Physics 152physics.valpo.edu/courses/p152/lectures/lecture36notes.pdf · Lecture Ch. 33:...

1

Physics 152Friday,

April 27, 2007

• LLenz s Law Practice

• IInductors

http://www.voltnet.com/ladder/

• Help sessions

• W 9 - 10 pm in NSC 119

• MasteringPhysics

• Hwk #5 due Fri., May 4

• Final Exam available Friday,

May 4 from Mrs. Wellsand.

AnnouncementsFriApr

.27.

Phys

152

Hint: Be able to do the homework(graded AND recommended) andyou’ll do fine on the exam!

Available Friday, May 4Due by our scheduled exam period

Monday, May 14 (10:30 am - 12:30 pm)

You may bring one 8.5”X11” index card (hand-written on both sides), a pencil or pen, and ascientific calculator with you.

I will put any constants and mathematicalformulas that you might need on a singlepage attached to the back of the exam.

AnnouncementsFriApr

.27.

Phys

152

Section 1: graphical problemSection 2: conceptual problem

Sections 3 - 6:One problem on Ch 32One problem on Ch 33

Two problems on material fromExams 1 and 2

REQUIRED Multiple Choice Section:10 questions covering all the above

material, 6 on previous material, 4 onChapters 32 - 34

AnnouncementsFriApr

.27.

Phys

152

Worksheet Problem #1

Ch. 33: Electromagnetic InductionLecture

36Phys

152

In which

direction will

the current

flow?

Two ways to answer this question:

1) The right-hand rule for moving charges

in a uniform magnetic field.

2) Lenz’s Law

x x x x x x x x x x x x




R Lv

x


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152

2

v to the right. B into the board.

Force on the positive charge carriers is up. Sothe current will go around counterclockwise.





R Lv

I

In which

direction will

the current

flow?

1) The right-hand rule for moving charges

in a uniform magnetic field.


36Phys

152

As the bar moves to the right, the flux throughthe loop into the board is increasing.

The current in the loop must create fluxout of the board. So the current goescounterclockwise.





R Lv

I

In which

direction will

the current

flow?

2) Lenz’s Law


36Phys

152

Worksheet Problems

#2, #3, and #4


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152

A circular loop is oriented with its planeperpendicular to a uniform magnetic fieldwith B = 1.5 T. At an instant when the radiusof the loop = 12.0 cm and is increasing at arate of 3.0 cm/s, what is the magnitude of theEMF induced in the loop?

Worksheet

Problem #5B

1. 0

2. 12 mV

3. 17 mV

4. 25 mV

5. 34 mV


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152

What happenswhen the switchis first closed?

Initially, there is no current in the circuit.

We know the final current in thiscircuit (from Ohm’s Law) will be I = V / R

So, for some period of time, the current ischanging from 0 to V / R.

V R

+

_


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152

As a result of the changing current, the magneticfield that the current creates changes too.

That magnetic field passes through the planeof this circuit.

There is therefore a changingmagnetic flux through the circuit.

V R

+

_x x x x x x

x x x x x x



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152

3

The magnetic flux is changingas the current increases.

An EMF is induced in the circuit!

By Lenz’s Law, the induced EMF results in a cur-rent that opposes the changes in magnetic flux.


V R

+

_x x x x x x

x x x x x x


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152

The very fact that the circuit is a closed

loop made of conducting material has

resulted in an induced EMF in the circuit

which opposes the current!

Of course, once the current attains its final

value, the induced EMF = 0 because the

magnetic field is no longer changing.

V R

+

_x x x x x x

x x x x x x


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152

Unless otherwise stated, we generally ignore

the self-inductance of the circuit itself.

However, you might observe its effect on thecircuits you construct in the laboratory.

When you put a potential differenceacross the circuit, the currentshould ramp up to its final value...

V R

+

_x x x x x x

x x x x x x


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152

We will pay attention to the self-inductance of asolenoid, which as a circuit element is called an

To power supply To power supply

INDUCTOR


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152

A bunch of current loopsconnected together!

So a magnetic flux exists through the inductor.

As was the case with the electrical circuit of the

last example, an induced EMF will be generated

across the inductor when the current through

the inductor changes.

What is aninductor (solenoid)?


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152

Again, by Lenz’s Law, the

induced EMF will generate

a current to oppose the

changing magnetic flux

through the inductor.

= N B A = Nμ0

N

LIA = μ

0

N2

LAI

What is the magnetic flux through aninductor of length L with N turns anda current I flowing through it?


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152

4

What is the inducedEMF through such acoil?

=d

dt= μ

0

N2

LA

dI

dt

= LdI

dt

where L is theinductance andis defined to be

L = μ0

2N

AL


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152

We can derive another definition by combining

AND

That is, equate these two statements of therelationship of EMF to changing fluxes andchanging currents.

TRUE ONLY FORA SOLENOID!!!!

= LdI

dt= N

d

dt

L = μ0

2N

AL


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152

L NI

=

[L] = [N][ ]

[I]= turns

Wb

A

[L] =

Tm2

A=

N

Amm

2

A

[L] =

Nm

A2=

Js

AC=

Vs

A= H

General expression forself-inductance

HENRY


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152


Two solenoids have the same cross-

sectional area. Solenoid B, however, is

twice as long and has twice the number

of turns as Solenoid A. The ratio of the

self-inductance of Solenoid B to that of

Solenoid A is

1. 1 / 4

2. 1 / 2

3. 1 / 1

4. 2 / 1

5. 4 / 1


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152

Inductors set up a back-EMF in a circuit asthe current in the circuit changes. Themagnitude of that EMF is given by

As a circuit element, therefore, inductorsaffect the rate at which the current in acircuit changes.

In some respects, their effect on a circuitis analogous to the role of a capacitor.

= LdI

dt


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152

Recall, capacitors are circuit elements

which store energy in an electric field

between a positively charged plate and

a negatively charged plate.

Inductors store energy in a magnetic field.

The amount of energy stored

in the magnetic field of an

inductor is given byU I=

1

2

2L


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152

5

Plugging in the voltage

across an inductor

The power consumed by a

circuit element is given by

P = I LdI

dt

P = I V

dU

dt= I L

dI

dt

Power is energy per

time, so we have

How do we go about deriving this?


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152

Now, integrating both sides

dU = L I dI

How do we go about deriving this?

dU

dt= I L

dI

dt

U I=1

2

2L


36Phys

152

V R

+

_

L

Let’s set up Kirchhoff’s loop equation:

V + VL + V

R = 0

V LdI

dtIR = 0

I tV

Ret L R( ) ( )/( / )

= 1

This differential equationhas a known solutiongiven by


36Phys

152

I

t

I=V/R

I=0.632V/R

I t I et( ) ( )max

/= 1

I V Rmax

/=

= the time constant = L / R


36Phys

152

V R

+

_

L

We can use the loop rule and Ohm’s law to

determine the potential difference across the

inductor:

V = VL + V

R

V

R(t) = I R = I

max(1 e

t / ) R =V

R(1 e

t / ) R


36Phys

152

V R

+

_

L

V t V eR

t( ) ( )/= 1

V t V V e Vet t

L( ) ( )/ /

= =1

Induced Back - EMF in the Inductor


36Phys

152

V R

+

_

L

6



36Phys

152

V R

+

_

L

In the circuit below, L = 7.00 H, R = 9.00 ,and V = 120 V. What is the self-inducedemf in the inductor 0.200 s after theswitch is closed?



36Phys

152

If we now open switch S1 and simultaneouslyclose switch S2, we once again have a changingcurrent in our circuit, inducing an EMF acrossour inductor.

VL + V

R = 0

LdI

dtIR = 0 I t I e

t L R( ) max

/( / )=

This time, the loop rule says:

V R

+

_

L

S2

S1

withsolution


36Phys

152

I

t

I=V/R

I=0.368V/RI t I e

t( ) max

/=

I V Rmax

/=

= the time constant = L / R

V R

+

_

L

S2

S1

If we allowed ourprevious circuit toreach its stable state…


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152

= =V V VeL R

t /

Using Ohm’s Lawand the loop rule

we know

VR = I R

VL + V

R = 0

V R

+

_

L

S2

S1


36Phys

152



36Phys

152

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