chapter 31 a

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Chapter 31A - Electromagnetic

Induction A PowerPoint Presentation by

Paul E. Tippens, Professor of Physics

Southern Polytechnic State University

© 2007

8/11/2019 Chapter 31 A


Objectives: After completing thismodule, you should be able to:

• Calculate the magnitude and direction ofthe induced current or emf in a conductormoving with respect to a given B-field.

• Calculate the magnetic flux through anarea in a given B-field.

•

Apply Lenz’s law and the right-hand rule

to determine directions of induced emf.• Describe the operation and use of ac and

dc generators or motors.

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Induced Current

When a conductor movesacross flux lines, magneticforces on the free electronsinduce an electric current.

Right-hand force rule showscurrent outward for down andinward for up motion. (Verify)

Down I

Down

vB

F

Up v

B

F

Up

I

B

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Magnetic Flux Densityf

Magnetic Fluxdensity:

A B A

F

• Magnetic flux lines

F are continuousand closed.

• Direction is thatof the B vector at

any point.

; = B BA A

F F

When area A is

perpendicular to flux:

The unit of flux density is the weber per square meter.

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Calculating Flux When Area is

Not Perpendicular to FieldThe flux penetrating thearea A when the normalvector n makes an angle

of q with the B-field is:

cos BA q F

The angle q is the complement of the angle a that theplane of the area makes with B field. (Cos q = Sin a)

n

A q

a

B

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Example 1: A current loop has an area of 40 cm2

and is placed in a 3-T B-field at the given angles.

Find the flux F through the loop in each case.

An

nn

A = 40 cm2 (a) q = 00 (c) q = 600 (b) q = 900

x x x xx x x xx x x xx x x x

(a) F = BA cos 00 = (3 T)(0.004 m2)(1); F 12.0 mWb

(b) F = BA cos 900 = (3 T)(0.004 m2)(0); F 0 mWb

(c) F = BA cos 600 = (3 T)(0.004 m2)(0.5); F 6.00 mWb

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Application of Faraday’s Law

Faraday’s Law:

- N t

F

E=

A change in flux F canoccur by a change in area orby a change in the B-field:

F = B A F = A B

n

n

n

Rotating loop = B A Loop at rest = A B

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Example 2: A coil has 200 turns of area 30 cm2.It flips from vertical to horizontal position in atime of 0.03 s. What is the induced emf if theconstant B-field is 4 mT?

S N

n

q

B

N = 200 turns

B = 4 mT; 00 to 900

A = 30 cm2 – 0 = 30 cm2

F = B A = (3 mT)(30 cm2)

F = (0.004 T)(0.0030 m2)

F = 1.2 x 10-5 Wb

-51.2 x 10 Wb(200)

0.03 s N

t

F

E E = -0.080 V

The negative sign indicates the polarity of the voltage.

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Lenz’s Law

Lenz’s law: An induced current will be in such a directionas to produce a magnetic field that will oppose themotion of the magnetic field that is producing it.

Flux decreasing by right moveinduces loop flux to the left.

N S

Left motion

I

Induced B

Flux increasing to left inducesloop flux to the right.

N S

Right motionI

Induced B

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Example 3: Use Lenz’s law to determine directionof induced current through R if switch is closed

for circuit below (B increasing).

R

Close switch. Then what isdirection of induced current?

The rising current in right circuit causes flux to increaseto the left, inducing current in left circuit that mustproduce a rightward field to oppose motion. Hencecurrent I through resistor R is to the right as shown.

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x x x x x x x x xx x x x x x x x



x x x x x x x x x

Directions of Forcesand EMFs

vL

vI

I

x

BI

v

Inducedemf

An emf E is induced by

moving wire at velocity vin constant B field. Note

direction of I.

From Lenz’s law, we seethat a reverse field (out) iscreated. This field causes a

leftward force on the wirethat offers resistance to themotion. Use right-handforce rule to show this.

x x xx x

x x x

x xx x xx x

x x x B

I

Lenz’s law

v

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Motional EMF in a Wire

Lv

I

I

x

x x x x x x xx x x x x x



x x x x x x x

BF

v

Force F on charge q in wire:

F = qvB; Work = FL = qvBL

Work qvBL

q qE =

EMF: BLvE=

If wire of length L moves withvelocity v an angle q with B:

Induced Emf E

v sin q v

q

B

sin BLv q E=

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Example 4: A 0.20-m length of wire movesat a constant speed of 5 m/s in at 1400 witha 0.4-T B- Field. What is the magnitude anddirection of the induced emf in the wire?

vq

B

North

South

sin BLv q E=0 (0.4 T)(0.20 m)(5 m/s)sin140E =

E = -0.257 V

Using right-hand rule, point fingersto right, thumb along velocity, andhand pushes in direction of inducedemf —to the north in the diagram.

v B

North

South

I

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The AC Generator

Rotating Loop in B-field• An alternating AC current isproduced by rotating a loopin a constant B -field.

• Current on left is outward

by right-hand rule.

• The right segment has aninward current.

• When loop is vertical, thecurrent is zero.

v

B

I

v

B

I

I in R is right, zero, left, and then zero as loop rotates.

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Operation of AC Generator

I=0

I=0

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Calculating Induced EMF

a

b

n

B

Area A = abx

. n

v

B

q

q

b/2

Each segment a

has constant

velocity v .

Rectangularloop a x b

x

n

vB

q

q

r = b/2

v sin q

v = w rBoth segments a moving withv at angle q with B gives emf:

sin ; Bav q E= 2bv r w w

22 sinT

b Ba w q E

sinT

BAw q E

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Sinusoidal Current of Generator

The emf varies sinusoidally with max and min emf

+E

-E

sin NBAw q EFor N turns, the EMF is:

x

.

x .

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Example 5: An ac generator has 12 turns ofwire of area 0.08 m2. The loop rotates in amagnetic field of 0.3 T at a frequency of 60

Hz. Find the maximum induced emf.

x

. nBq

f = 60 Hz

w = 2pf = 2p(60 Hz) = 377 rad/s

max ; Since sin 1 NBAw q E =

2

max (12)(0.3 T)(.08 m )(377 rad/s)E =

Emf is maximum when q = 900.

The maximum emf generated is therefore: E

max

= 109 V

If the resistance is known, then Ohm’s law (V = IR ) canbe applied to find the maximum induced current.

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The DC Generator

DC Generator

The simple ac generatorcan be converted to a dcgenerator by using a singlesplit-ring commutator toreverse connections twiceper revolution.

Commutator

For the dc generator: The emf fluctuates in magnitude,but always has the same direction (polarity).

tE

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The Electric Motor

In a simple electric motor, a current loop experiences atorque which produces rotational motion. Such motioninduces a back emf to oppose the motion.

Electric Motor

V

V – E b = IR

Applied voltage – back emf= net voltage

Since back emf Eb increases withrotational frequency, the startingcurrent is high and the operatingcurrent is low: Eb = NBA w sin q

E b

I

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Armature and Field Windings

In the commercial motor,many coils of wire aroundthe armature will producea smooth torque. (Note

directions of I in wires.)

Series-Wound Motor: Thefield and armature wiring

are connected in series.Motor

Shunt-Wound Motor: The field windings and thearmature windings are connected in parallel.

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Example 6: A series-wound dc motor has aninternal resistance of 3 W. The 120-V supplyline draws 4 A when at full speed. What is the

emf in the motor and the starting current?

V

E b

I

V – E b = IRRecall that:

120 V – Eb = (4 A)(3 W

Eb = 108 V The back emfin motor:

The starting current Is is found by noting that Eb = 0in beginning (armature has not started rotating).

120 V – 0 = I s (3 W I s = 40 A

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Summary

Faraday’s Law:

- N

t

F

E=

A change in flux F canoccur by a change in area orby a change in the B-field:

F = B A F = A B

cos BA q F ; = B BA A

F F

Calculating flux through an area in a B-field:

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Summary (Cont.)

Lenz’s law: An induced current will be in such a directionas to produce a magnetic field that will oppose themotion of the magnetic field that is producing it.

Flux decreasing by right moveinduces loop flux to the left.

N S

Left motion

I

Induced B

Flux increasing to left inducesloop flux to the right.

N S

Right motionI

Induced B

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Summary (Cont.)

sin BLv q E= Induced Emf E

v sin q v

q

B An emf is induced by a wiremoving with a velocity v at anangle q with a B-field.

sin NBAw q EFor N turns, the EMF is:

In general for a coil of N turns of area A rotatingwith a frequency in a B-field, the generated emfis given by the following relationship:

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Summary (Cont.)

DC Generator Electric Motor

V

The ac generator isshown to the right. Thedc generator and a dcmotor are shown below:

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Summary (Cont.)

V – E b = IR

Applied voltage – back emf

= net voltage

The rotor generates a backemf in the operation of amotor that reduces the

applied voltage. Thefollowing relationship exists:

Motor

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chapter 31 a

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