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Unformatted text preview: FIELDS: GRAVITATIONAL, COULOMB E,B
W V _) _, G : A mass m located in a Gravitational ﬁeld feels a force FG = mGF
—) —) —) ‘ Measure FG , map out GF.
—)  —> A mass M located at the origin creates a GF
,_). —GM
GF:— r2 7"" ” "anathema; QéFe’AZé'Zdr'yszM'
_, —> I' OF GF Through a closed surface is determined solely by the masses enclosed by the
surface. I E : A charge q located in an E ﬁeld feels a force FE '= qE . —> » —> ‘ L) —) ~Measure FE , map out E .
a _,
A stationary charge Q located at r=0 generates acoulornb YE ~ﬁeld ‘
' Q
E = 
—> 47rgar2 1
[+Q (source), —Q (Sink)] consequently, Z a EwAA = :ZQi.
I ~ V ' —) —) . 0 .
FLUX OF E THROUGH A CLOSED SURFACE IS DETERMINED SOLELY BY THE
L .
CHARGES ENCLOSED BY THE SURFACE. ' a TheDevices resulting from this are: (i) Battery *I g ii Capacitor C =. g which leads to energy density . = 18 E2
V . . "i; 2 0 That is the energy contained in 1m3 vol. of E —ﬁeld V
(iii) Resistor R = 7 which leads to i = 0E ,because I = J34; That isrif you apply E to a Cenductor it responds by setting up a current density 1 whose magnitude is determined by the electrical conductivity 0'. ' Magnetic (B): If a £3 ﬁeld is present, a charge q moving with velocity} will experience a
. —) ;
force E3 = q[va] 7
" —> —) Magnimdeong=4ijSin(3,§) E'EWvW/ey, 3’ ‘1' mac/’1‘”? MPEYV‘QMPA 6L Fae) w'ﬁI’C/h am ‘r’fD—aa/chlﬂ
+0 Vg/o “if? gang/ﬂ («QEma :3» mo. UI’ wills/«e Guyana? . out ate] ‘
W i}: a. t n, are, éﬁmje'muaz‘f Qua Direction of F}, 4) rt hand rule {LIEU Thumb,§l ﬁngersﬂig i Palm} Problem I: At i=0, charge q is at origin and has velocity v = we. Turn on a ﬁeld B = BE
") —>
Immediately, it experiences E, along3’). Makes )1 turn, but E3 turns also. Net result is as I
—> “’ —> shown in Figure. q goes around in circle,FB l v always so Kinetic Energy ﬁxed, magnitude of
. —> it *
v does not change.
a Particle moves under inﬂuence of EB = —quI7 . [v&§ are i to one another]
._) _) Note: Plane of orbit i to B ﬁeld.
 —> Note: Uniform circular motion needs a centripetal force. _ B .
angular velocity w = Til—z“ (see picture above)
_) Note: w independent of v. 9/8” Problem II: Force on Current Cm‘ng conductor of lenth 1; Cross. Sec A, charge density me each electron feels FB = (— e) [VD XE] —) # of electrons? neAAl
so total force on conductor F} = neAe A [[VDXE]
,  —+ _) electrons constrained to moVe alOng Also E = neAevD[Alj(B] ‘= I Al‘xg .
—) _ —> 7 —> _ _) ' Problem 1H Rectangular loop of wire suspended in a B —ﬁeld With’Current in loop as shown; Start with
—>
loop in
xypl, qt t=0.
F}. = 1132 an OA.
_, P} 4132 on O’A’ —) . W 4/? Net force is zero. Howe/ng £W%Q. 'z/ b
1131’“ Ile“ {I . 7‘ 52” f
2er 2y (1) 1131;)“; Rotatd 100k 11mm 313on ' F
.7." H T
.—) H ' Note I and B still at right angles to one another, Fl does not change but now rl : 581716 . [Direction of I? also ﬁxed by right hand rule] T = 1le Sin 6 )7 I (2) ta" 0' ‘
Rotate further: ———) r = O (3)
.—)
B
Hﬁ
£xf2=2
H; 2x3€=ﬁ Equations (1), (2), (3) combine to, give 7 = 1122ang Deﬁne Magnetic (Dipole) moment ,u = I l [9192: IAﬁ
3 = 5”“? If the coil has N turns ,u = IAN}?!
' '—) Note: The top (00’) and bottom (AA’) wires have equal and opposite Forces. They will make the
coil out of shape but have no other effect. We have seen that a stationary charge experiences a force in an E ~ﬁeld and a stationary
._) charge creates a (coulomb) E ~ﬁeld. Now We know that a moving chargevexperiences a force in
. a . V a B ~ﬁeld so it is natural to expect that a moving charge will generate a B ~ﬁeld. This is indeed
—) —) the content of the scicalled Biot Savart Law.  V ,X‘ r
_ ﬁ 1 —> —>
{3(2) _ 47: r3 ' ;7 T_ m o 0’ . . . . .
where yo = 475x10 A 1swun1versa1 constant. ThlS equation has the Immedlate
consequences! that for a current I in a conductor of length Al . ' A [X r
' _ & —> —>
g _ 4 ﬂ, r3 We will not use these equations in detail. . CASES OF SPECIAL INTEREST. Single current ~I in a long Wire: What can we say about B ﬂeld at a distance r from the wire?
_>
Notice that Al [+ y“. And the vector Aer is perpendicular to r so we can say that B must curl
—> —> —> —> : around the wire. W W [I out of page] V Looking end on (see picture) we have cylindrical symmetry so 3 can be a function of r only. It
—) turns out that B = L01
2717”
_ ii *
30’ B“ 2717' Thus, B is said to be Azimuthal, it is the direction which curls around I. [check with the sheet on right hand rules]. Next, take the wire and makea circular loop out of it, put it in xz—pl. with center at the origin.
What is the B ﬁeld at y or —y?
_) , .4: ‘J’
The B —ﬁeld lines are shown schematically, it turns out that
M, 21m:2 A
B = B — = — 3 _)( 472: (a2+y2)Ay Once again, we encounter IAﬁ so we can write using magnetic (dipole) moment # 2”
B = __Q___:_>_T
Far awa from , >> a
y I g y in At,
#0 2 ,u 7 ___i (a
B(y)= ——? 6 Magnetic Dipole ' r _, .. '
—> 47: y » e 3:
Recall that for an Electric Dipole W =€qaﬁ'**"**'* W ' ' "
and the E ﬁeld at y is a 4‘ CD
—’ A
1 4qa A
En») = z—Ly 2:» 15
—) 471750 _ a ) “a.
so that at y>>a , _ —? 6 Electric Dipole E00 = 47MB y. owever, there is a major difference here: along y the magnetic dipole has no “size” While 7 electric dipole has length (2a). You can Vsplit the latter but not the former. This has the
' eXtremely important consequence that Whereas electricﬁeld lines start at +9; and end at '—q, magnetic ﬁeld lines close on themselves there is no beginning and no end. CurrentCurrent force y 11911 I Two wires of lengths [1,12 carry currents 11,12. Separation d along y, wires parallel to x. Force on 12 due to [1. To calculate this ﬁrst. Write B1 at location of I2 I A
’50 3:7 69 Iowa mag: 3 jg; (mks “We
0B1 ext/mt L’s ﬁ'fb‘zw ‘0}  v
' E211: IzAlszl
_) —) > ’ _ _ #0111212 A “ 27rd y
Force is attractive
Force on [1 due to 12
“#01 A
7 7 W, , ,,,,,€2,: , , ﬂ, 7 ,7 , , , , , W , ,
El], = 11 Allez
_., —> —>
(“011121 A
_ 27rd y
k
_ 63 B2
a
[1
' — I I A
FOrce is attractive. If [1:12 = lmez‘er the forces/meter F = 2 d are an actionreaction pair.
._) The —sign with a? ensures force is attractive if 11 [2 parallel —> and repulsive when they are
._> p 5/5)“ Lac’tt aPo aw égpb‘ 60 Céeck 5373*
Incidentally, this is a very fundamental equation as it is used to deﬁne the unit of.Current— The
Ampere. That is, ' i 9F _ 11 = 12 ‘= lamp
QM d =1 meter antiparallel '—)
‘ (— Force per meter is 2x10“7 N And the claim is that I should be regarded as a “DIMENSION” in place of Q.
So we can write L) T, M, 49, [IT] ~
rather than   L, T, M 5’, Q ...
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This note was uploaded on 12/28/2011 for the course PHYSICS 122 taught by Professor Bhagat during the Fall '11 term at Maryland.
 Fall '11
 Bhagat
 Physics

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