MAGNETIC INDUCTION

MAGNETUIC FLUX:FARADAY’S LAW, INDUCED EMF:

MOTIONAL EMF:THE AC GENERATOR:

THE TRANSFORMER:INDUCTANCE:

STORED ENERGY:

Written by Dr. John Dayton

MAGNETUIC FLUX:

Magnetic flux is defined as the portion per unit area of the magnetic field penetrating perpendicularly through a surface.

= angle between B and A

A

B

The net magnetic flux through a closed surface is zero.

cosB A

EXAMPLE: A uniform magnetic field, B, exists in space oriented horizontally to the right. At a certain point is a cube positioned so that its left face is perpendicular to the magnetic field. Each face of the cube has an area A. What is the magnetic flux through each face of the cube and the net magnetic flux through the cube?

The flux through the left side is 1=-BA, A is outward and so is anti-parallel to B.

The flux through the right side is 2=+BA, A is parallel to B.

The flux through the other sides is 0, A is perpendicular to B.

The net flux is the sum of the flux through each side, net = 0.

The magnetic flux through any closed surface is always zero.

FARADAY’S LAW, INDUCED EMF:

If a conducting loop or coil has through its encircled area a changing magnetic flux, then there will be an induced emf in the loop or coil.

In general:

In most problems the changing flux will depend on only one term.

N = number of loops of wireN

t

E

cos cos sin( )B A

A B BAt t t

t

INDUCED ELECTRIC CURRENT:

IR

E

To determine the direction of the conventional current flow around the conducting loop first determine the direction of the area vector of the encircled area. Looking down on the area from above, traversing the perimeter in a counter-clockwise sense is considered positive and the

area vector would be pointing towards you. If /t is positive with this area vector, the current flow is in the negative direction, or clockwise. This is due to the “-” sign in Faraday’s law.

The induced emf is not localized as it is in a battery. It is spread out around the entire conducting path. The current flows towards decreasing potential. Every complete circuit it gains a U from the emf and looses U to the resistance, maintaining energy conservation.

conducting loop with enclosed area

magnetic field and flux through loop

B

choose a direction for the area vector

A

determine the positive direction around loop in relation to the area vector

+ direction

A

If B is decreasing, is negative and is oriented in the positive direction.

If B is increasing is positive and is oriented in the negative direction.

current

B decreasingA

1

2

3

4

5

DETERMINING THE DIRECTION OF AN INDUCED ELECTRIC CURRENT

EXAMPLE: A solenoid of length 10cm made with 1500 turns of wire carries a current of 12A that is steadily decreasing at the rate of 3A/s. Refer to the diagram for the initial direction of this current. Inside the solenoid is a coils whose center lies on the axis of the solenoid. The plane of the coil makes a 30O angle with the axis of the solenoid. The coil is made with 400 turns of wire, is circular with a radius of 3cm, and has a net resistance of 0.3 Ohms.

Calculate the following:

the induced emf in the coil:

the induced current in the coil:

the magnetic dipole moment of the coil:

the torque on the coil when Is is 2A:

IS

30o10cm

2 cos .03260oc c o s s s cN t N N l I t Vr E

0.107cI AR E

2 20.121Cc c cm mr AN I

.038 sin 6 00 . 04os o s s s c s NB N I l T m B m

BS

Ic

mc

mc

cos

cos

s c BA

o s s s c BA

t B t A

t N l I t A

EXAMPLE: Sliding rod. Two parallel, horizontal, frictionless, conducting tracks are connected together at their left end by a wire with a resistor of resistance R. The separation between the rods is l. A conducting rod rests on the rod and is free to move along the rods. A uniform magnetic field, B, fills space and is directed vertically upward. The rod is pulled to the right with a constant velocity v. A conducting loop is form on the left with the rod on one side. Calculate the following:

FM

Rl

x

v

B

Changing Area of Loop:

Induced emf in loop

Induced Current:

Power Loss:

Magnetic Force on Rod:

Applied Force on Rod:

Power Applied to Rod:

I

I FA

Tracks

Rod

Answers on next slide.

EXAMPLE: Sliding rod. Two parallel, horizontal, frictionless, conducting tracks are connected together at their left end by a wire with a resistor of resistance R. The separation between the rods is l. A conducting rod rests on the rod and is free to move along the rods. A uniform magnetic field, B, fills space and is directed vertically upward. The rod is pulled to the right with a constant velocity v. A conducting loop is form on the left with the rod on one side. Calculate the following:

FM

Rl

x

v

B

Changing Area of Loop:

Induced emf in loop

Induced Current:

Power Loss:

Magnetic Force on Rod:

Applied Force on Rod:

Power Applied to Rod:

(see diagram)A lxBlvtxBltABt

I

I

(see diagram)I Blv RRvlBRIPR

2222 2 2sin (see diagram)MF IlB B l v R

2 2 to maintain constant (see diagram)AF B l v R v

FA

RAA PRvlBvFP 222

Tracks

Rod

MOTIONAL EMF:

__

B

v

FBl

+

-EFE

A conducting rod is moved through a magnetic field. Mobile electrons experience a force FB that pulls them to one side of the rod. Thus one end of the rod becomes electrically negative and the other becomes electrically positive. An electric field builds between the ends of the rod. The electric field exerts a force FE on the electrons. Eventually FE equals FB and no further charge moves.

-

rod

electron

See next slide for more details.

E BF F

eE evB

e evBl

Bvl

E

E

v-

B

Moving Rod

MOTIONAL EMF:

Effective current of electrons

Magnetic force on electron

Charge separation in rod

Orientation of electric field

Electric force on electrons

Click to see each of the following:

IFB

+

-E

FE

Mobile Electrons

Uniform Magnetic Field

EXAMPLE: A metal bar of length 75cm is rolling down an incline at a steady speed of 3m/s. A uniform magnetic field of 0.5G exists in space perpendicular to the surface of the incline. What is the potential difference between the ends of the rod?

4 4.5 1 1.125 100 3 .75msBvl T m V E

THE AC GENERATOR:

B

A

Rotating Loop

The induced emf causes an alternating current in the loop. The interaction between this current and the magnetic field is a torque that opposes the rotation of the loop. Thus an external source of energy will be required to keep the loop in rotation.

max

sin

sin sino

o o

N NBAtNBA

t

E

E E

E E E

IS

EXAMPLE: A coil is made to rotate inside a solenoid. The solenoid consists of 2500 turns of wire, has a length of 30cm, and carries a current of 6A. The coil is made with 500 turns of wire, is circular with a radius of 2cm, and has a net resistance of .75 Ohms. If the coil is rotating at 500 revolutions per minute, what is the maximum current in the coil?

0

7

22 2

1

2 1

Need

Need , and

4 10 2500 6.0628

.3

.02 .00126

5002 2 52.360

60

500 .0628 .00126 52.360 2.072

oo

o C S C S C

TmAo S S

SS

C C

o

IRN B A B A

AN IB T

l m

A r m m

revf s

s

T m s V

EE

E

E

2.072

.72.76

53o A

VI

THE TRANSFORMER:

PNP NS S

Primary CoilSecondaryCoil

Ideally:

Magnetic field lines (in blue) from the primary coil are directed through the iron core to the secondary coil. The changing flux they cause in the secondary coil induces the emf in the secondary coil. This will not work if the current in the primary coil is constant.

Iron Core

A transformer consists of a primary coil (input voltage), a secondary coil (output voltage), and an iron core the coils are wrapped around.

P P S S

P S SS P

P S P

N Nt t

N

N N N

E E

E EE E

P P S S

P PS P P

S S

I I

NI I I

N

E E

EE

EXAMPLE: The primary coil of a transformer has 2000 turns of wire. (a) If the maximum emf in the primary coil is 24V and 75V is required at the secondary coil, how many turns of wire are required in the secondary coil? (b) If the maximum current in the primary coil is 5A, what is the ideal maximum current from the secondary coil?

625075 2000

24

51.6

24

75

P S S PS

P S P

P PP P S S S

S

VNN

N N V

A VII I I A

V

E E EE

EE E

E

A transformer whose secondary coil voltage is greater than that of the primary is called a step up transformer. If the secondary voltage is less than the primary voltage, it is called a step down transformer.

INDUCTANCE:

EO

VV

EO

When the switch is closed the voltage builds to o. The reason it does not instantly become o is because the change in current produces a changing magnetic field and an induced emf that opposes o. This emf is called a back-emf and the process is known as self-induction.

L is Inductance and has SI unit of Henry, H.

Back emf:

Inductor

0

2

0

2

0

B IN N A N n A

t t t

N I IA L

l t t

NL A

l

E

E

STORED ENERGY:

As current begins to flow through an inductor a magnetic field is created and increases in strength. Within the field energy is stored. In a short time t while the current is increasing an amount of energy U is added to the field.

As the current increases more and more energy is stored until the current stops increasing at Ifinal. The total energy stored is:

U P t I t

IU I L t IL I

t

E

212

212

U U IL I

U L I I LI

U LI

Energy density may also be calculated:

STORED ENERGY:

22 201 1

2 2

2 2 212 2

0

212

0

212

0

212

0

o

N AU LI I

l

N I lU A

l

AlU B

U U Bu

Vol Al

Bu

EXAMPLE: A circuit consists of a 20 resistor, 24V battery and an inductor in series. The inductor is a solenoid made of 100 turns of wire wrapped around a cylinder 2cm long with a radius of 5mm. Calculate the inductance of the solenoid and the energy density of the field inside.

3

2 272

22 5 51 12 2

5

22

7

21 1

5

2 2

4 10 100 .005

.0224

1.220

4.93 10 1.2 3.55 10

3.55 10also

.02 .005

4 10 100 1.2.

4.93 10

22.6

00754.02

. 0

0 5

7

Jm

TmAo

TmAo

o

mN AL

l mV V

I AR

U LI H A J

U U Ju

Vol l r m m

ANIB T

l

H

m

Bu

3

2

722.

106

4

4 TA

mmJ

T