Elec and Magnetics Formulas

7/25/2019 Elec and Magnetics Formulas

1/37

REVIEW:

ELECTRIC FORCE,ELECTRIC FIELD, ELECTRIC FIELD LINES, ELECTRIC FLUX, GAUSSS LAW, ELECTRIC POTENTIAL,

CONTINUOUS CHARGE DISTRIBUTIONS (ELECTRIC FIELD, ELECTRICPOTENTIAL, GAUSSS LAW), ELECTRIC CURRENT, MAGNETIC FORCE ON MOVING CHARGES AND WIRES, BIO-SAVART-LAW,

FORCE BETWEEN PARALLEL CURRENT CARRYING WIRES


2/37

ELECTRIC FORCE:

Coulombs law:

force on charge 1 due to charge 2 is

122

21e12 r

r

qqkF =

r

Net force on a charge due to several other charges:

VECTOR SUM of all forces on that charge due to other charges Called Principle of SUPERPOSITON

Each charge exerts a force on charge 1

Resultant force is 4131211 FFFF

rrrr

++=

says net force on charge 1 equals sum of force on 1from 2, force on 1 from 3, and force on 1 from 4


3/37

ELECTRIC FIELD

If the force on 0q at a point is Fr

, then electric field at that point is0q

FE

r

r

=

If the electric field at a point isr

, then the force on 0q at point is EqFrr

0=

Electric field at P due to a point charge is rr

qkE

2eP=

r

oUnit vector rpoints from q P

Electric field pointsaway from positive charge Electric field pointstoward negative charge


4/37

Superposition:

Totalr

at point P due to an arrangement of point charges is the VECTOR SUMof the electric field contributions from all charges around P

Total electric field at P is:

43212

EEEE

r

rqkE

i i

iieT

rrrrr

+++==

o iq is the charge at i

o iris the distance from iq P

o

ir

is the unit vector from iq

P

othe sum is a VECTOR SUM

Did example with electric dipole


5/37

ELECTRIC FIELD LINES:

r

vector at a point in space is tangent to the EFL through that point

Density of EFL is proportional to (magnitude) in that region

oLarger closer packing of lines

EFL start on positive charges and end on negative charges

Number of EFL starting/ending on charge is proportional to its magnitude

Electric field lines do not cross


6/37

Looked at motion of a particle in a uniform electric field:


7/37

ELECTRIC FLUX

General result for Electric Flux through element of area iA

iiiiii AEAErr

== cosE

Total flux through a closed surface:

== dAEAdE n

surfaceclosedover

E

rr


8/37

GAUSSS LAW (general statement):

0

enclosed

surfaceclosed

e

qAdE ==

rr

Powerful way to calculate electric field if we can factor nE out of integral

oTrick is to choose surface so that nE is uniform over all or part of surface

No charge inside: net number of lines leaving = 0 all lines go through

Positive charge inside: non-zero net number of lines leaving lines start on charge inside sphere

Used Gausss Law to calculate electric field around a point charge


9/37

ELECTRIC POTENTIAL

Difference in electric potential between points A and B is:

0q

UVVV AB

==

SO: ==B

A

AB sdEVVV r

r

oPotential difference between two points depends on electric fieldoNote sign and order of integration limits


10/37

EQUIPOTENTIALoAll points on plane perpendicular to uniform

r

field have same electric potential

SIGN OF V

o change in potential energy of 0q when movedfrom A B is

EdqVqU 00 ==

oSays that if A B is in same direction asr

, then 0


11/37

ELECTRIC POTENTIAL DUE TO A POINT CHARGE

Electric potential at distance rfrom a point charge q :

r

qkV e=

assumes that electric potential at infinity is 0

IMPORTANT:

Electric potential V is a scalar.o can just add contributions from different charges

ALLpoints at distance rfrom a point charge q have the same potentialoSpherical surface around point charge is an equipotential surface


12/37

Electric Potential due to multiple point charges: SUPERPOSITION Electric potential at P is

=i i

i

r

qkV e

oNot a vector sum. Contributions to Vadd

as scalars

Can get components ofr

from derivatives of V

Implies: dxdVEx = dy

dVEy = dzdVEz =

Says:r

is always perpendicular to equipotential surfaces


13/37

EQUIPOTENTIAL SURFACES:

like contour mapso valleys centred on negative chargeo hills centred on positive charge

Example: Electricdipole (contoursmarked in volts)

oElectric field linesstart on positive

charge, end onnegative chargeand areperpendicular toequipotentials at

crossing


14/37

Cange in potential energy of charge 0q+ moving from A B

( ) ( )( )V16V2400 == qVVqU AB

So: V400= qU

External force does V400ext =qW


15/37

CONTINUOUS CHARGE DISTRIBUTIONS:CHARGED OBJECTS WITH FINITE SIZE

Strategy:

Break distribution into small charge elements Find contribution to electric potential or electric field from a given element

Sum/integrate to find total Vor

r

ELECTRICAL POTENTIAL FROM A CONTINUOUS CHARGE DISTRIBUTION

total Vat Pis: =

=

chargeallover

ee0

limr

dqk

r

qkV

i i

i

qi

Identify r

dqkdV e= as the contribution to V from the charge element dq


16/37

CHARGE DENSITY

to convert sum/integral over charge elements into a sum/integral over spatialvariables (i.e.x, y, z)

oLinear charge density: L

Q

=

oSurface charge density:Q

=

oVolume charge density:V

Q=


17/37

ELECTRIC FIELDS DUE TO CONTINUOUS CHARGE DISTRIBUTIONS

Approach:

Break charge distribution into small elements (treat each as a point charge) Write vector sum of contributions from elements Take limit as elements become infinitesimally small INTEGRAL

o iris unit vectorpointing from iq toward P

o

iEr

is the contribution to

r

due to iq

=

=

rrdqk

rr

qkE

i

i

i

i

qi

lim

2e

2e0

r


18/37

Two Ways to use GAUSSS LAW:

0

insidee

q=

o relates flux to charge inside for surface of ANYshape

0

inside

surfaceclosedover

qAdE =

rr

ogives a way to calculater

for SPECIALcases

mostly useful if we normal component ofr

is constant over part ofthe surface and can be factored out of the integral


19/37

Using Gausss law to calculate Electric Field

r

Must know directionof electric field from symmetry of problem

o

radial (spherical symmetry) for point charge

o radial (cylindrical symmetry) for a long line ofcharge

ouniform for a large flat sheet of charge


20/37

Must choose Gaussian surface that allows us to calculate =surfaceclosedover

e AdE

rr

oMustbe able to factorr

out of fluxintegral in region of space where wewant to find electric field

Two cases for which we can evaluate surface

AdErr

for all or part of surface

r

uniform andperpendicular to part or all of Gaussian surface

othen flux is = AEAdE nrr

for that part of the surface

r

parallel (tangent) to part of the Gaussian surface


21/37

CONDUCTORS IN ELECTROSTATIC EQUILIBRIUM (19.11)PROPERTIES: ISOLATED CONDUCTOR IN ELECTROSTATIC EQUILIBRIUM

1st

: 0=Er

everywhere inside a conductor

Must be true or else charges would move until 0=E

r

2nd

:Any NET CHARGE on conductor must be on surface

can prove with Gausss Law

ANY NET CHARGE ON THE CONDUCTOR MUST BEON THE SURFACE


22/37

3rd

:Electric field JUST OUTSIDEa CHARGED CONDUCTOR:

MUSTbe perpendicular to the surface. If not, charge flowsalong surface

MUSThave magnitude0

n

=E

4th

:Surface charge density is HIGHESTwhere radius of surface curvature issmallest

Means highestr

at most pointed regions of surface


23/37

ELECTRIC POTENTIAL AT THE SURFACE OF AND INSIDE CHARGEDCONDUCTORS (text section 20.6)

r

surface means that the surface is an equipotential

0=Er

inside says potential Vinside conductor must be

same as potential Vat surface

oentire conductor is an equipotential


24/37

ELECTRIC CURRENT FLOWING CHARGE (Quick Review from Chap 21)

CURRENT: rate at which charge crosses a specifiedsurface

Average currentis t

Q

I

=ave

Q is amount of charge across surface in time t

DIRECTION OF CURRENT:

SAMEas direction of POSITIVEcharges crossingsurface

OPPOSITE direction of NEGATIVEchargescrossing surface

CONDUCTIVITY: current densityEvqnJ

d

==

dv is drift velocity (proportional to electric field)

nis density of charge carriers with charge q

+

+

-

-

Current I


25/37

RESISTIVITY: 1

=

IMPORTANT: resistivity and conductivityare properties of the material

RESISTANCE: relates current through particularobject to pot. diff. across it

Look at wire with cross-sectional area A

oPotential difference between point a and point b is VVV = ab

o l

VAI

= Define resistance

lR

= so that IRV =

Ohms Law: EEJ == Resistance unit is Ohm (): 1= 1 V/A

Resistivity unit is ohm-metre (m) Conductivity unit is (m)-1

A

I (current)a b

l


26/37

MAGNETIC FORCE & MAGNETIC FIELD (Chapter 22)

Magnetic Field Lines RULES

oLines start and stop on poles of a magnet OR lines form closed loops

around current-carrying wires No such thing as a magnetic monopole For magnet lines start at North pole and end at South pole

oMAGNITUDEof magnetic field ||Br

indicated by DENSITY of lines

oDIRECTION of magnetic field at a point is tangent to field lines at point

Direction of magnetic field at some location is the direction that acompass needle would point at that location.

N S


27/37

MAGNETIC FORCE BF

r

on a moving charge BvqF

r

r

r

=B

Cyclotron motion:

Charge with velocity perpendicular to magnetic field moves on

circular path (cyclotron orbit) in plane perpendicular to magnetic field

Radius of orbit is Bq

mvr=

angular frequency for motion (cyclotron frequency) m

qB

r

v==

Period of circular motion Bq

m

v

r

vT

2

2ncecircumfere===

Period independent of speed:

Fr


28/37

Devices that use magnetic force on a moving charge (22.4)

In space withr

andBr

, force on moving charge is Force)(LorentzBvqEqFr

r

rr

+=

VELOCITY SELECTOR

Contains region of space with uniform electric and magnetic fieldsoperpendicular to each otheroperpendicular to path of charged particle

In selector region:

o BvqFr

r

r

=B is up by RHR

o EqFrr

=E is down

Net force on charge is zero IF

EqBvq =

So particles with speed B

Ev= are undeflected

- - - - - - -

Barrier with holeE (down)

B(into screen)

+q

v


29/37

MASS SPECTROMETER

1st: ionize molecules and fragments

2nd: use velocity selector to pick out

fragments with sss BEv /=

3rd: inject ions with speed sv into uniform Br

with magnitude 0B

oRadius of path is 0Bq

vmr s=

SO:s

s

E

BrB

q

m 0=

- - - - - - -

velocity selectorBs,Es

+q

v

+q

v

Detector

r

B0


30/37

CYCLOTRON device to accelerate charged particles to high energy

for charge qin uniform Br

o found Bq

vmr=

operiod of orbit is: Bq

m

v

r

v

circumfT

22.===

NOTICE:period (i.e. timing of kick) does NOTdepend on speed

alternatingvoltage

source

Potential difference acrossgap changes each timeparticle reaches gap

okick raisesKby qV

+q

Bup


31/37

MAGNETIC FORCE ON A CURRENT-CARRYING CONDUCTOR

Force on each charge in wire is BvqF dBr

r

r

=

For cross-section areaAand ncarriers per unit volume,

force on wire segment of length lis ( ) lAnBvqF dBr

r

r

=

Current in wire is AvqnI d= so force on length l of wire is

BlIFBr

rr

=

oVector lr

points in direction of current; magnitude is length of segment

Force on segment sd

r

of arbitrary shaped wire is

BsdIFd Br

r

r

=

Bin

FB

l

A

ds


32/37

Magnetic Dipole Moment

For currentI circulating around loop of areaA

o AIr

r

= is the magnetic dipole moment of the loop

oMagnitude: IA=

oDirection:

vector

r

perpendicular to the plane of the loop direction by Right-hand rule:

fingers curl in direction of current

thumb shows direction ofr

Units of magnetic dipole moment: Am2

For a coil of n loops, nIA=

I A

=IA


33/37

Can express torque as vector product of magnetic dipole moment and field

Br

rr

=

Magnitude of torque is sinBAI=

Direction of Br

r

is into screen/page

True for any current loop in a magnetic field!

2

4

B

F2

F4

= IA


34/37

BIOT-SAVART LAW: 20

4 r

rsdIBd

=

r

r

For drawing, direction of rsd r

is out of screen/page

oSo Bd

r

at Pdue to sdr

points out of screen/page

Magnitude sin dsrsd =r

o

For a given r, contributionsBdr

fromsdr

are maximum for points onplane perpendicular to sdr

oCurrent in sdr

makes NOcontribution to Bdr

at points along direction sdr

ds

r

P

r


35/37

MAGNETIC FIELD AROUND A LONG (INFINITE) WIRE

Result of using Biot-Savart Law:

oMagnetic field lines circle wire no component of Br

parallel to wire

Magnitude of Br

inversely proportional to perpendicular adistance from wire

a

IB

2

0=r

(IMPORTANT RESULT)

Direction of magnetic field lines:

oAnother Right-Hand Rule: thumb alongI ; fingers curl in direction of Br

B

I

Iout


36/37

USE a

I

B

2

0

= TO FIND MAGNETIC FORCE BETWEEN PARALLEL WIRES

Field atI1due toI2isa

IB

2

20

2 =

Force onI1per unit length is a

II

l

F

2

2101 = towardI2 (for currents in same dir.)

Force onI2per unit length is aII

l

F

2

2102 = towardI1 (for currents in same dir.)

oBy Newtons 3rd

law.

a B2

I1

I2F1

F1

I2

I1B2

a


37/37

For parallel conductors, current in same direction:

oWires ATTRACTwith force per unit length a

II

l

F

2

210=

For parallel conductors, current in opposite direction:

oWires REPELwith force per unit length a

II

l

F

2

210=

Provides definition of AMPERE:

For 2 wires, 1 m apart, A121 ==II force per unit length is N/m102 7

Elec and Magnetics Formulas

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