30 Lecture Outline

8/13/2019 30 Lecture Outline

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Copyright © 2012 Pearson Education Inc.

PowerPoint® Lectures for

Univers i ty Physics , Thir teenth Edit io n

– Hugh D. Young and Roger A. Freedman

Lectures by Wayne Anderson

Chapter 30

Inductance




Energy through space for free??

• A puzzler!

Increasing

current in

time

Creates

increasing flux

INTO ring





• A puzzler!

Increasing

current in

time

Creates

increasing flux

INTO ring

Induce

counter-clockwise

current and

B field OUT

of ring





• But… If wire loop has resistance R, current

around it generates energy! Power = i2

/R!!

Increasing

current in

time

Induced currenti around loop of

resistance R





• Yet…. NO “potential difference”!

Increasing

current in

timeInduced current

i around loop ofresistance R





• Answer? Energy in B field!!

Increasing

current in

time

Increased fluxinduces EMF in

coil radiating

power


http://slidepdf.com/reader/full/30-lecture-outline 7/66Copyright © 2012 Pearson Education Inc.

Goals for Chapter 30 - Inductance

• To learn how current in one coil can induce anemf in another unconnected coil

• To relate the induced emf to the rate of changeof the current

• To calculate the energy in a magnetic field



Goals for Chapter 30

• Introduce circuit components calledINDUCTORS

• Analyze circuits containing resistors and

inductors

• Describe electrical oscillations in circuits and

why the oscillations decay



Introduction

• How does a coil induce a

current in a neighboring coil.

• A sensor triggers the traffic

light to change when a car

arrives at an intersection. How

does it do this?

• Why does a coil of metal

behave very differently from a

straight wire of the same metal?

• We’ll learn how circuits can be

coupled without being

connected together.



Mutual inductance

• Mutual inductance: A changing current in one coil

induces a current in a neighboring coil.

Increase current

in coil 1



Mutual inductance



Increase current

in coil 1

Increase B flux



Mutual inductance



Increase current

in coil 1

Increase B flux

Induce current in

loop 2

Induce fluxopposing change



Mutual inductance

• EMF induced in single loop 2 = - d (B2 )/dt

– Caused by change in flux through second loop of B field

– Created by the current in the single loop 1

• So… B2 is proportional to i1

• What does that proportionality constant depend upon?




Mutual inductance

• B2 is proportional to i1 and is affected by:

– # of windings in loop 1

– Area of loop 1

– Area of loop 2

• Define “M” as mutual inductance of coil 1 on coil 2




Mutual inductance

• Define “M” as mutual inductance on coil 2 from coil 1

• B2 = M21 I

• But what i f loop 2 also has many turns?

Increase #turns in

loop 2? =>

increase flux!




Mutual inductance

• IF you have N turns in

coil 2,

each with flux B ,

total flux is Nx larger

• Total Flux = N2 B2




Mutual inductance

• B2 proportional to i1

So

N2

B

= M21

i1

M 21 is the “Mutual Inductance”

[Units] = Henrys = Wb/Amp




Mutual inductance

• EMF2 = - N2 d [B2 ]/dt

and

• N2 B = M21i1

so

• EMF2 = - M21d [i1]/dt

• Because geometry is

“shared” M21 = M12 = M




Mutual inductance

• Define Mutual inductance:

A changing current in one

coil induces a current in a

neighboring coil.

• M = N2 B2/i1

M = N1 B1/i2




Mutual inductance examples

• Long solenoid with length l, area A, wound with N1 turns of wire

•

N2 turns surround at its center. What is M?





• M = N2 B2/i1

•

We need B2 from the first solenoid (B1 = m oni1 )

• n = N 1 /l

• B2 = B1 A

• M = N2 m oi1 A N 1 /l i1

• M = m oA N 1 N2 /l

• All geometry!





• M = m oA N 1 N2 /l

•

If N 1 = 1000 turns, N 2 = 10 turns, A = 10 cm2

, l = 0.50 m – M = 25 x 10-6 Wb/A

– M = 25 m H





• Using same system ( M = 25 m H)

•

Suppose i2 varies with time as = (2.0 x 106

A/s)t

• At t = 3.0 ms, what is average flux through each turn of coil 1?

• What is induced EMF in solenoid?





• Suppose i2 varies with time as = (2.0 x 106 A/s)t

•

At t = 3.0 ms, i2 = 6.0 Amps

• M = N1 B1/i2 = 25 m H

• B1= Mi2/N1 = 1.5x10-7 Wb

• Induced EMF in solenoid?

– EMF1 = -M(di2/dt)

– -50Volts

S lf i d




Self-inductance

• Self-inductance: A varying current in a circuit induces an emf

in that same cir cuit.

• Always opposes the change!

• Define L = N B/i

•

Li = N B

• I f i changes in time:

• d(L i)/dt = Nd B /dt = -EMF

or

• EMF = -Ldi/dt

I d i i l !




Inductors as circuit elements!

• Inductors ALWAYS oppose change:

• In DC circuits:

– Inductors maintain steady current flow

even if supply varies

• In AC circuits:

– Inductors suppress (filter) frequencies

that are too fast.

P t ti l i d t




Potential across an inductor

•

The potential across aninductor depends on the

rate of change of the

current through it.

• The self-induced emf

does not oppose current,

but opposes a change in

the current.

V ab = -Ld i/dt

M ti fi ld




Magnetic field energy

• Inductors store energy in the magnetic field:

U = 1/2 LI 2

• Units: L = Henrys (from L = N B/i )

• N B/i = B-field Flux/current through inductor that

creates that fluxWb/Amp = Tesla-m2/Amp

• [U] = [Henrys] x [Amps]2

• [U] = [Tesla-m2/Amp] x [Amps]2 = Tm2Amp

• But F = qv x B gives us definition of Tesla

• [B] = Teslas= Force/Coulomb-m/s = Force/Amp-m

M ti fi ld





• Inductors store energy in the magnetic field:

U = 1/2 LI 2

• [U] = [Tesla-m2/Amp] x [Amps]2 = Tm2Amp

• [U] = [Newtons/Amp-m] m2Amp

= Newton-meters = Joules = Energy!

M ti fi ld





• The energy stored in an inductor is U = 1/2 LI 2.

• The energy density in a magnetic field (Joule/m3) is

• u = B2/2m0 (in vacuum)

• u = B2/2m (in a magnetic material)

• Recall definition of m0 (magnetic permeability)

• B = m0 i/2pr (for the field of a long wire)

• m0 = Tesla-m/Amp

• [u] = [B2/2m0] = T2/(Tm/Amp) = T-Amp/meter



C l l ti lf i d t d lf i d d f




Calculating self-inductance and self-induced emf

• Toroidal solenoid with area A, average radius r, N turns.

• Assume B is uniform across cross section. What is L?

Calculating self inductance and self induced emf




Calculating self-inductance and self-induced emf

• Toroidal solenoid with area A, average radius r, N turns.

•

L = N B/i

• B = BA = (mo Ni/2pr)A

• L = mo N2A/2pr (self inductance of toroidal solenoid)

• Why N 2 ??

• If N =200 Turns, A = 5.0 cm2,

r = 0.10 m

L = 40 mH






• The potential across a

resistor drops in the

direction of current flow

Vab = Va-V b > 0

• The potential across an

inductor depends on the


current through it.









current through it.

• The self-induced emf

does not oppose current, but opposes a change in

the current.









current through it.

• The self-induced emf does

not oppose current, butopposes a change in the

current.

• The inductor acts like atemporary voltage source

pointing OPPOSITE to the

change.






•

The inductor acts like atemporary voltage source

pointing OPPOSITE to the

change.

• This implies the inductor

looks like a battery

pointing the other way!

• Note Va > V b!

Direction of current flow

from a battery oriented

this way






• What if current was

decreasing?

• Same result! The inductor

acts like a temporary voltage

source pointing OPPOSITE

to the change.

• Now inductor pushes current

in original direction

• Note Va < V b!

The R L circuit




The R-L circuit

• An R-L circuit contains aresistor and inductor and

possibly an emf source.

• Start with both switches open

• Close Switch S1:

• Current flows

• Inductor resists flow

• Actual current less thanmaximum E/R

• E – i(t)R- L(di/dt) = 0

• di/dt = E /L – (R/L)i(t)

The R L circuit




The R-L circuit

• Close Switch S1:

• E

–

i(t)R- L(di/dt) = 0• di/dt = E /L – (R/L)i(t)

Boundary Conditions

• At t=0, di/dt = E /L

• i( ) = E /R

Solve this 1 st order diff eq:

• i(t) = E /R (1-e -(R/L)t )

Current growth in an R-L circuit





• i(t) = E /R (1-e -(R/L)t )

• The time constant for an R-L circuit is = L/R.

• [ ]= L/R = Henrys /Ohm

• = (Tesla-m2/Amp)/Ohm

• = ( Newtons/Amp-m) (m2/Amp)/Ohm

• = (Newton-meter) / (Amp2-Ohm)

• = Joule/Watt

• = Joule/(Joule/sec)

• = seconds!






• i(t) = E /R (1-e -(R/L)t )

• The time constant for an R-L circuit is = L/R.

• [ ]= L/R = Henrys /Ohm

• EMF = -Ldi/dt

• [L] = Henrys = Volts /Amps/sec

• Volts/Amps = Ohms (From V = IR)

• Henrys = Ohm-seconds

• [ ]= L/R = Henrys /Ohm = seconds!

V ab = -Ld i/dt

The R-L circuit




The R-L circuit

• E = i(t)R+ L(di/dt)

• Power in circuit = E I

• E i = i2 R+ Li(di/dt)

• Some power radiated in resistor

• Some power stored in inductor

The R-L circuit example




The R L circuit example

• R = 175 W ; i = 36 mA; currentlimited to 4.9 mA in first 58 m s.

• What is required EMF

• What is required inductor

• What is the time constant?



Current decay in an R-L circuit




Current decay in an R L circuit

• Now close the secondswitch!

• Current decrease is opposed by inductor

• EMF is generated to keep

current flowing in the samedirection

• Current doesn’t drop to zeroimmediately






• Now close the secondswitch!

• – i(t)R - L(di/dt) = 0

• Note di/dt is NEGATIVE!

• i(t) = -L/R(di/dt)

• i(t) = i(0)e -(R/L)t

• i(0) = max current before

second switch is closed






• i(t) = i(0)e -(R/L)t






• Test yourself!

• Signs of Vab and Vbc whenS1 is closed?

• Vab >0; Vbc >0

• Vab >0, Vbc <0• Vab <0, Vbc >0

• Vab <0, Vbc <0






• Test yourself!


• Vab >0; Vbc >0

• Vab >0, Vbc <0• Vab <0, Vbc >0

• Vab <0, Vbc <0

• WHY?

• Current increases suddenly, so inductorresists change






• Test yourself!


• Vab >0; Vbc >0

• WHY?• Current still flows

around the circuitcounterclockwise

through resistor• EMF generated in L is

from c to b

• So Vb> Vc!





Cu e dec y c cu

• Test yourself!

• Signs of Vab and Vbc whenS2 is closed, S1 open?

• Vab >0; Vbc >0

• Vab >0, Vbc <0• Vab <0, Vbc >0

• Vab <0, Vbc <0





y

• Test yourself!

• Signs of Vab and Vbc whenS2 is closed, S1 open?

• Vab >0; Vbc >0

• Vab >0, Vbc <0

• WHY?

• Current still flowscounterclockwise

• di/dt <0; EMF generated in Lis from b to c!So Vb < Vc!

The L-C circuit




• An L-C circuit contains an inductor and a capacitor and is anoscillating circuit.

• Initially capacitor fully charged; close switch

• Charge flows FROM capacitor, but inductorresists that increased flow.

• Current builds in time.

• At maximum current, charge flow nowdecreases through inductor

• Inductor now resists decreased flow, andkeeps pushing charge in the original direction

i

The L-C circuit





• Initially capacitor fully charged; close switch

• Charge flows FROM capacitor, but inductorresists that increased flow.

• Current builds in time.

• Capacitor slowly discharges

• At maximum current, no charge is left on

capacitor; current now decreases throughinductor

• Inductor now resists decreased flow, andkeeps pushing charge in the original direction

i

The L-C circuit





The L-C circuit





• Now capacitor fully drained;

• Inductor keeps pushing charge in theoriginal direction

• Capacitor charge builds up on other sideto a maximum value

• While that side charges, “back EMF” from

capacitor tries to slow charge build-up

• Inductor keeps pushing to resist that change.

i

The L-C circuit





The L-C circuit





• Now capacitor charged on opposite side;

• Current reverses direction! System repeatsin the opposite direction

i

The L-C circuit





Electrical oscillations in an L-C circuit




• Analyze the current and

charge as a function of time.

• Do a Kirchoff Loop around

the circuit in the direction

shown.

• Remember i can be +/-

• Recall C = q/V

• For this loop:

-Ldi/dt – qC = 0





• -Ldi/dt – qC = 0

• i(t) = dq/dt

• Ld 2q/dt 2 + qC = 0

• Simple Harmonic Motion!

• Pendulums

• Springs

• Standard solution!

• q(t)= Qmax cos( w t+f )

where w = 1/(LC)½





• q(t)= Qmax cos( w t+f )

• i(t) = - w Qmax sin( w t+f )

(based on this ASSUMED

direction!!)

• w = 1/(LC)½ =

angular frequency

The L-C circuit





Electrical and mechanical oscillations




• Table 30.1 summarizes the analogies between SHM and L-C

circuit oscillations.

The L-R-C series circuit



• An L-R-C circuit exhibitsdamped harmonic motion

if the resistance is not toolarge.

30 Lecture Outline

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