This preview shows pages 1–8. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 ...
View Full
Document
 Fall '11
 Bhagat
 Physics

Click to edit the document details