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General Physics (PHY 2140)

Lecture 8Lecture 8Electricity and Magnetism

1. MagnetismApplication of magnetic forcesAmpere’s law

2. Induced voltages and inductionMagnetic flux

Chapter 19-20

http://www.physics.wayne.edu/~alan/2140Website/Main.htm

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Lightning ReviewLightning Review

Last lecture:1.1. MagnetismMagnetism

Magnetic fieldMagnetic fieldMagnetic force on a moving particleMagnetic force on a moving particleMagnetic force on a currentMagnetic force on a currentTorque on a current loopTorque on a current loopMotion in a uniform fieldMotion in a uniform field

Review Problem:

How does the aurora borealis (the Northern Lights) work?

sinF qvB θ=sinF BIl θ=

sinNBIAτ θ=/r mv qB=

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Magnetic Field of the EarthMagnetic Field of the Earth

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The Aurora compared to a CRTThe Aurora compared to a CRT

For more info see the aurora home page: http://sec.noaa.gov/pmap/

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19.6 Motion of Charged Particle in magnetic field19.6 Motion of Charged Particle in magnetic field

Consider positively charge Consider positively charge particle moving in a uniform particle moving in a uniform magnetic field.magnetic field.Suppose the initial velocity of Suppose the initial velocity of the particle is perpendicular to the particle is perpendicular to the direction of the field.the direction of the field.Then a magnetic force will be Then a magnetic force will be exerted on the particle and exerted on the particle and make follow a circular path.make follow a circular path.

× × × × × × ×

× × × × × × ×

× × × × × × ×

× × × × × × ×

Fv

qr

Bin

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The magnetic force produces a centripetal acceleration.2mvF qvB

r= =

mvrqB

=

The particle travels on a circular trajectory with a radius:

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Example: Proton moving in uniform magnetic fieldExample: Proton moving in uniform magnetic field

A proton is moving in a circular orbit of radius 14 cm in a unifA proton is moving in a circular orbit of radius 14 cm in a uniform orm magnetic field of magnitude 0.35 T, directed perpendicular to thmagnetic field of magnitude 0.35 T, directed perpendicular to the e velocity of the proton. Find the orbital speed of the proton.velocity of the proton. Find the orbital speed of the proton.

r = 0.14 mB = 0.35 Tm = 1.67x10-27 kgq = 1.6 x 10-19 C

mvrqB

=( )( )( )

( )19 2

27

6

1.6 10 0.35 14 10

1.67 10

4.7 10 ms

qBrvm

C T m

kg

− −

−

=

× ×=

×

= ×

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Application: Mass SpectrometerApplication: Mass Spectrometer

mvrqB

=

See prob. 30 in text

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Review Problem 2Review Problem 2How does your credit card work?

The stripe on the back of a credit card is a magnetic stripe, often called a magstripe. The magstripe is made up of tiny iron-based magnetic particles in a plastic-like film. Each particle is really a tiny bar magnet about 20-millionths of an inch long.

The magstripe can be "written" because the tiny bar magnets can be magnetized in either a north or south pole direction. The magstripe on the back of the card is very similar to a piece of cassette tape .A magstripe reader (you may have seen one hooked to someone's PC at a bazaar or fair) can understand the information on the three-track stripe.

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Review Example 1: Flying duckReview Example 1: Flying duck

8

13

5

sin(4.0 10 )(15 / )(5.0 102.6 1

)sin 600

F qvBx C m s x T

x N

θ− −

−

=

=

=

A duck flying horizontally due north at 15 m/s passes over Atlanta, where the magnetic field of the Earth is 5.0×10-5 T in a direction 60° below a horizontal line running north and south. The duck has a positive charge of 4.0×10-8C. What is the magnetic force acting on the duck?

B=5.0 x 10-5 T. q = 4.0×10-8Cv = 15 m/sθ = 60°F = qvBsinθ

60°

N

Ground

B

v

- to the west (into page)

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Review Example 2: Wire in EarthReview Example 2: Wire in Earth’’s B Fields B Field

A wire carries a current of 22 A from east to west. Assume that A wire carries a current of 22 A from east to west. Assume that at this location at this location the magnetic field of the earth is horizontal and directed from the magnetic field of the earth is horizontal and directed from south to north, south to north, and has a magnitude of 0.50 x 10and has a magnitude of 0.50 x 10--44 T. Find the magnetic force on a 36T. Find the magnetic force on a 36--m length m length of wire. What happens if the direction of the current is reverseof wire. What happens if the direction of the current is reversed?d?

B=0.50 x 10-4 T. I = 22 Al = 36 mFmax = BIl

( )( )( )max

4

2

0.50 10 22 36

4.0 10

F BIl

T A m

N−

−

= ×

=

= ×

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Application: The Velocity SelectorApplication: The Velocity Selector

electric

magnetic

F qEF qvB

==

Magnetic force is up…

But the electric force is down.

Since there is no deflection we can set these equal to each other. So we find:

thus we have/

q

v

E qvB

E B

=

=

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Example 2: Example 2:

Consider the velocity selector. The electric field between the plates of the velocity selector is 950 V/m, and the magnetic field in the velocity selector has a magnitude of 0.930 T directed at right angles to the electric field. Calculate the speed of an ion that passes undeflected through the velocity selector.

E = 950 V/mB = 0.93 Tv = ?

/v E B=

950 V/m = 0.9

1021 m/3

s T

EvB

= =

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19.7 Magnetic Field of a long straight wire19.7 Magnetic Field of a long straight wire

Danish scientist Hans Danish scientist Hans OerstedOersted (1777(1777--1851) discovered 1851) discovered somewhat by accident that an electric current in a wire somewhat by accident that an electric current in a wire deflects a nearby compass needle.deflects a nearby compass needle.In 1820, he performed a simple experiment with many In 1820, he performed a simple experiment with many compasses that clearly showed the presence of a compasses that clearly showed the presence of a magnetic field around a wire carrying a current.magnetic field around a wire carrying a current.

I=0 I

aligned with earth’s field

aligned in a circular pattern

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Magnetic Field due to CurrentsMagnetic Field due to CurrentsThe passage of a steady current in a wire produces a The passage of a steady current in a wire produces a magnetic field around the wire.magnetic field around the wire.

Field form concentric lines around the wireField form concentric lines around the wireDirection of the field given by the right hand rule.Direction of the field given by the right hand rule.

If the wire is grasped in the right hand with the thumb in the If the wire is grasped in the right hand with the thumb in the direction of the current, the fingers will curl in the directiondirection of the current, the fingers will curl in the direction of of the field.the field.

Magnitude of the field Magnitude of the field I

2oIBr

μπ

=

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Magnitude of the field

2oIBr

μπ

=

I

r

B

μo called the permeability of free space

74 10 /o Tm Aμ π −= ×

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Andre-Marie Ampere

Ampere’s Law

Consider a circular path surrounding a current, divided in segments Δl, Ampere showed that the sum of the products of the field by the length of the segment is equal to μo times the current.

I

r

BΔl

o encB l IμΔ =∑

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2oIBr

μπ

=

2 o encB l B l B r Iπ μΔ = Δ = =∑ ∑

Consider a case where B is constant and uniform.

Then one finds:

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19.8 Magnetic Force between two parallel conductors19.8 Magnetic Force between two parallel conductors

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22 2

oIBd

μπ

=

l

d

1

2 F1

B2 I1

I2

2 1 21 2 1 12 2

o oI I I lF B I l I ld d

μ μπ π

⎡ ⎤= = =⎢ ⎥⎣ ⎦

1 21

2oI IF

l dμπ

= Force per unit length( Attractive )

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Definition of the SI unit AmpereDefinition of the SI unit Ampere

If two long, parallel wires 1 m apart carry the same current, anIf two long, parallel wires 1 m apart carry the same current, and the d the magnetic force per unit length on each wire is 2x10magnetic force per unit length on each wire is 2x10--77 N/m, then the N/m, then the current is defined to be 1 A.current is defined to be 1 A.

1 21

2oI IF

l dμπ

= Used to define the SI unit of current called Ampere.

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Example 1: Levitating a wireExample 1: Levitating a wire

Two wires, each having a weight per units length of 1.0x10Two wires, each having a weight per units length of 1.0x10--44 N/m, are N/m, are strung parallel to one another above the surface of the Earth, ostrung parallel to one another above the surface of the Earth, one ne directly above the other. The wires are aligned northdirectly above the other. The wires are aligned north--south. When their south. When their distance of separation is 0.10 m what must be the current in eacdistance of separation is 0.10 m what must be the current in each in h in order for the lower wire to levitate the upper wire. (Assume theorder for the lower wire to levitate the upper wire. (Assume the two two wires carry the same current).wires carry the same current).

l

d

1

2I1

I2

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l

d

1

2

F1

B2I1

I2

Two wires, each having a weight per units length of 1.0x10-4 N/m, are strung parallel to one another above the surface of the Earth, one directly above the other. The wires are aligned north-south. When their distance of separation is 0.10 m what must be the current in each in order for the lower wire to levitate the upper wire. (Assume the two wires carry the same current).

mg/l

Weight of wire per unit length:mg/l = 1.0x10-4 N/m

Wire separation: d=0.1 m

I1 = I2

( )( )( )( )

21

7 24

24 10

1.0 10 /2 0.10

oIF mgl l d

Tm A IN m

m

μπ

π

π

−−

= =

×× =

7.1I A=

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Example 2: magnetic field between the wiresExample 2: magnetic field between the wires

The two wires in the figure below carry currents of 3.00A and 5.The two wires in the figure below carry currents of 3.00A and 5.00A in 00A in the direction indicated (into the page). Find the direction and the direction indicated (into the page). Find the direction and magnitude magnitude of the magnetic field at a point midway between the wires. of the magnetic field at a point midway between the wires.

3.00 A 5.00 A

20.0 cm

2o i

iIBd

μπ

=

( ) 65.00 3.00 4 102 0.1

onetB A A T

mμ

π−= − = ×

iUpwards

XX

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19.9 Magnetic Field of a current loop19.9 Magnetic Field of a current loop

Magnetic field produced by a wire can be enhanced Magnetic field produced by a wire can be enhanced by having the wire in a loop.by having the wire in a loop.

Δx1

I

Δx2

B

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19.10 Magnetic Field of a solenoid19.10 Magnetic Field of a solenoid

Solenoid magnet consists of a wire coil with multiple Solenoid magnet consists of a wire coil with multiple loops.loops.It is often called an electromagnet.It is often called an electromagnet.

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Solenoid MagnetSolenoid Magnet

Field lines inside a solenoid magnet are parallel, uniformly spaField lines inside a solenoid magnet are parallel, uniformly spaced ced and close together.and close together.The field inside is uniform and strong.The field inside is uniform and strong.The field outside is non uniform and much weaker.The field outside is non uniform and much weaker.One end of the solenoid acts as a north pole, the other as a souOne end of the solenoid acts as a north pole, the other as a south th pole.pole.For a long and tightly looped solenoid, the field inside has a vFor a long and tightly looped solenoid, the field inside has a value:alue:

oB nIμ=

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Solenoid MagnetSolenoid Magnet

n = N/l : number of (loop) turns per unit length.: number of (loop) turns per unit length.

I : current in the solenoid.: current in the solenoid.

oB nIμ=

74 10 /o Tm Aμ π −= ×

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( )( )( )7

100 turns 1000 turns/m0.10 m

4 10 Tm/A 1000 turns/m 0.500o

Nnl

B nI

A

μ

π −

= = =

=

= ×

Example: Magnetic Field inside a Solenoid.Example: Magnetic Field inside a Solenoid.

Consider a solenoid consisting of 100 turns of wire and length oConsider a solenoid consisting of 100 turns of wire and length of f 10.0 cm. Find the magnetic field inside when it carries a curren10.0 cm. Find the magnetic field inside when it carries a current of t of 0.500 A.0.500 A.

N = 100l = 0.100 mI = 0.500 A

74 10 Tm/Aoμ π −= ×

46.28 10B T−= ×

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Field Lines

Opposites Charges Attract Currents Repel

Comparison: Electric Field vs. Magnetic Field

Electric MagneticSource Charges Moving ChargesActs on Charges Moving ChargesForce F = Eq F = q v B sin(θ)Direction Parallel E Perpendicular to v,B

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Chapter 20Chapter 20 Induced EMF and InductionInduced EMF and Induction

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IntroductionIntroduction

Previous chapter: electric currents produce magnetic Previous chapter: electric currents produce magnetic fields (fields (OerstedOersted’’ss experiments)experiments)Is the opposite true: can magnetic fields create electric Is the opposite true: can magnetic fields create electric currents?currents?

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20.1 Induced EMF and magnetic flux20.1 Induced EMF and magnetic flux

Just like in the case of electric flux, Just like in the case of electric flux, consider a situation where the magnetic consider a situation where the magnetic field is uniform in magnitude and field is uniform in magnitude and direction. Place a loop in the Bdirection. Place a loop in the B--field.field.The flux, The flux, ΦΦ, is defined as the product of , is defined as the product of the field magnitude by the area crossed the field magnitude by the area crossed by the field lines.by the field lines.

where is the component of B where is the component of B perpendicular to the loop, perpendicular to the loop, θθ

is the angle is the angle between B and the normal to the loop.between B and the normal to the loop.Units: TUnits: T··mm22 or or WebersWebers ((WbWb))

cosB A BA θ⊥Φ = =

Definition of Magnetic Flux

B⊥

The value of magnetic flux is proportional to the total number of magnetic field lines passing through the loop.

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A square loop 2.00m on a side is placed in a magnetic field of A square loop 2.00m on a side is placed in a magnetic field of strength 0.300T. If the field makes an angle of 50.0strength 0.300T. If the field makes an angle of 50.0°° with the with the normal to the plane of the loop, determine the magnetic flux normal to the plane of the loop, determine the magnetic flux through the loop. through the loop.

Problem: determining a fluxProblem: determining a flux

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( )( )2

2

cos 0.300 2.00 cos(50.0 )

.386 0 Tm

BA T mθΦ = =

=

A square loop 2.00m on a side is placed in a magnetic field of strength 0.300T. If the field makes an angle of 50.0° with the normal to the plane of the loop, determine the magnetic flux through the loop.

Given:

L = 2.00 mB = 0.300 Tθ = 50.0˚

Find:

Φ=?

Solution:

From what we are given, we use

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20.1 Induced EMF and magnetic flux20.1 Induced EMF and magnetic flux

Two circuits are not connected: Two circuits are not connected: no current?no current?However, However, closing the switchclosing the switch we see that the compasswe see that the compass’’ needle needle movesmoves and then goes and then goes back to its previous positionback to its previous positionNothing happensNothing happens when the when the currentcurrent in the primary coil is in the primary coil is steadysteadyBut But same thingsame thing happens when happens when the the switch is openedswitch is opened, except , except for the for the needle going in the needle going in the opposite directionopposite direction……

Picture © Molecular Expressions

Faraday’s experiment

What is going on?