# Electric and Magnetic Fields/Forces Applied

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```					           Electric and Magnetic Fields/Forces
Applied
1700s – Nature Philosophers “seek unifying principles”
1800 - Volta discovers the _____________________.
SIGNIFICANCE?
- step #1 in the development of modern electrical
technology
1820 - Hans Christian Oersted (Danish Physicist)
seeks connection between electricity and
magnetism

+       -

B

Oersted discovered a Perpendicular Force !

“Instead of the compass needle pointing toward or away from
the wire, it pointed in a circular direction around the wire”

B

Physics 30 - Electric and Magnetic Fields       1
2-Finger Rule- Predicts the direction of the circular
magnetic field around the current
carrying wire.

- use your right hand for positive charges
- use your left hand for negative charges

number line
+ + + + + + +                      + +
-4 -3 -2 -1 0 1 2                   3 4

finger

By definition: Current = conventional current
= flow of positive charges

Electron current = flow of negative charges!

Physics 30 - Electric and Magnetic Fields   2
2 – finger rule                             Examples

x – represents cross section of                     e- out of the page
wire carrying e- into the page

Calculate the direction of B around
B
Eg #1                               eg #2
B
A

A

Eg # 3 Calculate the direction of B at A, B, C, and D.

Choose between
A                     A) into the page
B) out of the page
C) up the page
-        +                              C    D    D) down the page
E) left of the page
B                        B                   F) right of the page
G) not sure

Physics 30 - Electric and Magnetic Fields     3
2 – Finger Rule - predicts the direction of the magnetic field
( B ) around a current carrying wire
Thumb = v, voom, velocity
Index finger = B, bang, magnetic field

1820 – Ampere

“If magnets exert forces on magnets, and currents
exert forces on magnets, then could currents exert
forces on currents?”

*use 2 – finger rule

B

B

ATTRACT!                  REPEL!

B                                                             B

Definition of the Ampere of Amp         1c
s
: the amount of current in each of two long straight parallel
wires set one metre apart that causes a force of exactly
2 x 10-7 N to act on each metre of each wire.

Physics 30 - Electric and Magnetic Fields   4
3 – Finger Rule

- Oersted found that a current in a wire is surrounded
by a circular magnetic field.
- but any moving charge is surrounded by a magnetic
field.
- Therefore, a moving charge should be deflected by
an external magnetic field.
- The 3 – finger rule predicts the direction of
deflection of a moving charged particle through an
external magnetic field.

Use:          Left hand for negative charges
Right hand for positive charges
Thumb for V, voom, velocity

Index finger for B, bang, magnetic field
*do not bend it!
Third finger for FORCE

Velocity

Magnetic Field

Force

NOTE that velocity, magnetic field, and force are
mutually perpendicular!

V             B           F

Physics 30 - Electric and Magnetic Fields     5
3 - Finger Rule                         Examples

- determine the direction of deflection of the following
charged particles.

A)                               N                 B)                 N
e-
p+

S                                    S

C)                  N                              D)        N             S

2+

S                                             e-

E)                      p+                         F)        N             S

S                   N                                 2+

G)                 e-                              H)            N

e-

N                    S                                S

Physics 30 - Electric and Magnetic Fields     6
History of Lighting

1. ARC Lamp
- At the beginning of the 1800s, buildings and homes were
lit by candles and oil lamps.

- There was no street lighting except for a few lights hung
outside homes at night.

- 1801 – Sir Humphry Davy discovered the “arc lamp”.

Carbon tips
 very bright
 very hot
 burnt out very
fast

B

2. Incandescent
- 1865, Sprengel in Germany discovers a new improved
vacuum pump.

- when electricity is passed through a filament heat and
light are given off.

- Edison was not the first to invent incandescent lighting
but he did develop a practical bulb for use in the home.

- He recognized the need for parallel circuitry and found
a high resistant filament (carbonized cotton thread)

Physics 30 - Electric and Magnetic Fields       7
Series                          -2           -2   -2              - the same current flows
through each resistor but the
voltage is reduced by each
resistor.
6V        -                +    0V
3 amps                               3 amps
B

Parallel                                                            the same voltage for each
1 amp           A                                        resistor but the current is
1 amp           B                                        divided.
1 amp           C

6V                                    0V

3 amps -                +    3 amps

B
-    1882 – The Edison Electric light Company began installing lighting
systems and sold over 200,000 lamps.

2. Fluorescent                                    white powder

-                                   +
-             Gas               +
-                                   +
-                                   +

- electrons jump across the gap in the tube kicking up electrons in
the gas molecules on their journey.

- When the gas molecules’ electrons fall they give off U.V. light.

- When the U.V. light hits the white powder it is absorbed, kicking
up the electrons in the powder.

- When these electrons fall, white (visible) light is emitted.

- Fluorescent light is cool and efficient but initially expensive to
install.

Physics 30 - Electric and Magnetic Fields               8
Electromagnetic Induction
= the production of a current by means of magnetism

IRON
RING
-             +
B
primary                       secondary    Galvanometer
coil                         coil       measures the
magnitude and
direction of an
electrical current
-galvanometer needle
flickers when switch
is closed

B                                                G
IRON RING
not necessary!

- open switch and needle flickers in the other direction

Physics 30 - Electric and Magnetic Fields        9

Closed

Magnetic
lines of force

B                                       move out and set-up

Open

Magnetic
lines
collapse

B

Back on their source

Current is produced only when magnetic lines of force move
relative to a conductor

Physics 30 - Electric and Magnetic Fields    10
Electric Generator

- A device which converts mechanical (PE and KE) energy into
electrical energy

S

N

G

- solenoid =
coil of wire

S             N

Modern generators spin the electromagnets
inside large coils of wire! Ex. Hydroelectric Dam

Physics 30 - Electric and Magnetic Fields   11
Q. Which way should we turn the handle (from above) to
push electrons out the sliding point of contact?

Physics 30 - Electric and Magnetic Fields   12
- 2 – finger rule

PIVOT

N
N

Hg
Hg
-      +                                          -     +
S
S
B                                             B
PIVOT

Electric Motor = device which converts electrical energy into
mechanical
N                                           S

A
B

-              +

B                  3 – finger rule

Physics 30 - Electric and Magnetic Fields     13
TRANSFORMER
- device for changing voltage
(1832 – Joseph Henry)

5x                    10x

V        2V                                                          Step-up
transformer

G        1 Amp

Fe
-     +                   Primary               Secondary
Coil                    Coil
B                                                     V          G

Powerprimary = Powersecondary
V p * Ip = V s * Is
(2V)(1C/s) = (4V)(0.5C/s)
2 watts = 2 watts
*Assuming 100% Efficient

Physics 30 - Electric and Magnetic Fields        14
STEP – DOWN TRANSFORMER

10x             5x
Fe

V       2V

G 1 Amp

Primary           Secondary
-         +

B                                         V   G

Powerp = Powers
VpIp = VsIs
(2V)(1C/s) = (1V)(2C/s)
2w = 2w

Physics 30 - Electric and Magnetic Fields    15
Power Lines
P = I2R                                       P = VI

Power lost through                          Power
transmission                              delivered

So you want to keep current and
resistance low! But if you keep
current low, power delivered is
lower!!

Voltage must be HIGH to
compensate for the low
current.

To step up voltage and step
down the current, use a

STEP-UP TRANSFORMER!

Physics 30 - Electric and Magnetic Fields   16
Transalta, Utilicorp, Fortis Power Grid

Wabamum                               Edmonton        Sherwood
Coal Powered                                            Park
Generator

500,000V
144,000V
Down                                 14400V
220V

Step-up                                               HOME
Transformer

Physics 30 - Electric and Magnetic Fields     17
Resistor Board Examples

1.                            series        RT = R1 + R2 …
RT = 2 + 4
2                4       = 6

1 1 1
 
2.                                               RT R1 R2
1  1 1
 
2                         4              RT 2 4
RT = 1.3
parallel

OR – special case for two resistors in
parallel

“Product over sum” rule

RT = 2x4 = 8 = 4 = 1.3
2+4 6 3

Physics 30 - Electric and Magnetic Fields    18
3.                         2                RT = 2 + 4 + 6

series               4          = 12

6

1 1 1 1
  
4.                                                RT 2 4 6

2              4           6         RT = 1.09

parallel

5.                                    RT = 2 + 4 + 6 + 8
2                                 = 20
4

series                      6
8

6.
1  1 1 1 1
   
RT 2 4 6 8

2 4 6                        8        RT = 0.96

Physics 30 - Electric and Magnetic Fields     19
7.                                    4       series
2                             4 + 6 = 10
6       parallel
1  1  1
 
RT  2 10
RT = 1.6

8.              2                             parallel
1 1 1
     2.4
RT 4 6
4      6       series
RT = 2 + 2.4
= 4.4

9.                                             series
2 + 4 = 6
2             6       6 + 8 = 14
parallel
1 1 1
 
4             8       RT 6 14
RT = 4.2

Physics 30 - Electric and Magnetic Fields    20
10.                                     4                 series
R=4+6+8
2                                  6      = 18
parallel
1  1 1
8                    
RT 18 2
RT = 1.8

11.                                                         series
2                               8           2 + 4 = 6
10    8 + 6 = 14
4           6               parallel
1  1 1   1
   
RT 6 14 10

RT = 2.96

12.                2                  8                 series
8 + 12 + 10
4              12        = 30
parallel
6                     10             1

1 1

RT 30 4

RT = 3.53
series
RT = 2 + 3.53 + 6 = 11.53

Physics 30 - Electric and Magnetic Fields        21
13.                                                series
2 + 4 = 6
2                             6   6 + 8 = 14
parallel
1  1 1
 
4                             8   RT 6 14
RT = 4.2

14.
2

4        6   8     10

12
parallel
1  1 1 1 1
   
RT 4 6 8 10
RT = 1.56
series
RT = 2 + 1.56 + 12
= 15.56

Physics 30 - Electric and Magnetic Fields    22
Ex. Draw a circuit diagram containing a 2
and a 4 resistor in parallel hooked up to a
12V battery.
Place an ammeter beside the battery and a
voltmeter at the 4 resistor.

Note: ammeters are hooked up in series
voltmeters are hooked up in parallel

12V

2   4   V

A

Physics 30 - Electric and Magnetic Fields   23
Circuitry Example
light bulb light bulb
100        100
light bulb
100
120V

1. Calculate the total resistance
RT = 100 + 100 + 100
= 300
2. Calculate total amps
VT = ITRT
120 = IT(300)
ITotal = 0.40 amps
3. Calculate total power
P = IV     P = I2R    P = V2                   PT = 48 watts
R                   each bulb =
16watts
4. What would happen if one of the
resistors was eliminated?
RT = 200 IT = 0.60c/s PT = 72w
72w  2 bulbs = 36w/bulb
 Brighter!

Physics 30 - Electric and Magnetic Fields   24
Same light bulbs

120V                              100       100    100

1. Calculate RT
1   1   1   1
                             RT = 33.3
RT 100 100 100

2. Calculate IT
VT = ITRT
120V = I(33.3) IT = 3.6c/s

3. Calculate PT
P = I2R or P = IV or P = V2
R
PT = 432w 3 = 144w/bulb

4. What happens when you turn off one
of the bulbs?
Now RT = 50 
 IT = 2.4 amps
(still 1.2 amps/resistor)
PT = 288w 2 = 144w/resistor

Physics 30 - Electric and Magnetic Fields   25
wiring = 1.0

120V                                        Parallel   12
plug      120

1. Which lightbulb burns the brightest?
Less R, draws more amps  burns brighter
2. If both bulbs are on, how much voltage is
available to the bulbs?
a) Calculate RT
12 x 120 = 10.91 + 1.0 = 11.91
12 + 120
b) Calculate IT
VT = ITRT
120V = I(11.91)  IT = 10.076A
c) Calculate the IR drop due to the wiring.
V = IR
= (10.076)(1.0)  V  10V
d)  voltage available to the plugs is
120 – 10  110V

Physics 30 - Electric and Magnetic Fields   26
3. Calculate the voltage available to the
plug if only the 120 bulb is turned
on.

Step 1. Calculate RT
 RT = 121

Step 2. Calculate IT
VT = ITRT
120V = IT(121)
IT = 0.99A

Step 3. Calculate the IR drop due
to the wiring.
V = IR
= (0.99A)(1.0)
 1 volt

Step 4. Calculate the voltage
available to the light.
120 – 1 = 119V

Physics 30 - Electric and Magnetic Fields   27
Generator

N            S

B

* If we spin the generator
counterclockwise (at B) then the:

initial motion is . . .

induced motion is . . .

force caused by the induced
motion (AKA induced force) is . . .

Physics 30 - Electric and Magnetic Fields   28
Back Electro Motor Force
N                             S

B

e-        3-finger rule
B

Initial motion –

Induced motion –

Induced force –

= BACK EMF
Physics 30 - Electric and Magnetic Fields        29
Lenz’s Law
“similar to Newton’s 3rd Law”

“Forces caused by the induced motion of
changes, tend to oppose the original
motion”

Application for generators

“when the generator is producing current
it is harder to spin the wire”

Application for motors

When the motor begins spinning, its
RPMs are at a minimum  the back
EMF is at a minimum  the current into
the motor is at a maximum.

When the motor reaches maximum
RPM the back EMF is at a maximum and
therefore the current is at a minimum.

Physics 30 - Electric and Magnetic Fields   30
BLACK OUT = total loss of power

BROWN OUT = temporary loss of power
(in a home circuit) when an electric motor
in a circuit kicks in.

Why the brown out?

When the electric motor in the circuit
(household circuit = 15 amps) kicks in the
RPMs are at a minimum the back EMF
is at a minimum and lots of current enters
the motor depriving the other appliances
in the circuit their fair share of current or
power (the lights momentarily dim).

When the motor reaches maximum
RPMs, the back EMF is at a maximum
the current is at a minimum and the
rest of the appliances regain their fair
share of current (the lights shine bright
again).

Physics 30 - Electric and Magnetic Fields   31
Lenz’s Law                                       page 741 – Heath

- Forces caused by the
induced motion of charges,
tend to oppose the original
motion.

Ex. Which way do the
electrons move through A?

N        S                                            N   S

A

N      S                                 S      N  repel
Answer: left                                  2 – finger rule

Physics 30 - Electric and Magnetic Fields   32
DROP the bar magnets; which one hits
the ground first? (A, B, or the same time)

N                     N

A                            B

S                     S

G

*What is the direction of the electrons
through G? (up)

Lenz’s Law
- Forces caused by the induced
motion of charges, tend to oppose
the original motion.

Physics 30 - Electric and Magnetic Fields   33
The 3 – finger rule allows you to predict the
direction of deflection of a moving charged
particle through an external magnetic field,
while the formula

Fm = qvB

allows you to calculate the magnitude of this
magnetic deflecting force.

Examples
A) Calculate the deflecting force on an
electron moving 2.0 x 105 m/s
perpendicularly to a magnetic field of
strength 3.6 x 10-4T.

F = qvB
= (1.6 x 1019c)(2.0 x 105m/s)(3.6 x 10-4T)
= 1.2 x 10-17N

B) What is the electron’s acceleration?

F = mA
A = F = 1.2 x 10-17N
M 9.11 x 10-31kg
= 1.3 x 1013m/s2

Physics 30 - Electric and Magnetic Fields     34
Calculate the magnitude and
direction of the deflecting force on
the electron.

N

5m/s
e-                              3m/s   3.6 x 10-4T

4m/s
S

F = qvB
= (16 x 10-19c)(4 x 100m/s)(3.6 x 10-4T)
= 2.3 x 10-22N

direction – (3 – finger rule) out of the page

Physics 30 - Electric and Magnetic Fields     35
The magnitude of the magnetic force
on a moving charged particle
Fm = qvB
Fm = (1.6 x 10-19c)(3.0 x 106m/s)(4.0 x 10-3T)

N
Fm = 1.9 x 10-15N
e-              4.0 x 10-3T
3.0 x 106m/s
S

Direction? (out of the page)

Fm = qvB
Fm = (1.6 x 10-19c)(3.0 x 106m/scos 40)(4.0 x 10-3T)
Fm = 1.5 x 10-15N
N
40
4.0 x 10-3T

e-           3.0 x 106m/s       S

Direction? (helically out of the page)
Physics 30 - Electric and Magnetic Fields   36
Fm = IlB
= the magnitude of the magnetic
force on a conductor

I = current through the conductor A
l = length of conductor m
B = magnetic field strength T

Example
If a 10 cm wire carries 0.50 Amps of
current through a magnetic field of
2.0 x 10-5T, then what is the force on
the wire?

Fmag = IlB
= (0.50c/s) (0.10m)(2.0 x 10-5T)
= 1.0 x 10-6N

Where is this formula useful?
(start worksheet #1)

Physics 30 - Electric and Magnetic Fields   37
Transformers

Isecondary = Vp = Np Number of turns
Iprimary Vs Ns

Powerp = Powers
VpIp = VsIs

Assuming 100% efficient

Example
A step-up transformer converts 120 V
to 20,000 V in a television. If the
primary coil has 175 turns, how many
does the secondary have?

Vp = Np Ns = NpVs= 175(20,000V)
Vs Ns           Vp      120 V
= 2.92 x 104 X or turns

Physics 30 - Electric and Magnetic Fields   38
Example #2
A step-down transformer (Np = 888;Ns =
111) is connected to a 120 Volt power
line. If there is a 35 electrical device
placed in the secondary circuit, what is
the current in the primary coil?

1) Calculate Vs
Np = Vp Vs = NsVp = (111)(120V)
Ns Vs         Np       (888)
= 15.0 V

2) Calculate Is
Vs = IsRs
15V = Is(35)
= 0.429 A  0.43 A

3) Calculate Ip
Np = Is    Ip = NsIs = 111(0.43A)
Ns Ip            Np       888
= 5.4 x 10-2A

Physics 30 - Electric and Magnetic Fields   39
Generators
V = Blv                             V = voltage
B = magnetic field
l = length of wire
v = velocity of wire
Example
Given the information below,
a) Calculate the direction of the
(electron) current.
(3 – finger rule) counter clockwise
b) Calculate the magnitude of the
current.
V = Blv
= (4.4 x 10-3T)(0.44m)(3.3m/s)
= 6.4 x 10-3V
V = IR
 I = V = (6.4 x 10-3V) = 1.3mA
R      5.0

V = 3.3m/s

B = 4.4 x 10-3T
R = 5.0                  44 cm

Physics 30 - Electric and Magnetic Fields      40
Example
Given the information below,
a) What is the direction of the
electron current?
(3-finger rule) clockwise
b) What is the magnitude of the
current?
V = Blv
= (4.4 x 10-3T)(0.50m)(3.2m/s)
= 7.04 x 10-3v
V = IR
 I = V = (7.04 x 10-3v)
R       2.0
= 3.5 mA

2.0

x    x   x   x
x    x   x   x
50 cm                                      x    x   x   x B = 4.4 x 10-3T
v = 3.2 m/s               x    x   x   x
x    x   x   x
x    x   x   x

Physics 30 - Electric and Magnetic Fields   41
AC Generator                                              symbol of AC
power supply

+                90           MAX CURRENT

Average current (effective)
= max I (0.707)
Current                      45            180        360 time
(Amps)
minimum current = 0
sin 45 = O
H
-
If H = 1 then sin 45 = 0.707

*SAME CURVE FOR VOLTAGE

 Ieff = 0.707 Imax
In data book
Veff = 0.707 Vmax

Also
Pav = VeffIeff                                       Pmax = VmaxImax
Veff = IeffR                                         Vmax = ImaxR
Pav = Ieff2R                                         Pmax = Imax2R
P = V2
R

Physics 30 - Electric and Magnetic Fields    42
Example
The 120v supplied to a normal electrical outlet
is really an effective or average voltage.

a) What is the maximum voltage supplied?

Veff = 0.707Vmax                           Vmax = 120 = 170v
0.707

b) What is the minimum voltage supplied?

AC  0

Example
An AC generator produces a maximum current
of 10.0 Amps at a maximum voltage of 55.0
volts. What is the average (effective) power
dissipated in the circuit?

Pav = IeffVeff               or Pmax = ImaxVmax
Pav = Imax(0.707)Vmax(0.707)         = 10A(55v)
= 10.0A(0.707) 55v(0.707)        = 550vA
= 275 w                      Pav = 550vA  2
= 275 w

Physics 30 - Electric and Magnetic Fields   43
50 v

R1 = 4.0

R2 = 6.0

A) What is the average power dissipated in R2?
Pav = Veff2 = (50v)2 = 416.6 w                                   4.2 x 102w
R                    6.0
B) What is the peak power dissipated in the
circuit?
Veff = IeffR                     Pmax =VmaxImax                Pmax = (Vmax)2
 4 x6 
50v =        
46 
   I eff   Pmax =  050v  20..707A 
      
 .707  0
83


R
2
 50v 
50v = 20.83A                                                        =          
 0.707 

2.4 = 20.83A                                                             2.4

= 2.1 kw

Physics 30 - Electric and Magnetic Fields          44

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