Electromagnetic Induction. Induced EMF Faraday’s Law of Induction; Lenz’s Law EMF Induced in a...

Electromagnetic Induction

• Induced EMF

• Faraday’s Law of Induction; Lenz’s Law

• EMF Induced in a Moving Conductor

• Electric Generators

• Back EMF and Counter Torque; Eddy Currents

• Transformers and Transmission of Power

• A Changing Magnetic Flux Produces an Electric Field

Almost 200 years ago, Faraday looked for evidence that a magnetic field would induce an electric current with this apparatus:

Induced EMF

He found no evidence when the current was steady, but did see a current induced when the switch was turned on or off.

Induced EMF

Therefore, a changing magnetic field induces an emf.

Faraday’s experiment used a magnetic field that was changing because the current producing it was changing; the previous graphic shows a magnetic field that is changing because the magnet is moving.

Induced EMF

The induced emf in a wire loop is proportional to the rate of change of magnetic flux through the loop.

Magnetic flux:

Unit of magnetic flux: weber, Wb:

1 Wb = 1 T·m2.

Magnetic Flux

This drawing shows the variables in the flux equation:

Magnetic Flux

The magnetic flux is analogous to the electric flux – it is proportional to the total number of magnetic field lines passing through the loop.

Magnetic Flux

Magnetic FluxDetermining flux.

A square loop of wire encloses area A1. A uniform magnetic field B perpendicular to the loop extends over the area A2. What is the magnetic flux through the loop A1?

Faraday’s law of induction: the emf induced in a circuit is equal to the rate of change of magnetic flux through the circuit:

Faraday’s Law of Induction

or


A loop of wire in a magnetic field.

A square loop of wire of side l = 5.0 cm is in a uniform magnetic field B = 0.16 T. What is the magnetic flux in the loop (a) when B is perpendicular to the face of the loop and (b) when B is at an angle of 30° to the area A of the loop? (c) What is the magnitude of the average current in the loop if it has a resistance of 0.012 Ω and it is rotated from position (b) to position (a) in 0.14 s?

The minus sign gives the direction of the induced emf:

A current produced by an induced emf moves in a direction so that the magnetic field it produces tends to

restore the changed field.

or:

An induced emf is always in a direction that opposes the original change in flux that caused it.

Lenz’s Law

Magnetic flux will change if the area of the loop changes.


Magnetic flux will change if the angle between the loop and the field changes.



Induction stove.

In an induction stove, an ac current exists in a coil that is the “burner” (a burner that never gets hot). Why will it heat a metal pan but not a glass container?

Problem Solving: Lenz’s Law

1. Determine whether the magnetic flux is increasing, decreasing, or unchanged.

2. The magnetic field due to the induced current points in the opposite direction to the original field if the flux is increasing; in the same direction if it is decreasing; and is zero if the flux is not changing.

3. Use the right-hand rule to determine the direction of the current.

4. Remember that the external field and the field due to the induced current are different.

Lenz’s Law

Lenz’s Law

Practice with Lenz’s law.

In which direction is the current induced in the circular loop for each situation?

Lenz’s LawPulling a coil from a magnetic field.

A 100-loop square coil of wire, with side l = 5.00 cm and total resistance 100 Ω, is positioned perpendicular to a uniform 0.600-T magnetic field. It is quickly pulled from the field at constant speed (moving perpendicular to B) to a region where B drops abruptly to zero. At t = 0, the right edge of the coil is at the edge of the field. It takes 0.100 s for the whole coil to reach the field-free region. Find (a) the rate of change in flux through the coil, and (b) the emf and current induced. (c) How much energy is dissipated in the coil? (d) What was the average force required (Fext)?

This image shows another way the magnetic flux can change:

EMF Induced in a Moving Conductor


Force on the rod.

To make the rod move to the right at speed v, you need to apply an external force on the rod to the right. (a) Explain and determine the magnitude of the required force. (b) What external power is needed to move the rod?

A generator is the opposite of a motor – it transforms mechanical energy into electrical energy. This is an ac generator:

The axle is rotated by an external force such as falling water or steam. The brushes are in constant electrical contact with the slip rings.

Electric Generators

Electric Generators

.sincos BB tBA

dt

dtBA

If the loop is rotating with constant angular velocity ω, the induced emf is sinusoidal:

For a coil of N loops,

t

tNBA

sin

sin

0

Electric Generators

An ac generator.

The armature of a 60-Hz ac generator rotates in a 0.15-T magnetic field. If the area of the coil is 2.0 x 10-2 m2, how many loops must the coil contain if the peak output is to be E0 = 170 V?

150260m100.20.15T

V17022

0

0

BAN

NBA

A dc generator is similar, except that it has a split-ring commutator instead of slip rings.

Electric Generators

Electric GeneratorsAutomobiles now use alternators rather than dc generators, to reduce wear.

An electric motor turns because there is a torque on it due to the current. We would expect the motor to accelerate unless there is some sort of drag torque.

That drag torque exists, and is due to the induced emf, called a back emf.

Back EMF and Counter Torque


Back emf in a motor.

The armature windings of a dc motor have a resistance of 5.0 Ω. The motor is connected to a 120-V line, and when the motor reaches full speed against its normal load, the back emf is 108 V. Calculate (a) the current into the motor when it is just starting up, and (b) the current when the motor reaches full speed.


Motor overload.

When using an appliance such as a blender, electric drill, or sewing machine, if the appliance is overloaded or jammed so that the motor slows appreciably or stops while the power is still connected, the device can burn out and be ruined. Explain why this happens.

A similar effect occurs in a generator – if it is connected to a circuit, current will flow in it, and will produce a counter torque. This means the external applied torque must increase to keep the generator turning.


Induced currents can flow in bulk material as well as through wires. These are called eddy currents, and can dramatically slow a conductor moving into or out of a magnetic field.

Eddy Currents

Origins of Induced EMFs

dBvEdFqq

W)(

1

.dt

Bd

dE

Emf is the work done per unit charge by the source:

dt

ddBv B)(

Induced electric fields

Motional emf

The induced current is in a direction that tends to slow the moving bar – it will take an external force to keep it moving.

Motional EMF

vBEV

vBEevBeEFF

eEFEeF

evBFBveF

0

00BE

0E0E

BB

,00

reached, is mequilibriuWhen

downward ,

upward ,

The induced emf has magnitude


This equation is valid as long as B, l, and v are mutually perpendicular (if not, it is true for their perpendicular components).

Motional EMFDoes a moving airplane develop a large emf?

An airplane travels 1000 km/h in a region where the Earth’s magnetic field is about 5 x 10-5 T and is nearly vertical. What is the potential difference induced between the wing tips that are 70 m apart?

V.1s3600

m101000m70T105

35

vB

Induced Electric FieldsE produced by changing B.

A magnetic field B between the pole faces of an electromagnet is nearly uniform at any instant over a circular area of radius r0. The current in the windings of the electromagnet is increasing in time so that B changes in time at a constant rate dB/dt at each point. Beyond the circular region (r > r0), we assume B = 0 at all times. Determine the electric field E at any point P a distance r from the center of the circular area due to the changing B.

This microphone works by induction; the vibrating membrane induces an emf in the coil.

Applications of Induction: Sound Systems, Computer

Memory, Seismograph

Differently magnetized areas on an audio tape or disk induce signals in the read/write heads.


Memory, Seismograph

A seismograph has a fixed coil and a magnet hung on a spring (or vice versa), and records the current induced when the Earth shakes.


Memory, Seismograph

• Magnetic flux:

• Changing magnetic flux induces emf:

• Induced emf produces current that opposes original flux change.

Summary

• Changing magnetic field produces an electric field.

• General form of Faraday’s law:

• Electric generator changes mechanical energy to electrical energy; electric motor does the opposite.

Summary

.

Inductance AC Circuits

• Mutual Inductance

• Self-Inductance

• Energy Stored in a Magnetic Field

• LR Circuits

• LC Circuits and Electromagnetic Oscillations

• LC Circuits with Resistance (LRC Circuits)

• AC Circuits with AC Source

• LRC Series AC Circuit

• Resonance in AC Circuits

• Impedance Matching

• Three-Phase AC

Inductance

• Induced emf in one circuit due to changes in the magnetic field produced by the second circuit is called mutual induction.

• Induced emf in one circuit associated with changes in its own magnetic field is called self-induction.

Inductance

2221

222122

2221222

dt

dN

NNdt

Nd

Unit of inductance: the henry, H:

1 H = 1 V·s/A = 1 Ω·s.

Mutual inductance: magnetic flux through coil2 due to current in coil 1

Induced emf due to mutual induction:

Mutual Inductance

121212 IMN

dt

dIM

dt

dN 1

2121

221

Mutual InductanceSolenoid and coil.

A long thin solenoid of length l and cross-sectional area A contains N1 closely packed turns of wire. Wrapped around it is an insulated coil of N2 turns. Assume all the flux from coil 1 (the solenoid) passes through coil 2, and calculate the mutual inductance.

Mutual Inductance

Reversing the coils.

How would the previous example change if the coil with turns was inside the solenoid rather than outside the solenoid?

A changing current in a coil will also induce an emf in itself:

Self-inductance: magnetic flux through the coil due to the current in the coil itself:

Self-Inductance

2222 LIN

dt

dIL

dt

dN 222

222

Self-Inductance

Solenoid inductance.

(a) Determine a formula for the self-inductance L of a tightly wrapped and long solenoid containing N turns of wire in its length l and whose cross-sectional area is A.

(b) Calculate the value of L if N = 100, l = 5.0 cm, A = 0.30 cm2, and the solenoid is air filled.

Self-InductanceDirection of emf in inductor.

Current passes through a coil from left to right as shown. (a) If the current is increasing with time, in which direction is the induced emf? (b) If the current is decreasing in time, what then is the direction of the induced emf?

Self-InductanceCoaxial cable inductance.

Determine the inductance per unit length of a coaxial cable whose inner conductor has a radius r1 and the outer conductor has a radius r2. Assume the conductors are thin hollow tubes so there is no magnetic field within the inner conductor, and the magnetic field inside both thin conductors can be ignored. The conductors carry equal currents I in opposite directions.

A circuit consisting of an inductor and a resistor will begin with most of the voltage drop across the inductor, as the current is changing rapidly. With time, the current will increase less and less, until all the voltage is across the resistor.

LR Circuits

LR Circuits

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LR Circuits

RVyyIt

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y

y

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and 0 ,0At

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LR Circuits

R

Lτ

tR

V

tII

constant with time

exp1

exp1

Therefore,

0

0

If the circuit is then shorted across the battery, the current will gradually decay away:

LR Circuits

.

LR Circuits

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LR CircuitsAn LR circuit.

At t = 0, a 12.0-V battery is connected in series with a 220-mH inductor and a total of 30-Ω resistance, as shown. (a) What is the current at t = 0? (b) What is the time constant? (c) What is the maximum current? (d) How long will it take the current to reach half its maximum possible value? (e) At this instant, at what rate is energy being delivered by the battery, and (f) at what rate is energy being stored in the inductor’s magnetic field?

Solution:

ms.0.5693.0

,2

1exp ,

2

1 When

,exp1 d.

A.40.030

V12 c.

ms.3.730

mH220 b.

.0 zero, iscurrent the0,At t a.

21

max

max

0max

tt

tII

tIIR

VI

R

L

I

Solution:

W.2.130A20.0W4.2

, f.

W.42V12A40.02

1 e.

2

2RL

LR

RIIVPPP

PPP

.IVP

Remark: PR, the power dissipated from the resistor PL, the power stored in the inductor

Just as we saw that energy can be stored in an electric field, energy can be stored in a magnetic field as well, in an inductor, for example.

Analysis shows that the energy density of the field is given by

Energy Density of a Magnetic Field

Energy Stored in an Inductor

dt

diLiR

The equation governs the LR circuit is

Multiplying each term by the current i leads to

dt

diLiRii 2

inductor. thefrom dissipatedpower theis

battery. by the deliveredpower total theis 2Ri

i

Energy Stored in an Inductor

dt

diLi

dt

dUL

Therefore, the third term represents the rate at which the energy is stored in the inductor

The total energy stored from i=0 to i=I is

2

0L 2

1LI

dt

diLiU

I

Energy Density of a Magnetic Field

AB

LIU0

22

L 22

1

The self-inductance of a solenoid is L=μ0nA2l. The magnetic field inside it is B=μ0nI. The energy stored thus is

Since Al is the volume of the solenoid, the energy per volume is

0

2

B 2B

u

This is the energy density of a magnetic field in free space.

LC Circuits and Electromagnetic Oscillations

An LC circuit is a charged capacitor shorted through an inductor.

Electromagnetic Oscillations

The current causes the charge in the capacitor to decreases so I=-dQ/dt. Thus the differential equation becomes

LC Circuits

0dt

dIL

C

Q

Across the capacitor, the voltage is raised by Q/C. As the current passes through the inductor, the induced emf is –L(dI/dt). The Kirchhof’s loop rule gives

02

2

LC

Q

dt

Qd

The charge therefore oscillates with a natural angular frequency


022

2

xdt

xd

The equation describing LC circuits has the same form as the SHO equation:

.LC

1

Electromagnetic Oscillations

tQQ cos0

The charge varies as

The current is sinusoidal as well:

2cossin

sin

00

0

tItI

tQdt

dQI

Remark: When Q=Q0 at t=t0, we have φ=0.


The charge and current are both sinusoidal, but with different phases.


The total energy in the circuit is constant; it oscillates between the capacitor and the inductor:


LC circuit.

A 1200-pF capacitor is fully charged by a 500-V dc power supply. It is disconnected from the power supply and is connected, at t = 0, to a 75-mH inductor. Determine: (a) the initial charge on the capacitor; (b) the maximum current; (c) the frequency f and period T of oscillation; and (d) the total energy oscillating in the system.

Solution:

kHz,17Hz1017.0F102.1H10752

1

2

1

2 c.

mA63A103.6F102.1H1075

C100.6

b.

.C100.6V500F101200 a.

5

93

0

2

93

7

000max

7120

-

-

LCf

LC

QQI

CVQ

Solution:

J.105.1

F102.12

C100.6

2 d.

μs.59s109.51

c.

49

2720

5

C

QU

fT

LRC Circuits

Any real (nonsuperconducting) circuit will have resistance.

LRC CircuitsAdding a resistor in an LC circuit is equivalent to adding –IR in the equation of LC oscillation

0dt

dILIR

C

Q

Initially Q=Q0, and the switch is closed at t=0, the current is I=-dQ/dt. The differential equation becomes

02

2

LC

Q

dt

dQ

L

R

dt

Qd

LRC CircuitsThe equation describing LRC circuits now has the same form as the equation for the damped oscillation:

022

2

xdt

dx

dt

xd

The solution to LRC circuits therefore is

t

L

RtQQ cos

2exp0

LRC Circuits

where ω02=1/LC.

The system will be underdamped for R2 < 4L/C, and overdamped for R2 > 4L/C. Critical damping will occur when R2 = 4L/C.

The damped angular frequency is

220 2

L

R

LRC Circuits

This figure shows the three cases of underdamping, overdamping, and critical damping.

LRC Circuits

Damped oscillations.

At t = 0, a 40-mH inductor is placed in series with a resistance R = 3.0 Ω and a charged capacitor C = 4.8 μF. (a) Show that this circuit will oscillate. (b) Determine the frequency. (c) What is the time required for the charge amplitude to drop to half its starting value? (d) What value of R will make the circuit nonoscillating?

Solution:

Hz.3602

s102.5

2 b.

.oscillatescircuit theso ,0s102.5

H10402

3

F108.4H1040

1

2

1 a.

26

26

2

363

22

f

L

R

LC

Solution:

.180F108.4

H104044

,1

20

oscillate, not tocircuit for the d.

ms.182ln3

H104022ln

2

2ln22

1ln

2exp

2

1 c.

6

3

22

3

00

C

LR

LCL

R

R

Lt

L

Rt

L

RtQQ

AC Source• AC generator can produce– the emf– the current

varying sinusoidally in time

• The instantaneous current

• The instantaneous voltage

.sin0 tii

)sin(0 tvv

AC Source

• ω=2πf, the angular frequency; f, the frequency of the source.

• ψ, the phase difference between current and voltage.

• Resistors, capacitors, and inductors have different phase relationships between current and voltage when placed in an ac circuit.

The current through a resistor is in phase with the voltage.

A Resistor in an AC Circuit

.:peak value

sin

sin

,0 :rule loop

0R0

R0

0R

R

Riv

tv

tRiiRv

vv

A Resistor in an AC CircuitInstantaneous power dissipated in the resistor

RtiRip )(sin 220

2

average value over a complete cycle

22

)2sin(

2

)2cos(1)(sin

1)(

20

0

20

0

202

0

20av

2

itT

T

i

dtt

T

idtti

Ti

T

TT

A Resistor in an AC Circuit• root mean square (rms) current

• rms voltage

• rms power

.707.02

)( 00

av2 i

iiI

IRvV

vv

vV

av20RR

00

av2

)(

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)(

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)(

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R2

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av

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Therefore, the current through an inductor lags the voltage by 90°.

An Inductor in an AC Circuit

.:peak value

)90sin(

cos

cos

,0 :rule loop

0L0

L0

L0

0L

L

Liv

tv

tv

tLidt

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vv

An Inductor in an AC CircuitThe voltage across the inductor is related to the current through it in the form of Ohm’s law:

The quantity XL is called the inductive reactance, and has units of ohms:

LLL0L0 or IXVXiv

LX L


The instantaneous power supplied to the inductor is:

)cos()sin(L00L ttviivp

The average power over a complete circle is zero.

0)2cos(4

)2sin(2

1

)cos()sin(

0L00

0L00

0 L000 LL

T

T

TT

tvi

dttvi

dtttvidtpP


Reactance of a coil.

A coil has a resistance R = 1.00 Ω and an inductance of 0.300 H. Determine the current in the coil if (a) 120-V dc is applied to it, and (b) 120-V ac (rms) at 60.0 Hz is applied.

Solution:a. For dc current, there is no magnetic induction, so the current is determined by the resistance

A.12000.1

V120

R

VI

b. The inductive reactance is

113H300.0Hz6022L fLXThe current is

A.06.1113

V120

L

X

VI

A Capacitor in an AC CircuitThe current in the circuit is charging the capacitor, so i=+dq/dt. Thus dq=idt:

The quantity XC is called the capacitive reactance, and (just like the inductive reactance) has units of ohms:

.const)cos(

)sin(

0

0

t

i

tiidtq

The constant depends on the initial condition and we choose it to be zero.

Therefore, the current leads the voltage by 90°.

A Capacitor in an AC CircuitAccording to the loop rule, v-vc=0:

peak value ,

)90sin(

)cos(

)cos(

C0

C0

C0

C

C

iv

tv

tv

tC

i

C

qv

A Capacitor in an AC CircuitThe voltage across the inductor is related to the current through it in the form of Ohm’s law:

The quantity XC is called the reactance of the capacitor, and has units of ohms:

CCC0C0 or IXVXiv

CX

1

C

An Capacitor in an AC Circuit

The instantaneous power supplied to the capacitor is:

)cos()sin(C00C ttviivp

The average power over a complete circle is zero.

0)2cos(4

)2sin(2

1

)cos()sin(

0C00

0C00

0 C000 CC

T

T

TT

tvi

dttvi

dtttvidtpP

A Capacitor in an AC CircuitCapacitor reactance.

What is the rms current in the circuit shown if C = 1.0 μF and Vrms = 120 V? Calculate (a) for f = 60 Hz and then (b) for f = 6.0 x 105 Hz.

Solution:

A.44027.0

V120

,27.0F100.1Hz100.62

1 b.

mA,44A104.4k7.2

V120

,k7.2F100.1Hz602

1

2

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65C

2

C

6

C

I

X

X

VI

fCCX

AC Circuits with AC SourceThis figure shows a high-pass filter (allows an ac signal to pass but blocks a dc voltage) and a low-pass filter (allows a dc voltage to be maintained but blocks higher-frequency fluctuations).

Power in AC Circuits

The average power over a complete circle is

.sincossincossin

sinsin2

00

000

tttvi

tvtviivp

The instantaneous power delivered by the source emf is

.sincossincossin

sinsin11

0

200

0 0000av

dttttT

vi

dttvtviT

dtpT

p

T

TT

Power in AC CircuitsThe first term in the integrand contributes because the average of sin2(ωt) is ½ whereas the average of sin(ωt)cos(ωt) is zero. The average power is, therefore,

.cos2

100av vip

Rivv 0R00 cos since

RIIVP 2cos

The rms power becomes

Instantaneous and peak values of current and voltage

.sin0 tii )sin(0 tvv

Root mean square (rms) values of current and voltage

;20iI

20vV

Ohm’s law in LRC circuits

IZVZiv ;00

The impedance and reactance

2CL2 XXRZ

The phase angle

R

XX CLtan

The resonance angular frequency

LC

10

The rms power delivered by the source emf

RIIVP 2cos

The transformer is a device that can raise or lower the amplitude of ac potential differences. It consists of two coils, either interwoven or linked by an iron core. A changing emf in one induces an emf in the other.

The primary coil is connected to the source. While the secondary coil is connected to the load.

Transformers

The flux through both coils are the same. The emf’s are

Transformers

dt

dNV

dt

dNV

SSPP ,

Take the ratio of the emf’s, we find

P

S

P

S

N

N

V

V

Energy must be conserved; in the absence of losses, the power supplied by the primary coil and loaded to the secondary coil are the same. Therefore, the ratio of the currents must be the inverse of the ratio of turns:

Transformers

S

P

S

P

P

S

PPSS

N

N

V

V

I

I

VIVI

Transformers

Cell phone charger.

The charger for a cell phone contains a transformer that reduces 120-V ac to 5.0-V ac to charge the 3.7-V battery. (It also contains diodes to change the 5.0-V ac to 5.0-V dc.) Suppose the secondary coil contains 30 turns and the charger supplies 700 mA. Calculate (a) the number of turns in the primary coil, (b) the current in the primary, and (c) the power transformed.

Transformers work only if the current is changing; this is one reason why electricity is transmitted as ac.

Transformers

Transformers

Transmission lines.

An average of 120 kW of electric power is sent to a small town from a power plant 10 km away. The transmission lines have a total resistance of 0.40 Ω. Calculate the power loss if the power is transmitted at (a) 240 V and (b) 24,000 V.

Impedance MatchingWhen one electrical circuit is connected to another, maximum power is transmitted when the output impedance of the first equals the input impedance of the second.

The power delivered to the circuit will be a minimum when dP/dt = 0; this occurs when R1 = R2.

Three-Phase ACTransmission lines usually transmit three-phase ac power, with the phases being separated by 120°. This makes the power flow much smoother than if a single phase were used.

Three-Phase ACThree-phase circuit.

In a three-phase circuit, 266 V rms exists between line 1 and ground. What is the rms voltage between lines 2 and 3?

Solution:

V.460

2

V,650120180sin2

V,3762 V,266

32rms32

132

rms 1,1rms 1,

VVVV

VVV

VVV

• Mutual inductance:

• Energy density stored in magnetic field:

Summary

• Self-inductance:

Electromagnetic Induction. Induced EMF Faraday’s Law of Induction; Lenz’s Law EMF Induced in a...

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