mastering physics assignment 3...mastering physics assignment 2 chapters 19, 20 due monday, feb 11...

Mastering Physics Assignment 2

Chapters 19, 20

due Monday, Feb 11 at 11 pm


Chapters 21, 22

Available Monday, Feb 11 at 8 am

due Friday, Feb 29 at 11 pm

Week of February 11 - 15

Experiment 3: e/m of electron

1Wednesday, February 13, 2008

The next few weeks...

Ch. 18-22, 24, 25


Chapter 22: Electromagnetic Induction

• Induced emf and current

• Magnetic flux

• Faraday’s and Lenz’s laws

• Electric generators, back emf

• Omit 22.8, 22.9, (inductance and transformers)


!v

++

+

– – –

II

I

I

Charges inside the moving rod experience a

force due to the magnetic field...

Conductor

The moving conductor acts as a generator.

Electromagnetic Induction

The basis of electromagnetic induction.


!B

A charge Δq inside the wire moves with the coil relative to the magnetic

field. A component of field, B!

, is perpendicular to the velocity of the coil.

!B!

!F

x!F

I

!B!

Moving coil relative to magnet

!v

Motion of coil

toward the

magnet

The magnetic forces induce a current to flow around the coil.


!B

The charges in the coil are no longer moving as the coil is at rest, but

the induced current is the same as before...

There must be some more basic reason for the induced current.

→ Changing magnetic field at the position of the coil.

!B!I

!B!

Moving magnet relative to coil

!v


Induced emf

When the magnet moves

relative to coil, a current is

induced in the coil.

Reversing the magnet N and S

poles reverses the deflection.

Moving the coil to the magnet

produces the same deflection

as moving the magnet to the

coil – only the relative motion

of coil and magnet matters.

zero


Motional emf

The magnetic force Fm separates

the + and – charges in the conductor.

Fm = !q v B

Fm

The separated + and – charges

give rise to an electric field E in

the conductor.

!E

Fq = !q E

At equilibrium, the electrostatic force:

balances the magnetic force.

Fq = !q E

That is:

Also, E = V/L

The induced potential difference between the ends of the rod is: V = vLB

+!q

–!q

Fq = Fm

!qE = !qvB


Induced emf

The emf induced between the ends of a conductor that is moving in a magnetic field is:

V = vLB

The induced emf is the same whether the coil moves or the magnet moves, only the relative motion matters.

(V = vLBsin! when the angle between !B and!v is !)


Prob. 22.2/4: “Tethered Satellite Experiment”. A 20,000 m length

of wire is trailed behind the shuttle while in orbit around the

earth. The orbital speed of the shuttle is 7600 m/s.

If the earth’s magnetic field at the position of the shuttle is

5.1!10-5 T and the wire moves perpendicular to the field, what is

the induced emf between the ends of the wire?

V = vLB = 7600 ! 20,000 ! 5.1 ! 10-5 = 7752 V

Negative at the top.

!B!!v

Wire

+

–


Prob. 22.4/2: The drawing shows a type of blood flow meter. Blood is

conductive enough that it can be treated as a moving conductor. When it

flows perpendicular to a magnetic field, electrodes can be used to measure

the small voltage that develops across the vessel.

Suppose the speed of the blood is 0.3 m/s, the diameter of the vessel is 5.6

mm and B = 0.6 T. What is the magnitude of the voltage that is measured?

!vL+ + +

– – –

!B

Blood – a moving

conductor


Prob. 22.5: Each rod of length L = 1.3 m moves at speed v = 2.7 m/s

in a magnetic field, B = 0.45 T. Find the motional emf for each.

+

–


Prob. 22.C6: Initially the rod is at rest. Describe the rod’s motion

after the switch is closed. Be sure to account for the effect of any

motional emf that develops.

V0


The rod experiences a magnetic force to the right and accelerates.

I F = ILBV0

L


The moving rod now generates its own emf that opposes the emf of

the battery (a “back emf”). The current therefore decreases. The

rod continues to accelerate until the current is reduced to zero

(assuming no friction).

v

+

–

IV = vLBV0 F = ILB

Speed constant when vLB = V0


V = vLB

= “back emf”V0

I

I

Induced emf – equivalent circuit

R

Resistance of rails and bar

I = 0 when V = V0


60 W bulb, R = 240 !

Motional emf between ends of sliding rod, V = vLB

Power dissipated, W = VI = V2/R = 60 W, so (vLB)2/R = 60 W

B = 0.4 T L = 0.6 m

Therefore, v=!60R

LB=

!60" (240 !)

(0.6 m)" (0.4 T)= 500 m/s

In 0.5 s, the rod slides 250 m!

22.7/6: How long do

the rails have to be

to light the 60 W

bulb for 0.5 s?

!Fm

Fapplied


B = 0.4 T L = 0.6 m

Work done to light the lamp

There is an induced

current I in the bar

when the bar is moving

in the magnetic field.

Magnetic force on the

bar, Fm = ILB, opposes

the motion of the bar.

Fapplied

In 1 s, work done by the applied force in opposing Fm is W = Fm v

W = Fmv = (ILB) v = I (LBv) = I V = 60 W

That is, the power to light the bulb is supplied by doing work against

the magnetic force.

!Fm

60 W lamp


Fapplied

!Fm

• Sliding the bar along the rails generates an emf

• When the circuit is completed, a current flows and lights up the lamp

• A magnetic force acts on the current in the rod to oppose the motion

of the rod (a consequence of Lenz’s law – later)

• The work done in pushing the rod is equal to the electrical energy

dissipated in the lamp – mechanical energy is converted into electrical.


!B


The emf induced between the ends of the

falling rod is:

V = vLB

No current is flowing, so there is no magnetic

force on the rod.

The resistor R completes the circuit, so that

current flows and there is now a magnetic

force resisting the gravitational force that

accelerates the rod downwards.

The rod stops accelerating when the magnetic

force is equal to the weight of the rod.

Motional emf

L

L

!v

!v


Prob. 22.9: A conducting rod 1.3 m long slides down between two

frictionless vertical copper tracks at a constant speed of 4 m/s

perpendicular to a 0.5 T magnetic field.

a) What is the mass of the rod?

b) Find the change in gravitational PE in 0.2 s.

c) Find the electrical energy dissipated in the resistor in 0.2 s.

R = 0.75 !

L = 1.3 m


Induced emf

Changing the area of the loop also

induces an emf.

An emf is induced in the coil

whenever the number of field lines

passing through the coil changes.

The number of field lines is a

measure of “magnetic flux”.

→ There is an induced emf

whenever there is a change of

magnetic flux passing through the

coil.



Chapters 19, 20

due today at 11 pm!


Chapters 21, 22



Experiment 3: e/m of electron


Fapplied

!Fm

• Sliding the bar along the rails generates a motional emf: V = vLB

• When the circuit is completed and a current flows, the resulting

magnetic force opposes the motion of the rod.

• The work done in pushing the rod is equal to the electrical energy

dissipated in the lamp – mechanical energy is converted into electrical.


!B


Induced emf

The emf induced between the ends of the moving rod is: V = vLB

Between time t0 and t, the rod moves a distance x – x0 = v(t – t0), so

A0 = x0LA = xL

With B perpendicular to the loop, (BA) is the “magnetic flux” passing

through the loop. The induced emf is equal to the rate of change

of magnetic flux passing through the loop – Faraday’s Law.

LA0 A

L

V = vLB =(x! x0)(t! t0)

LB =!

A!A0

t! t0

"B =

(BA)! (BA)0t! t0

=!(BA)

!t


Magnetic Flux

Magnetic flux, != Bcos"!A= BAcos"

Faraday’s Law: the induced emf is equal to the rate of change of magnetic flux

Unit of Magnetic Flux:

1 Weber (Wb) = 1 T.m2


Magnetic Flux and Field Lines

!= BAcos0! = BA != BAcos60

! = BA/2 != BAcos90! = 0

The magnetic flux is proportional to the number of magnetic field lines

passing through the coil.


B = 0.4 T L = 0.6 m

Fapplied

!Fm

Induced emf and rate of change of flux

As the rod is moved to the right, the area of the closed loop increases and

the magnetic flux passing through the loop increases. There is increasing

magnetic flux passing through the loop and pointing into the page.

R

!BI

!BI

!BI

The induced current, I, produces a magnetic field, BI, pointing out of

the page, that opposes the change in magnetic flux. This is Lenz’s law.

Faraday’s Law:

V =!"

!t= IR

= B!A

!t

= BLv


Faraday’s Law

The induced emf is equal to the

rate of change of magnetic flux.

The direction of the induced current is such that the

magnetic field produced by the current opposes the change in

magnetic flux that generated the current in the first place.

Lenz’s Law

Magnetic flux: != BAcos"


Prob. 22.C11: Use Lenz’s law to verify

that the induced current is in the

direction in the diagram.

• The flux through the loop is into

the page and is decreasing as the

area of the loop decreases.

• The induced current produces a

magnetic field that opposes the

decreasing flux.

→ The magnetic field produced

by the induced current must point

into the page.

→ The current flows clockwise in

the loop.


Conducting ring falling through a magnetic field

The magnetic flux passing through the

ring is constant (zero), so there is no

induced emf or current.

!BI

I

I = 0

!Fm

A magnetic force is generated that opposes the motion of the ring.

The magnetic flux passing through

the ring is increasing and is directed

into the page.

The induced current produces a

magnetic field, BI, that opposes the

increase of flux.


The magnetic flux passing

through the coil is constant, so

there is no induced emf or

current and no magnetic force.

The induced emf generates a current that

opposes the change in magnetic flux,

producing a magnetic field, BI, into the page.

!BIx

I

!Fm = 0

!Fm

A magnetic force opposes the motion of the coil.


through the coil is decreasing

and points into the page.


Eddy Currents

Move an electrical conductor into or out of a magnetic field. The

changing magnetic flux in the conductor produces a current that

circulates inside the conductor - an “Eddy current”.

Fm

Magnetic force

opposes motion

of metal !

electromagnetic

braking


Prob. 22.32/70: A bar magnet is falling through a metal ring. In part (a)

the ring is solid all the round around, but in part (b) it has been cut

through.

Explain why the motion of the magnet in part (a) is retarded, whereas it

is not in part (b).


Prob. 22.33: A circular loop of wire rests on a table. A long, straight

wire lies on this loop over its centre.

The current I in the straight wire is increasing. In what direction is

the induced current, if any, in the loop?

! !!

!

• What is the total magnetic flux through the loop?

• Does it change?

!B


A wire is bent into a circular loop as shown. The radius of the circle is

2 cm. A constant magnetic field B = 0.55 T is directed perpendicular

to the plane of the loop. Someone grabs the ends of the wire and

pulls it taut, so the radius shrinks to zero in 0.25 s.

Find the magnitude of the average induced emf between the ends of

the wire.

B

I


Prob. 22.70/32: Indicate the direction of the electric field between

the plates of the capacitor if the magnetic field is decreasing in

time.

++++

– – – –E

I

I

I

BI

BI

BI

Induced magnetic field


"

Prob. 22.26: A 0.5 m copper bar, AC, sweeps around a conducting circular track at 15 rad/s.

A uniform magnetic field points into the page, B = 0.0038 T.

Find the current in the loop ABC.

The loop forms a closed circuit of increasing area, so the magnetic flux passing through the loop increases and an emf is generated.

and I = V/R = 0.00713/3

= 2.4 mA

IBI

# = 15 rad/s

r = 0.5 m

The flux passing through the loop is: != BA= B

!"

2#

"!#r2 = Br

2!/2

The induced emf is: V =!"

!t=Br

2

2! !#

!t=Br

2$

2

V =0.0038!0.52!15

2= 7.13!10"3 V

θI


Guitar pickup

• The strings are magnetizable

• A permanent magnet magnetizes them

• The vibration of a string changes the magnetic flux through the

coil at the frequency of vibration of the string

• An emf is induced in the coil at that frequency


Playback of tape recording

• Recorded tape is magnetized in N-S patches

• The tape passes by the playback head which channels and concentrates

the magnetic field through an iron core

• The changing magnetic flux induces an emf in the coil


Moving coil microphone

• Sound waves cause the diaphragm of the microphone to move in/out

• A coil moves with the diaphragm relative to a permanent magnet,

causing the magnetic flux through the coil to change in step with the

pressure variations of the sound wave

• An emf is set up in the coil at the frequency of the sound wave


Ground fault detector

If the return current (green) is equal to the supply current (red), the

magnetic fluxes around the iron ring are equal and opposite and cancel.

If the currents differ, the fluxes do

not cancel and there is a net flux

varying at 60 Hz, which induces a

current in the sensing coil.



Chapters 21, 22



Midterm Break


Tutorial and Test 2

Chapters 19, 20, 21


Faraday’s Law

The induced emf is equal to the

rate of change of magnetic flux.

The direction of the induced current is such that the

magnetic field produced by the current opposes the change in

magnetic flux that generated the current in the first place.

Lenz’s Law

Magnetic flux: != BAcos"


Electric Generator


through the coil varies as the coil

rotates – an emf is generated.

Flux, != BAcos"= BAcos(#t)


V0

V =V0 sin(!t)

EMF from electric generator,

using rate of change of flux

Flux, != BAcos"= BAcos(#t)

V =V0 sin(!t), V0 = BA!

(diff. calculus)Then, the induced emf is V = (!)!"!t

= BA! sin(!t)

Reminder of Lenz’s Law


V = BLvsin!

V = BLvsin!

Electric GeneratorEMF generated in moving conductors

Total emf generated: Vtot = 2BLvsin!

Therefore, Vtot = BLW!sin"= BA!sin"

If the coil has N turns, then: Vtot = NBA!sin"=V0 sin"

W

Area, A = LW

v= r!=!W

2

"!, != angular frequency of rotating coil

VV ++–

–

2V

= V0 sin!t


Prob. 22.63/39: The cross-

sectional area of the coil is 0.02

m2 and the coil has 150 turns.

Find the rotation frequency of

the coil and the magnetic field.

The period is T = 0.42 s

Therefore, B=V0

NA!=

28

150!0.02!14.96 = 0.624 T

!=2"

T=

2"

0.42 s= 14.96 rad/s != 2" f , so f =

14.96

2"= 2.38 Hz

Peak voltage, V0 = 28 V = NBA#.

Alternating Current Electric Generator

V0 = 28 V

f = 1/T = 2.38 Hz

V = NBA!sin!t(Could also be V = NBA! cos !t)


TimeTime

Root Mean Square (rms)

Power $ V2

V0 = NBA#V0

V2 = V02 sin2 #t

Time

V02

Mean = V02/2

Mean power $ Vrms2 = V0

2/2

Vrms = V0/!2 = rms voltage

V = V0 sin #t


Prob. 22.40: A generator uses a coil that has 100 turns and a 0.5 T

magnetic field. The frequency of the generator is 60 Hz and its

emf has an rms value of 120 V.

Assuming that each turn of the coil is a square, determine the

length of the wire from which the coil is made.

• What is the peak voltage generated?

• What is the area of the coil?

A = a2...


Electric Generator

Electric Motor

Just like a generator, but use a current to cause the coil to rotate.

Once the coil is rotating, it acts as a generator, producing an emf that opposes the rotation of the coil! – the “back emf.”

L

L As the coil rotates and a current is induced in

it, a magnetic force is generated that opposes

the rotation of the coil.

! Work has to be done to rotate the coil

against this torque.


Back emf

Vback

A power supply drives the motor The motor acts as a generator

Kirchhoff’s loop law: V !Vback = IR

Symbol for

AC

generator


Vback

Prob. 22.38/36: A vacuum cleaner is plugged into a 120 V socket and

draws 3 A of current in normal operation (motor running at full speed,

back emf at maximum value) and the back emf is then 72 V.

Find the coil resistance of the motor.

V !Vback = IR

So, 120 – 72 = 3R

R = (120 – 72)/3 = 16 "

When the motor is first switched on, while it’s still spinning slowly, the

back emf is small and: V – 0 = IR,

Then, I = (120 V)/(16 !) = 7.5 A ⇒ the motor draws extra current while

it is speeding up. This is the time it’s most likely to trip the breaker.


Prob. 22.36/38: A 120 volt motor draws a current of 7 A when

running at normal speed. The resistance of the armature wire is

0.72 !.

a) Determine the back emf generated by the motor.

b) What is the current at the instant the motor is turned on and

has not begun to rotate?

c) What series resistance must be added to limit the starting

current to 15 A?


22.43/74: A motor is designed to operate on 117 V and draws a

current of 12.2 A when it first starts up. At its normal operating

speed, the motor draws a current of 2.30 A.

Obtain:

a) the resistance of the armature coil,

b) the back emf developed at normal speed,

c) the current drawn by the motor at one-third normal speed.


Prob. 22.25/71: A conducting coil of 1850 turns is connected to a

galvanometer. The total resistance of the circuit is 45 !. The area

of each turn is 4.7 ! 10-4 m2.

The coil is moved into a magnetic field, the normal to the coil being

kept parallel to the magnetic field. The amount of charge that is

induced to flow around the circuit is 8.87 ! 10-3 C.

Find the magnitude of the magnetic field.


Summary of Chapter 22

• EMF induced in a moving conductor: V = vLB sin"

• Mechanical work has to be done to produce

electrical energy – you don’t get something for

nothing

• Magnetic flux, % = BA cos&, proportional to

number of field lines.

• Faraday’s law: V = N !%/!t,

Lenz’s law: the magnetic field

produced by the induced current opposes

the changing magnetic flux.

• Electric generator, back emf


mastering physics assignment 3...mastering physics assignment 2 chapters 19, 20 due monday, feb 11...

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