Magnetic forces on a moving charge. EE Lecture 26. Lorentz Force Law and forces on currents. Laws of magnetostatics

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1 Mgnetic forces on moving chrge o fr we ve studied electric forces between chrges t rest, nd the currents tht cn result in conducting medium 1. Mgnetic forces on chrge 2. Lws of mgnetosttics 3. Mgnetic field of current EE Lecture 26 The force on chrge q test due to n electric field E is F = q test E A mgnetic force on chrge results when chrge q test moves t velocity u through mgnetic field: F mg = q test u Here is new field clled the mgnetic flux density. The eqution shows tht its units re N s/(c m); this unit is defined s Tesl (T) or Weber per m 2 (Wb/m 2 ) The mgnetic force is proportionl to u, q test, nd The direction of the force is perpendiculr to both nd u 1 Lorentz Force Lw nd forces on currents The Lorentz force lw combines both the electric nd mgnetic forces on chrge q test : F = q test ( E + u ) Note becuse the mgnetic force is perpendiculr to the velocity of q test, mgnetic forces cnnot result in the mgnitude of u chnging, only the direction of u Consider current flowing through smll piece of wire with vector length l. The units of l re A m = C m/s, the sme s those of q test u. This is becuse current represents moving chrges The force on smll piece of wire crrying current is then df = l For complete circuit we just dd this up over the circuit to get the totl force 2 Lws of mgnetosttics Just s we introduced E to help with describing forces between chrges, helps in describing the mgnetic forces on moving chrges (currents) Just s the behvior of E ws fundmentl to electrosttics, the behvior of is fundmentl to mgnetosttics Just s we needed to introduce D in electrosttics to cpture mteril medium effects, we will introduce new field H in mgnetosttics for this purpose. For simple medi, we use = µh where µ is the permebility of the medium, in Henries/m. The units of H then work out to be Amps/m t turns out most medi hve µ = µ 0. Only iron or iron lloys hve lrge µ vlues. Mgnetic behviors of mterils (when importnt) re usully more complicted thn dielectric behviors 3 4

2 Fundmentl lws of mgnetosttics The lws of mgnetosttics in integrl form re: d = 0 H dl = J d = tot C The first eqution is Guss lw for mgnetic fields ; it looks like Guss lw for electric fields, but there is no right hnd side. This is equivlent to sying tht there re no mgnetic chrges; mgnetic fields do not strt or stop, they re continuous The second eqution is Ampere s Lw ; similr to the pth-independent eqution for electrosttic fields, but in this cse, there is right hnd side Thus line integrls of H fields re not pth independent. The right hnd side is the totl current pssing through the surfce bounded by curve C. Remember right-hnd-rule reltionship between dl nd d for n open surfce Differentil forms Differentil forms of the lws of mgnetosttics re obtined by using the divergence theorem nd tokes theorem. Results re: = 0 H = J The bsence of divergence of gin confirms tht there re no mgnetic chrges; the next flux out of or into ny point is zero The second eqution shows tht mgnetic fields tend to circulte round currents These equtions re exctly dul to our equtions of electrosttics; lthough our procedures re nlogous in studying electro- or mgneto- sttics, the bsic behviors of the fields re different due to these lws n mgnetosttics, we tlk bout finding the mgnetic field produced by currents, nd the mgnetic flux produced in circuit. These effects led to the inductor circuit element 5 Mgnetic field of smll wire n electrosttics, we used Coulomb s Lw to find the electric field produced by point chrge; the nlog in mgnetosttics is the mgnetic field produced by smll current-crrying wire The eqution tht describes this process is clled the iot-vrt Lw: (R) = µ l (R R ) 4π R R 3 The smll wire crries current through vector length l, nd is locted t position R : different from electrosttics becuse currents hve directions while chrges don t This field is observed t position R in medium with permebility µ (usully µ 0 ) Field inversely proportionl to distnce squred, but direction is perpendiculr to both l nd the line between source nd observer 6 Mgnetic fields of currents Mgnetosttics is lso liner, so if we wnt to find the mgnetic field produced by lrger wire, chop it up into mny smll pieces nd dd up the previous eqution for ech everl exmples of this in the book, but not covered in detil in this clss t is not very relistic to think of n isolted smll piece of current, becuse currents relly flow through closed circuits However, understnding the effects produced by one smll piece of current helps with eventully understnding those of more complicted currents Now tht we cn find the mgnetic flux density produced by smll current-crrying wire 1, we cn use our force lw (df = 2 l 2 ) to tlk bout forces between smll wires crrying 1 nd 2 This is the mgnetosttic nlog of the force between point chrges in the originl form of Coulomb s Lw 7 8

3 Exmple mgnetic field clcultion A smll wire crries 10 ma in the ẑ direction through 1 cm distnce. The current is centered t (x = 1 m, y = 0 m, z = 0 m) in Crtesin coordinte system in free spce. Find the mgnetic flux density t the origin produced by this smll wire, nd the force on second 10 ma current flowing the ˆx direction through 1 cm distnce, plced t the origin. olution: Use expression for smll wire field. Observtion point is the origin, so R = 0. The source current l = ẑ(1 cm)(10 ma) is locted t (1,0,0), so R = ˆx. Thus R R = ˆx, with mgnitude 1. Then (R = 0) = µ 0 ẑ ( ˆx) 4π (1) 3 (1) = ŷ (2) EE Lecture Ampere s Lw 2. Using Ampere s Lw to find fields 3. Exmples 4. Field of olenoid Tesl. to both the current direction (ẑ) nd R R = ˆx. Force on the second current t the origin is ˆx(10 4 ) = ẑ Newtons. Wek, nd to dl 2 nd! 9 Ampere s Lw Ampere s Lw in integrl form is: H dl = J d = tot C From this eqution it is cler tht the units of H (the mgnetic field intensity) re A/m; recll = µh The closed contour C nd ssocited open surfce bounded by the contour re completely rbitrry Right-hnd-rule reltionship between dl nd d: curl right hnd fingers in direction dl circultes, then thumb points in direction to choose for d This is n integrl eqution tht describes the line integrl of H over closed contour; lwys true, but generlly not much use for finding the field produced by current However, s for Guss Lw in electrosttics, there re few highly symmetric problems for which Ampere s lw mkes it very esy to find mgnetic fields produced by currents 10 Using Ampere s Lw to find fields To mke Ampere s Lw useful we need the following from symmetry: 1. H hs only dl component long some pth C 2. H l is constnt over pth C f we cn find contours C ( Amperin contours ) for which this is true, then for these contours H dl = LH l where L is the circumference of the pth Ampere s Lw then sttes: so tht H l = tot /L LH l = tot tot is the totl mount of current flowing through surfce bounded by curve C 11 12

4 Ex: Mgnetic field of n infinite thin wire Find the mgnetic field produced by n infinitely long current flowing long the z xis. Think bout this problem in terms of cylindricl coordintes Think bout possible coordinte vritions: results cnnot vry in z since line cn be moved up or down nd nothing chnges. Results cnnot vry in φ since line cn be spun nd nothing chnges. Results cn vry in r Think bout possible field components: No ẑ component since fields re to current direction. No ˆr component since fields re to R R, Only ˆφ component is possible: H = ˆφH φ (r) This hs only dl component tht is constnt if we choose circulr Amperin contours, rbitrry rdius r 0 Line integrl of H is then 2πr 0 H φ (r 0 ). Current enclosed is (note consistent with rh rule with dl = ˆφ, d = ẑd) Mgnetic field is then H = ˆφ 2πr Ex: Mgnetic field of n infinite thick wire Find the mgnetic field produced by n infinitely long current flowing long the z xis. The current is uniformly spred through wire of rdius z C 2 r 2 C 1 r 1 13 olution ymmetries re identicl to previous problem: H = ˆφH φ (r) Agin choose circulr Amperin contours, rbitrry rdius r 0, line integrl of H is still 2πr 0 H φ (r 0 ). However the mount of current enlosed vries s the contour rdius chnges. The current density J in the wire is /(π 2 ) since it is uniformly distributed. For r 0 < we enclose current J d = πr2 0 π, so tht 2 H φ (r 0 ) = r πr 0 = r0 2π 2 For r 0 > we enclose ll the current nd H φ (r 0 ) = H(r) 2πr 0 14 Mgnetic field of solenoid A more complicted exmple involves the mgnetic field produced by n infinite solenoid. A solenoid is mde from wire wrpped tightly round the body of cylinder. Current flows through wire in the ˆφ direction f the solenoid is infinite, we still hve lot of symmetry: no φ or z vritions. Ruling out field components is hrder here - we ll hve to try some new tricks to mke this work N H() = 2π H 1 H 2 r 15 16

5 olenoid Guss Lw for mgnetic fields sys d = 0. Let s choose closed cylinder for our Gussin surfce Cylinder hs two end cps nd its body to produce flux. However, flux out of end cps will cncel since fields cnnot vry in z Flux out of body involves ˆr component of field tht does not vry in φ. Only wy this cn vnish for every surfce is if field ˆr component vnishes. Next use Ampere s Lw with circulr Amperin contour s in our infinite line current cses. ymmetry is the lmost the sme s those cses, except we still hven t ruled out field ẑ component. However circulr Amperin contour involves ˆφ component only; it is constnt over the circle ecuse these contours enclose no current, we find the ˆφ component of the field vnishes olenoid mgnetic field Finl summry of this informtion is H = ẑh z (r) Now we need n Amperin contour tht will involve the ẑ component of the field. Choose rectngulr pths: four sides re two in the ±ẑ direction nd two in the ±ˆr direction ince the field hs no ˆr component, only the sides with the ẑ dependencies mtter First choose the pth to be entirely outside the solenoid, no current enclosed so line integrl is zero. Move one end very fr wy. Field must vnish on other end to mke totl line integrl zero. Result is tht field outside the solenoid is zero everywhere Next choose pth of length L with one leg inside the solenoid, the other outside. Current enlosed is N = nl where n is the number of turns per unit length of the solenoid ecuse field outside the solenoid is zero, we find H dl = H in z L = nl Finish solenoid olve to find Hz in = n; this nswer is correct t every point inside the solenoid. Finl nswer: H = 0 outside solenoid, H = ẑn inside the solenoid The solenoid is very useful for creting lrge mgnetic fields; cn increse field by incresing current or by wrpping wires more tightly (increse number of turns per unit length=n) The mgnetic flux density produced is = µh, which cn be mde lrger by wrpping the solenoid round mteril with lrge µ (iron). Lrger s led to lrger mgnetic forces 1. Mgnetic dipole 2. Mgnetic forces nd torques 3. A simple motor EE Lecture 28 This device is lso known s n electromgnet : use electric currents to crete strong mgnetic effects Rel solenoid will not be infinite: field not completely uniform inside, not completely zero outside; pproximtion is ok for length much lrger thn rdius of solenoid 19 20

6 Mgnetic dipole n our study of electrosttics, we tlked bout n electric dipole, mde from two opposite but equl chrges seprted by distnce d ecuse there re no mgnetic chrges, it is not immeditely cler how to crete mgnetic dipole However, it cn be shown tht smll loop of current produces mgnetic field tht is very similr to tht produced by the electric dipole; thus smll loop of current is mgnetic dipole Consider smll loop of wire crrying current. The wire contour bounds n re A. The (vector) mgnetic dipole moment of this wire is defined s Mgnetic dipole properties As long s the loop is smll (i.e. we re not interested in wht hppens inside the loop), the shpe of the wire doesn t mtter; ll mgnetic effects cn be described in terms of the mgnetic dipole moment Permnent mgnets re mterils where the electron orbits of mny molecules with the mteril become ligned. Net effect is tht there re lrge number of non-cnceling mgnetic dipoles in the medium A permnent mgnet then lso produces mgnetic dipole field E H H + N m = Aˆn where ˆn is vector norml to A, chosen by the rh-rule with the direction of. Units of m re A-m 2 - () Electric dipole (b) Mgnetic dipole (c) r mgnet 21 Mgnetic forces nd torques We know bout the mgnetic force on smll piece of current in mgnetic field: df = dl Now consider mgnetic dipole in uniform = ˆx x field: choose simple squre shpe in x y plne for simplicity. Add up the force on ech piece of wire: we get b(±ŷ ˆx x ) on sides 1 nd 3, (±ˆx ˆx x ) on sides 2 nd 4 y 2 22 Mgnetic torque Force on sides 2 nd 4 vnishes, force on sides 1 nd 3 is b x (±ẑ) These cncel, so the totl force on the loop is zero; the loop will not trnslte However, there is torque exerted on the loop tht will cuse the loop to rotte bout the pivot xis ; recll from mechnics tht torque T = R F; direction of T is xis of rottion Here the two forces produce totl torque of T = ( ˆx/2 ˆx/2) b x ẑ = x (b)ŷ z 1 Pivot xis O b x F 1 d 1 O /2 d 3 F z

7 Torque on mgnetic dipole Thus smll loop of current plced in uniform field experiences no net force, but does experience torque tht will cuse rottion unless the loop is restrined We cn simplify our previous nswer in terms of the mgnetic dipole moment of the loop. Here m = (b)ẑ; using this T = m A simple motor Consider loop of wire tht is llowed to rotte bout pivot xis. The loop is plced in region where = ˆx x, with x constnt This uniform field could be creted either by permnent mgnet or by n electromgnet, for exmple solenoid Very similr to our previous cse, but notice if we llow the loop to rotte, the direction of m is chnging Although we only used this one exmple, this is true for ny mgnetic dipole in uniform field This fct mkes mgnetic forces useful for cusing rottionl motion; used in design of motors, genertors, etc. Mgnetic forces re preferred in these systems becuse iron hs very lrge µ nd is esily vilble, in contrst to high ǫ mterils 25 z Pivot xis F 1 F 2 1 n^ 2 θ 4 b 3 F 4 F 3 26 y x Motor continued Let s cll the rottion ngle of the loop θ; then m = (b) (ẑ sinθ + ˆxcos θ) The torque on the loop is T = m = (b) x [(ẑ sinθ + ˆxcos θ) ˆx] implifying this is T = ŷ(b) x sin θ F 1 m n^ θ /2 /2 sin θ O θ F 3 Finish motor Notice this torque cn be negtive or positive depending on the sign of θ. This will cuse the loop to oscillte bove the pivot xis until it remins stedy t θ = 0 where there is no torque This is kind of like bringing two permnent mgnets close to ech other; they will try to lign f we wnt to keep the loop rotting to crete motor, we ve got to do something bout this. Trick is to reverse current direction in the loop when the torque would like to reverse sign n this mnner the torque cn be kept lwys positive, nd the rottion sustined The reversl of current is ccomplished using broken circulr connector t the end of the loop; polrity of voltge source then gets reversed hlfwy through every rottion 27 28