarly, B used to be defined as the FB magnetic force per unit pole B= qv "  NOW we define B in terms of the force exerted on a moving charged particle.

€

Chapter 19

Section 3 Magnetic Force

Charged Particles in a Magnetic Field, continued •  The direction of the magnetic force on a moving charge is always perpendicular to both the magnetic field and the velocity of the charge. •  An alternative right-hand rule can be used to find the direction of the magnetic force.

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Chapter 19

Section 3 Magnetic Force

Right-Hand Palm Rule: Force on a Moving Charge (“Alternative” Right Hand Rule)

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CONVENTIONS Certain CONVENTIONS have been adopted in order to represent the direction of the magnetic field and the current in a wire. A magnetic field directed into the paper is represented by a group of x's, while a magnetic field out of the paper is represented by a group of dots.

Magnetic field line in plane of paper

In which direction will the magnetic force point?

. v

-e

v

-e

B

B

Note: Force is in opposite direction if charge is negative (instead of conventional positive charge).

.

F

v

F

-e

v

-e

B

B

Note: Force is in opposite direction if charge is negative (instead of conventional positive charge).

Determine the direction of the magnetic force in the following diagrams:

-

Determine the direction of the magnetic force in the following diagrams: Force into page

Force up page

No force, since v is not perpendicular to B

(a proton)

B

B

(a proton)

Force up page Force into page Center-seeking (centripetal) Force

B

B

Force on Electric Charge Moving in a Magnetic Field If a charged particle is moving perpendicular to a uniform magnetic field, its path will be a circle.

Units of Magnetic Field  

 

The SI unit of magnetic field is the Tesla (T) €

Another (non-SI) unit is a Gauss (G) 4   1 T = 10 G

FB B= qv

A Few Typical B Values  

Conventional laboratory magnets  

 

Superconducting magnets  

 

2.5 T 30 T

Earth’s magnetic field  

5 x 10-5 T

For each of the following 3 particles, state whether the particle’s charge is positive, negative, or zero.

For each of the following 3 particles, state whether the particle’s charge is positive, negative, or zero.

Solution Particle 3 follows the path given by the righthand rule. So #3 is positive. Particle 1 is opposite that, so #1 is negative. Particle 2 is un-deflected so there must be no force on it, which means the charge is 0.

Centripetal Force •  Note that the force on a moving charged particle through a uniform magnetic field is a centripetal force. Remember, centripetal force causes a change in direction, not a change in speed (so magnitude of v stays the same).

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Lesson 3 (will cover this in the future).

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Chapter 19

Section 3 Magnetic Force

Example Problem #1 Particle in a Magnetic Field

A proton moving east experiences a force of 8.8 × 10–19 N upward due to the Earth’s magnetic field. At this location, the field has a magnitude of 5.5 × 10–5 T to the north. Find the speed of the particle.

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Chapter 19

Section 3 Magnetic Force

Sample Problem, continued Particle in a Magnetic Field Given: q = 1.60 × 10–19 C B = 5.5 × 10–5 T Fmagnetic = 8.8 × 10–19 N Unknown: v=?

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Chapter 19

Section 3 Magnetic Force

Sample Problem, continued Particle in a Magnetic Field Use the definition of magnetic field strength. Rearrange to solve for v.

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Example Problem #2 A beam of electrons travels to the right at 3.0 x 106 m/s

through a uniform magnetic field of 4.0 x 10-2 T at right angles to the field (coming out of the page). What is the force acting on each electron (magnitude & direction)?

Example Problem #2 A beam of electrons travels at 3.0 x 106 m/s through a uniform magnetic field of 4.0 x 10-2 T at right angles to the field. What is the force acting on each electron (magnitude & direction)?

Example Problem #2 A beam of electrons travels at 3.0 x 106 m/s through a uniform magnetic field of 4.0 x 10-2 T at right angles to the field. What is the force acting on each electron (magnitude & direction)?

θ = angle between v and B

Force on a Current-Carrying Wire If conducting charges move through a wire with an average speed v, the time required for them to move from one end of the wire to the other (length, l) is   Δt = l /v (since velocity = distance/Δt = l/Δt).   The amount of charge that flows through the wire in this time is q = IΔt = Il /v. Therefore, the force exerted on the wire is

  F = q vB = (Il /v) vB   so

F = Il B

Chapter 19

Section 3 Magnetic Force

Magnetic Force on a Current-Carrying Wire •  A current-carrying wire in an external magnetic field undergoes a magnetic force. •  The force on a current-carrying conductor perpendicular to a magnetic field is given by:

magnitude of magnetic force = (magnitude of magnetic field) × (current) × (length of conductor within B)

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Force on an Electric Current in a Magnetic Field; Definition of B The force on the wire depends on the current, the length of the wire, the magnetic field, and its orientation.

This equation defines the magnetic field B.

Chapter 19

Section 3 Magnetic Force

Force on a Current-Carrying Wire in a Magnetic Field

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Magnetic Force Moving charges in an electric current experience a force due to magnetic field.

N

N

Physics 1 (Garcia) SJSU

Magnetic Force on Charges Moving negative electric charges deflected by magnetic fields. Force is up on negative charges.

Chapter 19a CONVENTIONAL POSITIVE Force on

Current-Carrying Wire in a Magnetic Field

Force on an Electric Current in a Magnetic Field; Definition of B The force on the wire depends on the current, the length of the wire, the magnetic field, and its orientation.

This equation defines the magnetic field B.

Chapter 19

Section 3 Magnetic Force

Example Problem 3 Force on a Current-Carrying Conductor A wire 36 m long carries a current of 22 A from east to west. If the magnetic force on the wire due to Earth’s magnetic field is downward (toward Earth) and has a magnitude of 4.0 × 10–2 N, find the magnitude and direction of the magnetic field at this location.

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Chapter 19

Section 3 Magnetic Force

Example Problem, continued Force on a Current-Carrying Conductor Given: I = 22 A Fmagnetic = 4.0 × 10–2 N Unknown: B=?

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Chapter 19

Section 3 Magnetic Force

Example Problem, continued Force on a Current-Carrying Conductor Use the equation for the force on a current-carrying conductor perpendicular to a magnetic field. Rearrange to solve for B.

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Chapter 19

Section 3 Magnetic Force

Example Problem, continued Force on a Current-Carrying Conductor Using the right-hand rule to find the direction of B, face north with your thumb pointing to the west (in the direction of the current) and the palm of your hand down (in the direction of the force). Your fingers point north. Thus, Earth’s magnetic field is from south to north.

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Example #4: Magnetic Force on a current-carrying wire.

A wire carrying a 30-A current has a length  of 12 cm between the pole faces of a magnet at an angle θ = 60°, as shown. The magnetic field is approximately uniform at 0.90 T. We ignore the field beyond the pole pieces. What is the force on the wire?

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= 30 A (0.12 m) (0.90 T) (0.866) = 2.8 N Direction: into the page

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Chapter 19

Section 3 Magnetic Force

Force Between Parallel Conducting Wires

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Force Between Two Wires I I2

I1

d 92

Force Between Two Wires II I1

I2

93

Current same direction—force attractive.

Current opposite direction —force repulsive.

94

Chapter 19

Section 3 Magnetic Force

Magnetic Force on a Current-Carrying Conductor, continued •  Two parallel current-carrying wires exert a force on one another that are equal in magnitude and opposite in direction. •  If the currents are in the same direction, the two wires attract one another. •  If the currents are in opposite direction, the wires repel one another.

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Force between Two Parallel Wires Parallel currents attract; antiparallel currents repel.

Chapter 19

Standardized Test Prep

Multiple Choice •  Wire 1 carries current I1 and creates magnetic field B1. •  Wire 2 carries current I2 and creates magnetic field B2. Q. What is the direction of the magnetic field B1 at the location of wire 2? F. to the left G. to the right H. into the page J. out of the page

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Chapter 19

Standardized Test Prep

Multiple Choice, continued •  Wire 1 carries current I1 and creates magnetic field B1. •  Wire 2 carries current I2 and creates magnetic field B2. Q. What is the direction of the magnetic field B1 at the location of wire 2? F. to the left G. to the right H. into the page J. out of the page

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Chapter 19

Standardized Test Prep

Multiple Choice, continued •  Wire 1 carries current I1 and creates magnetic field B1. •  Wire 2 carries current I2 and creates magnetic field B2. Q2. What is the direction of the force on wire 2 as a result of B1? A. to the left B. to the right C. into the page D. out of the page

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Chapter 19

Standardized Test Prep

Multiple Choice, continued •  Wire 1 carries current I1 and creates magnetic field B1. •  Wire 2 carries current I2 and creates magnetic field B2. Q2. What is the direction of the force on wire 2 as a result of B1? A. to the left B. to the right C. into the page D. out of the page

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Chapter 19

Standardized Test Prep

Multiple Choice, continued •  Wire 1 carries current I1 and creates magnetic field B1. •  Wire 2 carries current I2 and creates magnetic field B2. Q3. What is the magnitude of the magnetic force on wire 2?

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Chapter 19

Standardized Test Prep

Multiple Choice, continued

•  Wire 1 carries current I1 and creates magnetic field B1. •  Wire 2 carries current I2 and creates magnetic field B2. Q3. What is the magnitude of the magnetic force on wire 2?

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Example 5: Measuring a magnetic field. A rectangular loop of wire hangs vertically as shown. A magnetic field B is directed horizontally, perpendicular to the wire, and points out of the page at all points. The magnetic field is very nearly uniform along the horizontal portion of wire ab (length l = 10.0 cm) which is near the center of the gap of a large magnet producing the field. The top portion of the wire loop is free of the field. The loop hangs from a balance which measures a downward magnetic force (in addition to the gravitational force) of F = 3.48 x 10-2 N when the wire carries a current I = 0.245 A. What is the magnitude of the magnetic field B? Copyright © 2009 Pearson Education, Inc.

Solution The magnetic force on the left vertical section of wire points to the left; the force on the vertical section on the right points to the right. These two forces are equal and opposite, so they cancel.

Copyright © 2009 Pearson Education, Inc.

Solution The magnetic force on the left vertical section of wire points to the left; the force on the vertical section on the right points to the right. These two forces are equal and opposite, so they cancel.

−2

F 3.48 × 10 B= = = 1.42 T Il .245(.100)

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Chapter 19

Section 2 Magnetism from Electricity

Magnetic Field Due to a Long Straight Wire •  Right Hand Rule

–  Grasp the wire in your right hand

–  Point your thumb in the direction of the current

–  Your fingers will curl in the direction of the field

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Magnetic Field Due to a Long Straight Wire The field is inversely proportional to the distance from the wire:

The constant µ0 is called the permeability of free space, and has the value:

Magnetic Field Due to a Straight Wire Experimentally determined equation.

Current in wire

µ I 0 B= 2π

r

Perpendicular distance from wire to point at which B is to be determined.

The value of the constant µ0, which is called the permeability of free space, is µ0/2π = 2 x 10-7 T m/A exactly. 108

Example 6 Calculation of B near a wire: A vertical electric wire in the wall of a building carries a dc current of 25 A upward. What is the magnetic field at a point 10 cm due north of this wire?

109

r

110

. r

111

Using the formula for the magnetic field near a straight wire

112

Using the formula for the magnetic field near a straight wire

So we can obtain the magnetic field at 10cm away as −7

(4 π × 10 T⋅ m / A)(25A) = (2π )(0.1m)

113

€

Using the formula for the magnetic field near a straight wire

So we can obtain the magnetic field at 10cm away as −7

(4 π × 10 T⋅ m / A)(25A) = (2π )(0.1m)

= 5.0 x 10-5 T €

114

Force Between Two Parallel Wires

115

Force between Two Parallel Wires The magnetic field produced at the position of wire 2 due to the current in wire 1 is:

The force this field exerts on a length l2 of wire 2 is:

Force Between Two Parallel Wires Force per unit length of wire

F µ0 I1I2 = 2π

d l (d or r can be used).

Distance between wires 117

d

118

Current same direction—force attractive.

Current opposite direction —force repulsive.

119

Example 7 Force between two current-carrying wires: The two wires of a 2.0-m-long appliance cord are 3.0 mm apart and carrying a current of 8.0 Amps (dc) in opposite directions. Calculate the force between the wires.

120

121

−7

2

(2.0 × 10 T⋅ m / A)(8.0A) (2.0m) F= −3 3.0 × 10 m where

µ0 −7 = 2.0 × 10 Tm/A 2π

122

−7

2

(2.0 × 10 T⋅ m / A)(8.0A) (2.0m) F= −3 3.0 × 10 m

= 8.5 x 10-3 N

123

Example 8 Suspending a current with a current: A horizontal wire carries a current I1 = 80 Amps, dc. A second parallel wire 20 cm below it must carry how much current, I2, so that it doesn’t fall due to gravity? The lower wire has a mass of 0.12 g per meter of length.

124

Magnetic field due the current I1

125

This downward force must be balanced by the magnetic force exerted on the wire by the first wire.

126

This downward force must be balanced by the magnetic force exerted on the wire by the first wire.

Fg mg FB µ0 I1I2 = = = l l l 2πd

€

127

This downward force must be balanced by the magnetic force exerted on the wire by the first wire.

Fg mg FB µ0 I1I2 = = = l l l 2πd Solving for I2…

128

This downward force must be balanced by the magnetic force exerted on the wire by the first wire.

Fg mg FB µ0 I1I2 = = = l l l 2πd Solving for I2…

mg2πd I2 = lµ0 I1 129

€

This downward force must be balanced by the magnetic force exerted on the wire by the first wire.

Fg mg FB µ0 I1I2 = = = l l l 2πd Solving for I2…

mg2πd I2 = lµ0 I1 −3

2

(0.12 × 10 kg /m)(9.8m /s )2π (0.20m) = −7 (4 π × 10 T⋅ m / A)(80A) 130

€

This downward force must be balanced by the magnetic force exerted on the wire by the first wire.

µ0 I1I2 l2 Fg = mg = FB = 2πd Solving for I2…

mg2πd I2 = l2 µ0 I1 −3

2

(0.12 × 10 kg /m)(9.8m /s )2π (0.20m) = −7 (4 π × 10 T⋅ m / A)(80A) = 15 A €

131