REVIEW FOR FINAL - E & M 2016.pptx

Electricity

and

FORCE is inversely proportional to the square of the distance.

FORCE is directly proportional to the product of the charges.

F

α

1

R

2

FORCE is inversely proportional to the square of the distance.

FORCE is directly proportional to the product of the charges.

F

α

q

1

·

q

2

R

2

Coulomb’s Law

F

= k

q

1

·

q

2

R

2

k = 8.9875 x 10

N·m

/ C

k ~ 9 x 10

N·m

/ C

F

α

q

1

·

q

2

F

α

1

-+ -+ + +

-+ +

+

+ +

+ +

+ -+

The Quantity of Charge

The

quantity of charge

(q)

can be defined in

terms of the number of electrons, but the

Coulomb (C)

is a better unit for later work. A

temporary

definition might be as given below:

The Coulomb: 1 C = 6.25 x 10

electrons

The Coulomb: 1 C = 6.25 x 10

electrons

Which means that the charge on a single electron is:

1 electron: e

= -1.6 x 10

C

Example 3. A –6 mC charge is placed 4 cm from a +9 mC charge. What is the resultant force on a –5 mC charge located midway between the first

charges?

-

+

2 cm

+9 mC -6 mC

q_{1 r} q₂

2

2 cm

-r₁

1. Draw and label.

q₃

2. Draw forces.

F₁

1 nC = 1 x 10

C

3. Find resultant;

right is positive.

F

=

675 N

F

=

1013 N

9 -6 -6

1 3

1 _{2 2}

1

(9 x 10 )(6 x 10 )(5 x 10 ) ; (0.02 m) kq q F r  

9 -6 -6

2 3

2 _{2 2}

1

Example 4. Three charges, q1 = +8 mC, q2 = +6 mC and q3 = -4 mC are arranged as shown below. Find the resultant force on the –4 mC charge due to the

others.

Draw

free-body diagram.

-53.1o

-4 mC q₃

F₁

F₂

Note the directions of forces F

and F

on

q

based on attraction/repulsion from

q

and

q

.

+

-4 cm

3 cm

5 cm

53.1o

+6 mC

-4 mC

+8 mC q₁

q₂

q₃

The Electric Field

1. Now, consider point

P

a

distance

r

from

+Q

.

2. An electric field

E

exists

at

P

if a

test

charge

+q

has a force

F

at that point.

3. The

direction

of the

E

is

the same as the direction of

a

force

on

+ (pos)

charge.

E

4. The

magnitude

of

E

is

given by the formula:

Electric Field

+

+

+

+ +

+

+

+

Q

.

P

r

+q

+

F

N

; Units

C

F

E

q

Field Near a Negative Charge

Note that the field

E

in the vicinity of a

negative

charge

–Q

is

toward

the charge—the direction that a

+q

test charge would move.

Force on

+q

is with

field direction.

Force on

-q

is

against field

direction.

E

Electric Field

.

r

+

+

q

F

--

-

---Q

E

Electric Field

.

r

--q

F

--

-

The Magnitude of E-Field

The

magnitude

of the electric field intensity at a

point in space is defined as the

force per unit

charge

(N/C)

that would be experienced by any

test charge placed at that point.

Electric Field

Intensity E

Electric Field

Intensity E

The

direction

of

E

at a point is the same as the

direction that a

positive

charge would move

IF

placed at that point.

N

; Units

C

F

E

q

 



 

The E-Field at a distance r from a

single charge Q

+

+

+

+ +

+

+

+

.

r

Consider a test charge

+q

placed

at

P

a distance

r

from

Q

.

The outward force on +q is:

The electric field

E

is therefore:

+

+

q

F

+

+

+

+

+

+

+

+

.

r

E

F

kQq r

E

q

q





E

kQ

The Resultant Electric Field.

The resultant field

E

in the vicinity of a number

of point charges is equal to the

vector sum

of the

fields due to each charge taken individually.

Consider E for each charge.

+

-

·

q

q

q

-A

E₁

E₃

E₂

E_R

Vector Sum:

E = E

+ E

+ E

Vector Sum:

E = E

+ E

+ E

Directions are based

on positive test charge.

Magnitudes are from:

2

kQ

E

r

Electric Field Lines

+

+

+

+ +

+

+

+

Q

-

- -

-

-

---Q

Electric Field Lines

are imaginary lines drawn in

such a way that their direction at any point is the

same as the direction of the field at that point.

Field lines go

away

from

positive

charges and

-- - -- -- -- - -- -- -- -- - -- _- - -- - - -- - -- -- - --- - -

-Field lines are in the direction of the force that

would be put on a

POSITIVE

test charge.

+ + + + + + -+ -+ + + + + -+ -+ -+ -+ + + + ₊ ++ + + + + + + + + -+ -+ + + -+ -+ + - +-+ + + ₊ + - _{- +} + + ₊ + + + ₊ + + -+ + -++ -+ + + + + + + ++ + + + + + + + + ₊ + + + + ++ + + + + + + + + + + + + + ++ + ₊ + + + + + + + + + + + ++ + + + + + +

Field lines are in the direction of the force that

would be put on a

POSITIVE

test charge.

http://son.nasa.gov/tass/images/electric_fields3.jpg

+

-+

-+

-+

-+

-+

-+

-+

-+

-+

-+

-The Density of Field Lines

D

N

Gaussian Surface

s

Gauss’s Law: The field E at any point in space is

proportional to the line density

s

at that point.

Gauss’s Law: The field E at any point in space is

proportional to the line density

s

at that point.

D

A

Radius

r

r

N A

s

Line Density and Spacing Constant

Consider the field near a positive point charge q:

Gaussian Surface

Radius r

r

Then, imagine a surface (radius r) surrounding q.

E

is proportional to D

N/

D

A

and

is equal to

kq/r

at any point.

Define e

as spacing constant. Then:

;

N

kq

E

E

A

r

D





D

0

Where is:

N

E

A

e

e

D



D

1

4

k

e



Permittivity of Free Space

The proportionality constant for line density is

known as the

permittivity

e

and it is defined by:

Recalling the relationship with line density, we have:

Summing over entire area

A gives the total lines as:

N =

N =

e

e

EA

EA

2 -12

0 ₂

1

C

8.85 x 10

4

k

N m

e









0

N

E or

N

E A

A

e

e

D



D 

D

Positive

Work done

by

an

external force

increases

the Potential Energy.

Positive

Work done

by

the

field

decreases

the Potential Energy.

d

F

d

F

Negative

Work done

by

an

external force

decreases

the Potential Energy.

Negative

Work done

by

the

field

increases

the Potential Energy.

d

F

d

F

SIGN CONVENTIONS

E

E

E

Work to Move a Charge

+

+

+

+ +

+

+

+

¥

qE

F

+

Work to move

+q

from

A

to

B

.

·

· A

B

r

r

At A:

At B:

Avg. Force:

Distance:

r

- r

2 a a kqQ F r  avg a b kqQ F r r  2 b b kqQ F r 

(

)

a b

kQq

Work

Fd

r

r

r r







1

1

Absolute Potential Energy

+

+

+

+ +

+

+

+

¥

qE

F

+

·

· A

B

r

r

Absolute P.E. is

relative to ¥

.

It is work to bring

+q

from infinity to

point near

Q

—i.e.,

from

¥

to r

Properties of Space

E

Electric Field

+

+

+

+ +

+

+

+

Q

.

r

An

electric field

is a property of

space allowing prediction of the

force on a charge at that point.

The field E exist independently of

the charge q and is found from:

E is a Vector

E

=

kQq

r

q

;

F

E

F

qE

q





2

:

kQ

Electric Field

E

r

Electric Potential

Potential

+

+

+

+ +

+

+

+

Q

.

r

Electric potential

is another property of

space allowing us to predict the P.E. of

any

charge q at a point.

Electric

Potential:

The units are:

joules per coulomb

(J/C)

For example, if the potential is

400 J/C

at point

P

, a

–2 nC

charge at that point would have P.E. :

U = qV = (-2 x 10

C)(400 J/C);

U = -800 nJ

U = -800 nJ

P

V

U

q



;

U

V

U

qV

q

Calculating Electric Potential

Potential

+

+

+

+ +

+

+

+

Q

.

r

P

Electric Potential Energy and

Potential:

Substituting for

U, we find V:

The potential due to a positive charge is positive; The potential due to a negative charge is negative.

(Use sign of charge.)

The potential due to a positive charge is positive; The potential due to a negative charge is negative.

(Use sign of charge.)

kQ

V

r



;

kQq

U

U

V

r

q





 

kQq

r

kQ

V

q

r

Potential For Multiple Charges

The Electric Potential

V

in the vicinity of a number

of charges is equal to the algebraic sum of the

potentials due to each charge.

The Electric Potential

V

in the vicinity of a number

of charges is equal to the algebraic sum of the

potentials due to each charge.

+

-

·

Q

Q

Q

-A

r

r

r

Potential is + or – based on sign of the charges Q.

Potential is + or – based on sign of the charges Q.

3

1 2

1 2 3

A

kQ

kQ

kQ

V

r

r

r



+

+

kQ

V

r

Parallel Plates

V

+ + + +

-V

E

+q

F = qE

Consider Two parallel plates of equal

and opposite charge, a distance

d

apart.

Constant E field:

F = qE

Constant E field:

F = qE

Work =

Fd

= (qE)d

Also, Work =

q(V

– V

)

So that:

qV

= qEd

and

V

V

= Ed

= Ed

The potential difference between two oppositely

charged parallel plates is the product of

E

and

d

.

+

+

+ +

+

+

Q

+ +

Electric Field

weakens with distance.

Electric Field

UNIFORM

+ + + + + + + + + + + + + +

-Equipotential

Lines

g - Field

(J/kg)

Equipotential

Lines

E - Field

(J/C)

(N/C)

Lake Tahoe

1,645 ft deep

Equipotential

Lines

g - Field

-+ + + + + + + + +

Q V

∝

-+ + + + + + + + + + + + + + + + + + + + + +

Q

∝

V

Q V

∝

Q = C·V

C = Q

V

C = Capacitance

U = (qE)d

U/q = Ed

V = Ed

Work = Fd

More Charge, More Lines in a given Area, Larger Electric Field, More Voltage. Voltage (V) is directly proportional to Charge (Q).

0

Capacitance of Spherical Conductor

+Q

r

E and V at surface.

At surface of sphere:

Recall:

And:

Capacitance:

Capacitance, C

2

;

kQ

kQ

E

V

r

r





0

1

4

k

e



0

4

kQ

Q

V

r

e

r





C

Q

V



0

4

Q

Q

C

V

Q

e

r

Parallel Plate Capacitance

d

Area A +Q

-Q

You will recall from Gauss’ law that

E

is also:

For these two

parallel plates:

Q

is charge on either

plate.

A

is area of plate.

And

and

Q

V

C

E

V

d





0 0

Q

E

A

s

e

e





0

V

Q

E

d

e

A





C

Q

A

V

e

d

-+ + + + + + + + + -+ + + + + + + + + + + + + + + + + + + + + + -+ + + + + + + + +

0

Q

A

C

V

e

d

-+ + + + + + + + + -+ + + + + + + + + -+ + + + + + + + +

0

Q

A

C

V

e

d

Dielectric Constant

K

K

The Permittivity of a Medium

The capacitance of a parallel plate capacitor

with a dielectric can be found from:

The constant

e

is the

permittivity

of the medium

which relates to the density of field lines.

0

or

or

A

A

C KC

C K

C

d

d

e

e







2 2

-12 C 0

;

8.85 x 10

K

Q = .0008 V

C = .8 mF

Q = C·V

V ≡ U/q

U = q·V

U

= ½ q·V

U

= ½ Q·V

C ≡ Q/V

-+ - + -+ - + -+ - + -+ - + -+ - + -+ - + -+ - + -+ - + -+ - + ---

-+

Q

=

Q

=

Q

V

V

V

V

+

V

+

V

=

V

C

C

C

Q

C

Q

C

Q

C

+

+

=

Q

C

1

C

+

C

1

+

C

1

=

-+ - + -+ - + -+ - + -+ - + -+ - + -+ - + -+ - + -+ - + -+ - + ---

-+

Q

=

Q

=

Q

V

V

V

V

+

V

+

V

=

V

C

C

C

Q

C

Q

C

Q

C

+

+

=

Q

C

1

C

+

C

1

+

C

1

=

Example 1. Find the equivalent capacitance of the three

capacitors connected in series with a 24-V battery.

+

+ -- ++ -- ++ -

-2 mF

C

C

C

24 V

4 mF 6 mF

C

for

series:

C

=

1.09 mF

C

= 1.09

m

F

1

1

1

i

e i

C





C

1

1

1

1

2

4

6

e

C



m

F

+

m

F

+

m

F

1

0.500 0.250 0.167

e

C



+

+

1

1

0.917 or

0.917

e e

C

Example 1 (Cont.): The equivalent circuit can be shown as

follows with single C

+

+ -- ++ -- ++ -

-2 mF

C₁ C₂ C₃

24 V

4 mF 6 mF 1.09 mF

C_e

24 V

C

=

1.09 mF

C

= 1.09

m

F

Note that the equivalent capacitance

C

for capacitors in

series

is always

less than

the least

in the circuit. (1.09

m

F < 2

m

F)

1

1

1

i

e i

1.09 mF

C_e

24 V +

+ -- ++ -- ++ -

-2 mF

C₁ C₂ C₃

24 V

4 mF 6 mF

C

=

1.09 mF

C

= 1.09

m

F

Q

= C

V = (1.09

m

F)(24 V);

Q

Q

= 26.2 mC

= 26.2

m

C

For series circuits:

Q

= Q

= Q

= Q

Q

Q

= Q

= Q

= Q

= Q

= 26.2

=

26.2

m

mC

C

Example 1 (Cont.): What is the total charge and the charge

on each capacitor?

Q

C

V



-+

-+

-+

-+

-+

-+

-+

-+

-+

-+

-+

-+

-+

Q

Q

Q

+

+

Q

Q

Q

Q

=

C

V

+

C

V

+

C

V

C

V

=

C

+

C

+

C

C =

Example 2. Find the

equivalent capacitance

of the three

capacitors connected in

parallel

with a 24-V battery.

C

for

parallel:

C

=

12 mF

C

= 12

m

F

C

C

C

2 mF 4 mF 6 mF

24 V

Q = Q

+ Q

+ Q

V

= V

= V

= V

C

= (2 + 4 + 6)

m

F

Note that the equivalent capacitance

C

for

capacitors in

parallel

is always

greater than

the largest

in the circuit. (12

m

F > 6

m

F)

1

n

e i

i

C

C



Example 2 (Cont.) Find the

total

charge Q

and

charge

across

each capacitor.

C

=

12 mF

C

= 12

m

F

C

C

C

2 mF 4 mF 6 mF

24 V

Q = Q

+ Q

+ Q

V

= V

= V

=

24

V

Q

= (2

m

F)(24 V) =

48

m

C

Q

= (4

m

F)(24 V) =

96

m

C

Q

= (6

m

F)(24 V) =

144

m

C

Q

= C

V

Q

= (12

m

F)(24 V)

Q

= 288 mC

Q

= 288

m

C

;

Q

C

Q CV

V

Electric Current

Electric current

I

is the rate of

the flow of charge

Q

through

a cross-section

A

in a unit of

time

t

.

One ampere A is charge flowing at

the rate of one coulomb per second.

One

ampere A

is charge flowing at

the rate of one

coulomb per second

.

A

+

-Wire

+Q

t

Q

I

t



1 A

1C

1 s

Electromotive Force

A

source of electromotive force (emf)

is a

device that uses chemical, mechanical or

other energy to provide the potential

difference necessary for electric current.

Ohm’s Law

Ohm’s law

states that the current

I

through a

given conductor is directly proportional to the

potential difference

V

between its end points.

Ohm’s law allows us to define

resistance R

and to write the following forms of the law:

'

:

Ohm s law

I



V

;

;

V

V

I

V

IR

R

R

I

Resistivity of a Material

The

resistivity

r

is a property of a material

that determines its electrical resistance

R

.

Recalling that

R

is directly proportional

to length

L

and inversely proportional

to area

A

, we may write:

The unit of resistivity is the

ohm-meter (

W

m)

or

L

RA

R

A

L

r

r

Electric Power

Electric power

P

is the rate at which electric

energy is expended, or work per unit of time.

V

q

V

To charge C: Work = qV

Substitute q = It , then:

I

and

Work

qV

q

P

I

t

t

t







VIt

P

t

Calculating Power

Using Ohm’s law, we can find electric

power

from any two of the following parameters:

current

I

,

voltage

V

, and

resistance

R

.

Ohm’s law:

V = IR

2 2

;

;

V

P VI

P I R

P

R

Parallel Circuit

6

A

1

A

2

A

3

A

6

V

6

V

6

V

6

V

2 W

3 W

6 W

12 V 3 Ω3 Ω

2 Ω

2 Ω

6 Ω

Series Circuit

12 V

2 Volts

4 Volts

6 Volts

1 Ω

1 Ω

2 Ω

2 Ω

3 Ω

3 Ω

2 Amps

6 V

3 V 2 V

1 V

Resistances in Series

Resistors are said to be connected in

series

when there is a

single path

for the current.

The current

I

is the same for

each resistor

R

, R

and

R

.

The energy gained through

E

is lost through

R

, R

and

R

.

The same is true for voltages:

For series

connections:

For series

connections:

I = I

V

= I

= I

T

= V

+ V

+ V

I = I

= I

= I

V

= V

+ V

+ V

R₁ I

V_T

R₂ R₃

Sources of EMF in Series

The

output direction

from a

source of emf is from

+

side:

a

-

+

E

b

Thus, from

a

to

b

the

potential increases

by

E

;

From

b

to

a

, the

potential decreases

by

E

.

Example: Find DV for path

AB and then for path BA. R

3 V

+

-+

-9 V

A

B

AB: DV = +9 V – 3 V = +6 V

Parallel Connections

Resistors are said to be connected in

parallel

when there is more than one path for current.

2 W 4 W 6 W

Series Connection:

For Series Resistors:

I

= I

= I

= I

V

+ V

+ V

= V

Parallel Connection:

6 W

2 W 4 W

For Parallel Resistors:

V

= V

= V

= V

Equivalent Resistance: Parallel

V

= V

= V

= V

I

= I

+ I

+ I

Ohm’s law:

The equivalent resistance

for Parallel resistors:

The equivalent resistance

for Parallel resistors:

Parallel Connection:

R₃

R₂ V_T

R₁

V

I

R



3 1 2

1 2 3

T

e

V

V

V

V

R



R

+

R

+

R

1

1

1

1

e

R



R

+

R

+

R

1

1

1

i

e i

Kirchhoff's Law

I

I

I

I

I

I

50 Ω

35 Ω 62 Ω

6 V

4 V

1 V

Sign Conventions for Emf’s



When applying Kirchhoff’s laws you must

assume a consistent, positive

tracing direction.



When applying the

voltage rule

, emf’s are

positive

if normal output direction of the emf is

with

the assumed tracing direction.



If tracing from

A to B

, this

emf is considered

positive

.

A

E

B

+



If tracing from

B to A

, this

emf is considered

negative

.

A

E

B

-Signs of IR Drops in Circuits



When applying the

voltage rule

,

IR drops

are

positive

if the assumed current direction is

with

the assumed tracing direction.



If tracing from

A to B

, this

IR drop is

positive

.



If tracing from

B to A

, this

IR drop is

negative

.

I

A

+

B

I

Example 4.

Use Kirchhoff’s laws to find the currents in the

circuit drawn to the right.

10 W

12 V

6 V

20 W

5 W

Junction Rule:

I

+ I

=

I

Junction Rule:

I

+ I

=

I

12 = 5

I

+ 10

I

Voltage Rule:



E

=



IR

Consider

Loop I

tracing

clockwise

to obtain:

I

I

I

+

Example 4 (Cont.)

Finding the currents.

6 = 20

I

- 10

I

Voltage Rule:



E

=



IR

Consider

Loop II

tracing

clockwise

to obtain:

10 W

12 V

6 V

20 W

5 W

I

I

I

+

Loop II

12 = 5

I

+ 10

I

6 = 20

I

- 10

I

I

= I

+

I

10 W

12 V

6 V

20 W

5 W

I

I

I

Example 4 (Cont.)

Finding the currents.

12 = 5(

I

+

I



+ 10

I

12 = 5

I

+ 15I

48 = 20

I

+ 60I

42 = + 70

I

I

= +0.6

6 = 20

I

- 10

I

-Example 2. A 3-V battery has an internal resistance of

0.5

W

and is connected to a load resistance of 4

W

.

What current is delivered and what is the terminal

potential difference

V

?

I

= 0.667 A

I = 0.667 A

V

=

E

– Ir

V

= 3 V – (0.667 A)(0.5

W

)

V

= 2.67 V

V

= 2.67 V

R = 4 W

I

+

-E = 3 V

r

R

3 V

4

0.5

I

R r

+



W +

W

Capacitors and Resistors in Parallel

•

The capacitor in the figure is initially uncharged

when the switch S is closed.

•

Immediately after the switch is closed, the

potential is the same at points c and d.

-+

+ -+

+ -+

Capacitors and Resistors in Parallel

•

After some time, the capacitor is fully charged.

-+ ₊

-+

-+

-+ -+

-+

-+

-+ -+

-+

-+

Capacitors and Resistors in Parallel

•

After some time, the capacitor is fully charged.

-+ -+

-+

-+ -+

-+

-+

-+ -+

-+

-+

-+

A₃ A₁ A₂ V₁ V₂ V₄ V₃ 12 V

Charge = ? ? ?

A_{1 ___________} A_{2 ___________} A_{3 ___________}

Charge _{______}

V_{1 ___________} V_{2 ___________} V_{3 ___________} V_{4 ___________}

A_{1 ___________} A_{2 ___________} A_{3 ___________}

Charge _{______}

V_{1 ___________} V_{2 ___________} V_{3 ___________} V_{4 ___________}

8 Amps 4 Amps 4 Amps

0 Coulombs

8 Volts 4 Volts 4 Volts 0 Coulombs

6 Amps 0 Amps 6 Amps

0 Coulombs

6 Volts 0 Volts 6 Volts 6 Coulombs

Im

m

e

d

ia

te

ly

Lo

n

g

ti

m

e

A_{1 ___________} A_{2 ___________} A_{3 ___________}

Charge_{1 _______} Charge_{2 _______} Charge_{3 _______}

V_{1 ___________} V_{2 ___________} V_{3 ___________} V_{4 ___________} V_{5 __________} V_{6 ___________}

2.31Amps 1.39 Amps 0.923 Amps 0 Coulombs

9.23 Volts 2.77 Volts 2.77 Volts

0 Coulombs

A₃ A₁ V₁ V₂ V₃ 12 V A₂

V₄ V₆

6.0 μF

4.0 Ω

3.0 Ω

2.0 Ω

4.0 μF

Charge₂

Charge₁ Charge₃

1.5 μF

0 Volts 0 Volts 0 Volts 0 Coulombs

A_{1 ___________} A_{2 ___________} A_{3 ___________}

Charge_{1 _______} Charge_{2 _______} Charge_{3 _______}

V_{1 ___________} V_{2 ___________} V_{3 ___________} V_{4 ___________} V_{5 __________} V_{6 ___________}

Right Hand Rule

MAGNETIC FIELD

+

(-)

(+)

Curre nt Magnetic

Left Hand Rule

MAGNETIC FIELD

-

(-)

(+)

Current

(-)

(+)

Magnetic Field Lines

Convention

Magnetic Field Lines

are drawn

FROM

NORTH

The Density of Field Lines

Magnetic Field B is sometimes called the flux

density in Webers per square meter (Wb/m

).

Magnetic Field B is sometimes called the flux

density in Webers per square meter (Wb/m

).

D

N

Line density

D

A

Electric Field – “E”

Df

Line density

D

A

Magnetic field flux lines

f

N S

Magnetic Field – “B”

N

E

A

D



D

B

A

D



Magnetic Flux Density

Df

Magnetic Flux density:

D

A

• Magnetic flux lines are

continuous and closed.

• Direction is that of the B

vector at any point.

• Flux lines are NOT in

direction of force but ^.

When area A is perpendicular to flux:

When area A is perpendicular to flux:

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

B

A

 

; =

B

BA

A



Calculating Flux Density When Area is

Not Perpendicular

The flux penetrating the

area A when the normal

vector n makes an angle

of q with the B-field is:

The angle q is the complement

of the angle a that the plane of the area makes with the B

field. (Cos q = Sin a

B

A

a

B

A

q

cos

BA

q

Magnetic

Field Ve

loci ty

Force on Charges

Magnetic Force on Moving Charge

Imagine a tube that

projects charge

+q

with velocity

v

into

perpendicular

B

field.

Magnetic force F on charge moving in B field.

Experiment shows:

Each of the following results in a greater magnetic

force F

: an increase in

velocity

v

, an increase in

charge

q

, and a larger

magnetic field B

.

+

q

Deflection

force

greatest

when path perpendicular

to field. Least at parallel.

+

q

sin

+ + + + + + + + + + + + + +

-E

+

+

· · · · · · · · · · · · · · · ·

· · · · · · · · · · · · · · · · · · · ·

· · · · · · · · · · · ·

· · · · · · · · · · · · · · · ·

B

+

Force is always down

regardless of

motion

!

Force is always

perpendicular to

motion

!

Crossed E and B Fields

The motion of charged particles, such as electrons, can be controlled by combined electric and magnetic fields.

The motion of charged particles, such as electrons, can be controlled by combined electric and magnetic fields.

x x x x x x x

x

+

-e

-v

Note: F_E on electron is upward and

opposite E-field.

But, F_B on electron is

down (left-hand rule).

Zero deflection

when F_B = F_E

B

v

F

E e

-

-B

v

F

-The Velocity Selector

This device uses crossed fields to select only those velocities for which F_B = F_E. (Verify directions for +q)

This device uses crossed fields to select only those

velocities for which F_B = F_E. (Verify directions for +q)

x x x x x x x

x

+

-+q

v Source

of +q

Velocity selector

When F_B = F_E :

By adjusting the E and/or B-fields, a person can select only those ions with the desired velocity.

By adjusting the E and/or B-fields, a person can select only those ions with the desired velocity.

qvB qE



E

v

B

Circular Motion in B-field

The magnetic force F on a moving charge is always perpendicular to its velocity v. Thus, a charge moving in a B-field will experience a centripetal force.

The magnetic force F on a moving charge is always perpendicular to its velocity v. Thus, a charge moving in a B-field will experience a centripetal force.

X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X

+

+ +

+

Centripetal F_c = F_B

R

F_c

The radius of path is:

The radius of path is:

2

;

;

C B

mv

F

F

qvB

R





mv

qvB

R



F



F

mv

R

qB

Mass Spectrometer

+q

R

+

-x -x -x -x -x -x -x -x -x x x x x x x x x x x x x x x x x x x x x x

x x x x x x x x x x x x Photographic plate m₁ m₂ slit

Ions passed through a velocity selector at

known velocity emerge into a magnetic field as shown. The radius is:

The mass is found by measuring the radius R:

Force on a Conductor

Since a current I is charge q moving through a wire, the magnetic force can be given in terms of current.

Since a current I is charge q moving through a wire,

the magnetic force can be given in terms of current.

I = q/t

L

x x x x x x x x x x x x x x x x x x x x x x x x

x x x x x x x

F

Right-hand rule: force F is upward.

F = qvBF = qvB

Since v = L/t, and I = q/t,

we can rearrange to find:

The force F on a conductor of length L

and current I in perpendicular B-field: F = IBLF = IBL

F = BIL

F = BIL

L

q

F

q

B

LB

t

t

+

-B

B

B

B

B

0 1 2 3 4 5 6 7

∝

0

2

I

B

r

m



.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

+

x x x x x x x x x x x

x x x x x x x x

x x x x x x x x x x x

x

x x x x x x x x

Force Between Parallel Wires

I₁

Recall wire with I₁

creates B₁ at P:

Out of paper!

d P

I₂ d

Now suppose another wire with current I₂ in same direction

is parallel to first wire. Wire 2 experiences force F₂ due to B₁.

From right-hand rule,

what is direction of F₂? Force F_DownwardForce F_Downward2 2 is is

F₂

I₂

F₂ B

0 1 1

2

I

B

d

m



Calculating Force on Wires

The field from current in wire 2 is given by:

The force F₁

on wire 1 is: FF11 = I = I11BB22LL

I₂ d _F

1 I1

Attraction

F₂

1 2

L

The same equation results

when considering F₂ due to B₁

The force per unit length for two wires separated by d is:

The force per unit length for

two wires separated by d is:

0 2 2

2

I

B

d

m





2

I

F

I

L

d

m







 







0 1 2

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

x

+

+

x x x x x x x x x

x x x x x x x x x

x x x x x x x x x

x x x x x x x x x

x x x x x x x x x

x x x x x x x x x

x x x x x x x x x

x x x x x x x x x

x x x x x x x x x

Increasing Magnetic Field

FORCE

C

U

R

R

E

N

T

Decreasing Magnetic Field

+

FORCE

C

U

R

R

E

N

T

Mag FIELD

x x x x x x

x x x x x x

x x x x x x

x x x x x x

x x x x x x

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

DC

+

x x x x x x x x x x x

x x x x x x x x

x x x x x x x x x x x

x

x x x x x x x x

+

+

AC

x x x x x x x x x x x

x x x x x x x x

x x x x x x x x x x x

x

x x x x x x x x

Lenz’s Law

Lenz’s law: An induced current will be in such a direction as to produce a magnetic field that will oppose the

motion of the magnetic field that is producing it.

Lenz’s law: An induced current will be in such a direction

as to produce a magnetic field that will oppose the

motion of the magnetic field that is producing it.

Flux decreasing by right move induces loop flux to the left.

N S

Left motion

I

Induced B

Flux increasing to left induces loop flux to the right.

N S

Right motion

I

Example 3: Use

Lenz’s law

to determine direction of induced

current through

R

if switch is closed for circuit below

(

B

increasing

).

R

Close switch. Then what is direction of induced current?

The rising current in right circuit causes flux to increase to the left, inducing current in left circuit that must

produce a rightward field to oppose motion. Hence current I through resistor R is to the right as shown.

The rising current in right circuit causes flux to increase

to the left, inducing current in left circuit that must

produce a rightward field to oppose motion. Hence

Faraday’s Law

The induced voltage in a coil is

proportional to the product of the

number of loops and the rate at

which the magnetic field changes

Magnetic Flux Density

Df

Magnetic Flux density:

D

A

• Magnetic flux lines

 are continuous

and closed.

• Direction is that

of the B vector at

any point.

When area A is perpendicular to flux:

When area A is perpendicular to flux:

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

B

A

 

; =

B

BA

A



Calculating Flux When Area is

Not Perpendicular to Field

The flux penetrating the

area A when the normal

vector n makes an angle

of q with the B-field is:

The angle q is the complement of the angle a that the

plane of the area makes with B field. (Cos q = Sin a)

n

A q

a

B

cos

BA

q

Example 1: A current loop has an area of

40 cm

and is

placed in a

3-T

B-field at the given angles. Find the

flux



through the loop in each case.

A n

n n

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

q

x x x x x x x x x x x x x x x x

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

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

x x x x x x x x x

x x x x x x x x x

x x x x x x x x x

x x x x x x x x x

x x x x x x x x x

x x x x x x x x x

x x x x x x x x x

x x x x x x x x x

x x x x x x x x x

Increasing Magnetic Field

Faraday’s Law:

-

N

t

D

x x x x x x

x x x x x x

x x x x x x

x x x x x x

x x x x x x

Decreasing Magnetic Field

Faraday’s Law:

-

N

t

D

Decreasing Magnetic Field

FORCE

C

U

R

R

E

N

T

Mag FIELD

x x x x x x

x x x x x x

x x x x x x

x x x x x x

x x x x x x

Decreasing Area

Faraday’s Law:

-

N

t

D

Decreasing Magnetic Field

x x x x x

x x x x x

x x x x x

x x x x x

x x x x x

Decreasing Area

Faraday’s Law:

-

N

t

D

D

Decreasing Magnetic Field

x x x x x x

x x x x x x

x x x x x x

x x x x x x

x x x x x x

Decreasing Area

Faraday’s Law:

-

N

t

D

D

Motional EMF in a Wire

L v

I

x

x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x

x x x x x x

B

F

v

EMF:

If wire of length L moves with

velocity v an angle q with B:

Induced Emf E

v sin q q

v

B

ε =

ΔФ

=

=

= BLv

Δt

BΔA

Δt

BLΔx

Δt

BLv

E =

sin

BLv

q

E =

; =

B

BA

A







t

D D

Motional EMF in a Wire

L v I

I

x

x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x

B

F

v

Force F on charge q in wire:

F = qvB; Work = FL = qvBL

EMF:

If wire of length L moves with

velocity v an angle q with B:

Induced Emf E

v sin q q

v

B

Work qvBL q  q

E =

BLv

E =

sin

BLv

q