chapter 31 induction and inductance · 31-3 faraday’s law of induction magnet in motion ammeter s...

Aljalal-Phys.102-26 May 2004-Ch31-page 1

Chapter 31

Induction and Inductance


31-1 Two Symmetric Situation

Getrotation

Put current

current loop + magnetic field =

torque

Electric motor

RotateGet

current

torque + magnetic field=

current

Electric generator


31-2 Two Experiments

Ammeter

No current

S

Stationarymagnet

N

Wire

current in one direction

Magnetin motion

current in the other direction

Magnetin motion

Ammeter

SN

Wire

Ammeter

SN

Wire

i i

A current appears only if there is a relative motionbetween the loop and the magnet


31-2Two Experiments

Small current

Magnetin slow motion

Ammeter

SN

Wire

i

Large current

Magnetin fast motion

Ammeter

SN

Wire

i

Faster motion produces a greater current



Magnetin motion

Ammeter

S

N

Wire

i

Magnetin motion

Ammeter

SN

Wire

i

current in one direction

current in the other direction

If magnet is reversed, current is also reversed



Magnetin motion

Ammeter

S

N

Wire

i

Induced current

Induced emfwork done per unit charge to produce an induced current

Inductionprocess of producing an induced current and emf



S

ER

S

ER

Just after the switch is closed

current briefly appearsin the blue coil

Switch is open long time ago

No currentin the blue coil

Ammeter

Wire

Ammeter

Wire

i


31-1 Two Symmetric Situation

S

ER

S

ER

Switch is closed long time ago

No currentin the blue coil

Just after the switch is open

current briefly appearsin the blue coil

Ammeter

Wire

Ammeter

Wire

i


31-3 Faraday’s Law of Induction

Magnetin motion

Ammeter

S

N

Wire

iS

ER

Just after closing or opening the switch

Ammeter

Wire

i

An emf induced in a loop when the number of magnetic field lines that pass through the loop is changing

The value of the induced emf is determined by the rate at which the number of magnetic field lines changes



Magnetin motion

Ammeter

S

N

Wire

i

Magnetin motion

Ammeter

S

N

Wire

i

the number of magnetic field lines are increasing

the number of magnetic field lines are decreasing



S

ER

Ammeter

Wire

i

Just after the switch is open

S

ER


Ammeter

Wire

i





Magnetic Flux trough area A

= dABBA

→ →Φ ∫ i

Area vectorMagnitude: area of dADirection: perpendicular to area dA



Special case

Loop lies in a plane perpendicular to the magnetic field

B→

Uniform magnetic field

= dABB

→ →Φ ∫ i = B dA∫ = B dA∫ = B A

A = BBΦ



SI unit for magnetic flux ΦΒ

WeberWb

2Tesla1 Webe =1 (m eter)r

= dABB

→ →Φ ∫ i



Faraday's Law

The magnitude of the emf E induced in a conducting loop is equal to the rate at which the magnetic flux ΦΒ through that loop changes with time

d = -dt

BΦE

Induced emf tends to oppose the flux change



d = -dt

BΦE

Total emf is the sum of individual induced emf

d = - Ndt

BΦE

dAn emf of is induced in every turn. dtΦ

Loop of N turnsOne-turn loop



= dABBA

→ →Φ ∫ i

Magnetic flux through a loop can be changedby

changing the magnitude of the magnetic field Bor

changing the size of the loop in the magnetic fieldor

changing the angle between B and the loop.



Checkpoint 1

a b c d

B

e

Rank the time intervals according to the magnitude of the emf induced in a conducting loop perpendicular to B, greatest first.

uniform magnetic field

= dABB

→ →Φ ∫ i = B A

d = -dt

BΦE

d = - AtB

dE

t

b,then d and e tie,then a and c tie

Slope of B-t curve


31-3 Faraday’s Law of InductionSample Problem 31-1

Long Solenoid S220 turns/cm

= dABB

→ →Φ ∫ i

= B A

d = - Ndt

BΦE

Current in the solenoid is reduced at steady rate from 10 A to zero in 25 ms. What is the emf induced in coil C?

Coil C130-trun coil

D = 3.2 cm d = 2.1 cm

Flux through coil C

2dA = ( )2

π

02( n d d = - i) N ( ( ) )

dt 2πµE

Number of turns of coil C

Cross sectional area of coil C

Produced by Solenoid S

0B = i n µ

Number of turns per unit length of the solenoid S

02( n i) d= ( )

2B πµΦ


31-3 Faraday’s Law of InductionSample Problem 31-1

Coil C130-trun coil

D = 3.2 cm d = 2.1 cm

02( n d d = - i) N ( ( ) )

dt 2πµE 2

0 d = - N n dt

( )2

d iµ π

final initiall

final initiall

i - id idt t - t

= -30 -10

25 x 10=

-7-

23

.021 = - (130)(4 x10 )(220) 0 -10( )25 x

( )2 10

π πE = 75 mV

i

t25 ms

10 A

Long Solenoid S220 turns/cm

Current in the solenoid is reduced at steady rate from 10 A to zero in 25 ms. What is the emf induced in coil C?


31-4 Lenz’s Law

An induced current has a direction such that the magnetic field due to the current opposes the change in the magnetic flux that induces the current

d = -dt

BΦE

Induced emf tends to oppose the flux change


31-4 Lenz’s Law

Increasing B

Induced Biopposes the change in the flux

Induced current

Magnetin motion

Ammeter

S

N

Wire

i


Use the right-hand rule to find the direction of

the current


31-4 Lenz’s Law

Decreasing B


Induced current


the current

Magnetin motion

Ammeter

S

N

Wire

i



31-4 Lenz’s Law

Increasing B


Induced current

Magnetin motion

Ammeter

S

N

Wire

i

the number of magnetic field lines are increasing Use the right-hand rule

to find the direction of the current


31-4 Lenz’s Law Induced Biopposes the change in the flux

Magnetin motion

Ammeter

S

N

Wire

i


Decreasing B

Induced current


the current


31-4 Lenz’s Law


S

ER


Ammeter

Wire

i

0i

Increasing B


Induced current


the current


31-4 Lenz’s Law Induced Biopposes the change in the flux


S

ER

Ammeter

Wire

i

Just after the switch is openDecreasing B

Induced current

0i


the current


Checkpoint 231-4 Lenz’s Law

Rank according to the magnitude of the current induced in the loop, greatest first.Magnetic fields change at identical rates

Decreasing B

Increasing B

a b ci

i

i

i

i

i

a and b tie,then c (zero)


31-4 Lenz’s Law

r

r/2

batE

B→

Uniform magnetic

fieldSample Problem 31-2

2= 4.0 t +2.0 B t+3.0B in Tesla and t in seconds

bat = 2.0 VE

r = 0.2 mResistance of the loop R = 2.0 Ω

What is the magnitude and direction of the emf Eind induced around the loop by the magnetic field at t = 10 s?

indd=dt

BΦE

d(BA)=dt

dB= Adt

2r dB=2 dt

π

22r d= (4.0 t +2.0 t+3.0)

2 dtπ 2r= (8.0 t+2.0)

2π

2(0.2)= (8.0 (10) +2.0)2

π = 5.15 V


31-4 Lenz’s Law

Sample Problem 31-2

indi

batE

B→

ddt

BΦ= 5.15 V > 0.

indB→

indE

Uniform magnetic

field

At 10 s, Flux is increasing.

The induced Bind must be into the page to oppose this increase.From the right-hand rule, the induced current must be clockwise.The induced emf Eind must be clockwise.


31-4 Lenz’s Law

Sample Problem 31-2

r

batE

B→Uniform

magnetic field

bat = 2.0 VE

Resistance of the loop R= 2.0 Ω

indAt t = 10 s, = 5.15 VE

indi

indE

neti

What is the current in the loop at t = 10 s?

ind batnet

- i =R

E E 5.15 - 2.0 = 1.6 A2.0

=

Since Eind > Ebat, the net current is clockwise


Non-uniform magnetic field

B→

x

y

H

W

dA

dx

31-4 Lenz’s Law

Sample Problem 31-3

B = 4 t2 x2

B in Tesla and t in secondsW = 3.0 mH = 2.0 mWhat is the magnitude and direction of the emf Eind induced around the loop by the magnetic field at t = 0.10 s?

indd=dt

BΦE

= dABB

→ →Φ ∫ i = B dA∫

W

0

B = H dx∫ 2 2W

0

4 t x H= dx∫2

W

0

24 t = H dxx ∫3

2 W = 4 t H3

32 3 = 4 t 2

32 = 72 t

2ind

d d= = (72 t )dt dt

BΦE = 144 t = 144 (0.1) = 14.4 V


B→

x

y31-4 Lenz’s Law

Sample Problem 31-3

indB→

indi

indE

ddt

BΦ= 14.4 V > 0.At 0.10 s, Flux is increasing

The induced Bind must be out of the page to oppose this increase.From the right-hand rule, the induced current must be counterclockwiseThe induced emf Eind must be also counterclockwise


31-5 Induction and Energy Transfer

B→

Conducting loop


You pull at a constant velocity

v→

Work that you do to pull a conducting loop out of a uniform magnetic field appears as thermal energy In the loop

We will showFapp v = i2 R

Rate at which work is done to pull a loop out of a uniform magnetic field

Rate at which thermal energy is dissipated in the loop



B→

Conducting loop


v→

x

L

indd=dt

BΦE

= dABB

→ →Φ ∫ i = B dA∫

= B dA∫ = B A = B L x

indd=dt

BΦE d = (

dtB L x)

d x = LB dt

= B L v

Induced emf

ind = vB L E



B→

Conducting loop

The flux is decreasing, the induced Bind must be into the page to oppose this decrease.From the right-hand rule, the induced current and emf Eind must be clockwise

v→

x

L


Induced current

ind B L vi = =R RE

indE

indB→

i

Collective resistance of the wire

i

R

ind = vB L E



B→

appF→

L

Induced currentind B L vi = =R RE

i

To pull at constant velocity,

1F→

2F→

3F→

1appF = m a = 0F →→

+

1appF = F

10 = i L B sin 90F = i L B

appF v = i L B v B L v= L B vR

2 2 2 B L v=R

2 2 B L vi R= ( ) RR

2 2 2 B L v=R

Balance each other

Work that you do to pull a conducting loop out of a uniform magnetic field appears as thermal energy In the loop



Ammeter

Wire

Magnetin motion

i

Opposing force from the induced

magnetic field

Magnetin motion

Opposing force

S

N

Loop acts like a magnet

S

N

S

N

Whenever you move a magnet toward a conducting closed loop, a magnetic force resists the motion, requiring you to do positive workThe energy you transfer to the system (magnet + loop), appears as

thermal energy in the loop



Ammeter

Wire

Magnetin motion

i

Opposing force from the induced

magnetic field

Magnetin motion

Opposing force

S

N

Loop acts like a magnet

N

S

S

N

Whenever you move a magnet away from a conducting closed loop, amagnetic force resists the motion, requiring you to do positive workThe energy you transfer to the system (magnet + loop), appears as

thermal energy in the loop



Checkpoint 3Loops with edge lengths either L or 2L, move at the same constant speedRank according to the maximum magnitude of the emfinduced in the loops as the loops move into the magnetic field, greatest first.

B→


a b

c d

v→

ind=B L vE

ind=B (2L) vEc and d tie,then a and b tie.

chapter 31 induction and inductance · 31-3 faraday’s law of induction magnet in motion ammeter s...

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