Fields_Gravitational

Fields_Gravitational - FIELDS: GRAVITATIONAL, COULOMB E,B W...

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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 («Q-Ema :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, along-3’). 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 xy-pl, 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 sci-called 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 electric-field lines start at +9; and end at '—q, magnetic field lines close on themselves there is no beginning and no end. Current-Current 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 action-reaction 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 anti-parallel '—) ‘ (— 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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Fields_Gravitational - FIELDS: GRAVITATIONAL, COULOMB E,B W...

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