electric current gives rise to magnetic fields b i can a...

Chapter 20.1 Induced EMF and magnetic flux

Electric current gives rise to magnetic fields

Can a magnetic field give rise to a current?

The answer is yes as discovered by Michael Faraday.

I

B

Michael Faraday1791-1867

B× Moving magnetic fields induce

currents

×B

Change the strength of B:Induced current

Induced EMF!

vB×


Moving current loop induces currents

Fig. 20.2


SI units: weber (Wb)

20.2&20.4 Faraday’s law of induction

Michael Faraday1791-1867

Faraday’s law:The induced emf, ɛ, in a closed wire is equal to the time rate of change of the magnetic flux, ΦB.

An induced emf, ɛ, always gives rise to a current whose magnetic field opposes the original change in magnetic flux.

Lenz’s law

Faraday’s Law and Lenz’s Law bothstate that a loop of wire will want its magnetic flux to remain constant.

2. A circular loop with a radius of 0.20 m is placed in a uniform magnetic field B=0.85 T. The normal to the loop makes an angle of 30o with the direction of B. The field increases to 0.95 T what is the change in the magnetic flux through the loop?

B=0.85T

B’=0.95T

30o

Change in flux?

' 'cos ' cosB B A BAθ θ∆Φ = −2'

' 30o

A A Rπθ θ

= =

= =

2

2

2

cos ( ' ) cos30( ' )

(0.2) cos30(0.95 0.85)

1.1 10

B

B

B

A B B R B B

x Wb

θ π

π−

∆Φ = − = −

∆Φ = −

∆Φ =

8. A circular coil with a radius of 20 cm is in a field of 0.2 T with the plane of the coil perpendicular to the field. If the coil is pulled out of the field in 0.30 s find the average emf during this interval

B

cos 0B BAN Nt t

θε ∆Φ −= =

∆ ∆

2 2

2

0.2 (0.2)0.3

8.4 10

B Rtx V

π πε

ε −

= =∆

=

N=cosθ=A=

11

πR2

B=0

20.3 Motional emf

x x x x x

x x x x x

x x x x x

x x x x x

A voltage is produced by a conductor moving ina magnetic field

B into the page

v

Charges in the conductor experience a force upward

F

L

The work done in movinga charge from bottom to top

W FL qvBL= =

F qvB=

The potential difference isWV vBLq

∆ = =

V∆

x x x x x

x x x x x

x x x x x

x x x x x

A voltage is produced by a conductor moving ina magnetic field

B into the page

v

Charges in the conductor experience a force upward

F

L W FL qvBL= =

F qvB=

The potential difference isWV vBLq

∆ = =

V∆

Voltage

velocity

20.3 Motional emf

20.3 Motional emf

x x x x x

x x x x x

x x x x x

x x x x x

B into the page

vFL

V∆

The potential difference can drive a current through a circuitThe emf arises from changing flux due to changing area according to Faraday’s Law

wire

R

I

Bx ABV vLB LBt t t

ε∆Φ∆ ∆∆ = = = = =

∆ ∆ ∆

BLvIR Rε

= =

Changing Magnetic Flux

x x x x x

x x x x x

x x x x x

x x x x x

B into the page

vF

LV∆

18. R= 6.0 Ω and L=1.2 m and B=2.5 T. a) What speed should the bar be moving to generate a current of 0.50A in the resistor? b) How much power is dissipated in R? c) Where does the power come from?

wire

R

I 0.5(6.0)2.5(1.2)

1.0 /

BLvIR RIRvBL

v m s

ε= =

= =

=

a)

b) 2 2(0.5) (6.0)1.5

P I RP W= ==

c) Work is done by theforce moving the bar

Lenz’s LawThe polarity of the induced emf is such that it induces a current whose magnetic field opposes the change in magnetic flux through the loop. i.e. the current flows to maintain the original flux through the loop.

NS

V

B Bin

B increasing in loop

Bin acts to oppose thechange in flux

Current direction thatproduces Bin is as shown (right hand rule)I

+

-

Emf has the polarity shown. ε drives current inexternal circuit.

ε

20.4 Lenz’s law revisited

Now reverse the motion of the magnet

NS

V

B

Bin

B decreasing in loop

Bin acts to oppose thechange in flux

Current direction thatproduces Bin is as shown (right hand rule)I

+

-

Emf has the polarity shown.

ε

The current reverses direction


N S

Lenz’s Law and Reaction Forces

NS

V

B Bin

I

A force is exerted by the magnet on loop to produce the current

A force must be exerted by the current on the magnet to oppose the change

The current flowing in the direction shown induces a magnetic dipole in the current loopthat creates a force in the opposite direction

FmcFcm


B

εThe flux through the loop

cosB BA θΦ =tθ ω=

ω = angular velocity (radians/s)

Normal to the plane

θ

Flux through a rotating loop in a B field20.5 Generators

B

t∆Φ∆

tBΦ

tB

t∆Φ∆

sinB BA tt

ω ω∆Φ= −

∆

cosB BA tωΦ =

BA

-BA

BAω

-BAω

BRelation between ΦB and

proportional to ω

20.5 Generators

The emf generated by a loop of N turns rotating at constantangular velocity ω is

sinNBA tε ω ω=

t

εNBAω

-NBAω

0

BNt

ε ∆Φ= −

∆

20.5 Generators

35. In a model ac generator, a 500 turn rectangular coil8.0 cmx 20 cm rotates at 120 rev/min in a uniform magneticfield of 0.60 T. a) What is the maximum emf induced in the coil?

sinNBA tε ω ω=The maximum value of ε

max

max(120 2 )(500)(0.6)(0.08 0.2) 60

60

NBAxx V

ε ωπε

=

= =

20.5 Generators

Rotational

Work

Alternating Current (AC) generator

20.5 Generators

Direct Current (DC) generator

20.5 Generators

DC motor

ε drives rotation

A generator is motor acting in reverse

I

20.5 Generators

20.6 Self-Inductance• a property of a circuit carrying a current

• a voltage is induced that opposes the change in current

• used to make devices called inductors

Self- inductance of a circuit a reverse emf is produceby the changing current

B

tε ∆Φ= −

∆

Self-inductance of a coil

I

Current increasing

BB increases,

changes magnetic flux inthe coil,

B A Bt t

∆Φ ∆=

∆ ∆

Bt

∆∆

Produces emf in coilB A BN N

t tε ∆Φ ∆= − = −

∆ ∆

The direction of the induced emf opposes the change in current.

ε+ -

20.6 Self-Inductance

A changing current in a coil induces an emf that opposes the change

I I

I increasing

ε +-

I decreasing

ε+ -

induced emfopposes I

induced emfsupports I


Inductance L is a measure of the self-induced emf

I

Current increasing

BNt

ε ∆Φ= −

∆

ILt

ε ∆= −

∆

The self-induced emf is

L is a property of the coil, Units of L , Henry (H) VsA

ε

proportionality constant is L

but B It t

∆Φ ∆∝

∆ ∆


Inductance of a solenoid with N turns and length ℓ, wound around an air core (assume the length is much larger than the diameter).

2

oNL Aµ=l

B oNBA IAµΦ = =l

Bo

N I At t

µ∆Φ ∆=

∆ ∆l2

Bo

N I IN A Lt t t

ε µ∆Φ ∆ ∆= − = − = −

∆ ∆ ∆l

lA

It

∆∆

Bt

∆∆

inductance proportional to N squared x area/length


An air wound solenoid of 100 turns has a length of 10 cm and a diameter of 1 cm. Find the inductance of the coil.

2 2 2

7 2 25

44 10 (100) (0.01) 1.0 10

0.1(4)

o oN N dL A

L x H

µ µ π

π π−−

= =

= =

l l

I

l= 10 cm

d=1 cm


20.7 RL circuits

The inductor prevents the rapid buildup of current

But at long time does not reduce the current,

ILt

ε ∆= −

∆

0It

∆=

∆at t=∞

t

oI I (1 e )−τ= −

LR

τ =Applications of Inductors:

Reduce rapid changes of current in circuits

Produce high voltages in automobile ignition.

20.8 Energy stored in a magnetic fieldEnergy is stored in a magnetic field of an inductor.

I I=Io

B=0 BoB increasing

ε

Work is done against ε to produce the B field.

This produces a change in the PE of the inductor

212LPE LI=

This stored PE can be used to do work

electric current gives rise to magnetic fields b i can a...

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