ELECTRODYNAMICS FOR DAVES IES

Transcription

1 ELECTODYNAMICS FO DAVES IES A guide made for paing the coure of electrodynamic The ued book i Griffith, Introduction to electrodynamic, 3 rd edition I have had thi coure three time, and I till manage to hate it even more - Dave Hozowki My advice: Don t think about what you re doing, jut ue what you think i the eaiet method (Something that Jeppe Juul told me in one leon) Made by Dave Hozowki third year on B.Sc Univerity of Copenhagen 1

2 Content Section 1: The mathematic... 5 Section 2: Electrotatic... 6 Find E z, a circle; λ... 6 Find E z, a charged circle; ς... 6 Find E outide, infinite wire... 6 Find E region, two wire; ρ, ς... 6 Find V inide and outide, Sphere, uniform charge denity, ref. at... 6 Find V at ditance P above line, L... 7 Find V at ditance P above urface charged circle... 7 If we imagine having two circle with radiu a and b with urface charge, with ref point... 7 Find V at ditance P above circle... 7 Find V, ytem of 2 phere, +Q,-Q, with known dipole moment of ytem... 7 Find W, uniform phere... 8 E out and E in for uniform phere... 8 Find C, 2 cylinder... 8 Section 3: Harder Mathematic... 9 Find the force on the charge q (xy plane conducter)... 9 Find V in region, λ- wire, above conducting plane... 9 Find ς, λ- wire on conducting plane... 9 Find p, urface charged pherical hell... 9 Find V approx at point far from phere...1 Find V at arbitrary point r>, ref at, phere, Q...1 Section 4: D-field / Polarization / Surface and denity charge...11 Calculate the bound charge, phere with P...11 Calculate E inide and outide, phere with P...11 Find E at arbitrary, cylinder with P...11 Find E inide and outide, cylinder with P...12 Find D inide and outide, cylinder with P...12 Calculate D inide and outide, phere with P...12 Find D,E,P,V diff, ς b and C in each lap, parallel-plate capacitor with ε r...12 Find E, ς inner, outer, charged phere, Q, Shell of conducting material

3 Find D, E, P, charged phere with Q (a), dielectric layer (a b), conducting layer (b c), vacuum (other)...14 Find F, parallel-plate capacitor...14 Verify that the calculated ρ b and ς b are right in a dielectric material...15 A lat dave-thumb rule...15 Section 5: B-field...16 Find K, when I i uniform, wire...16 Find J, wire...16 Find B in region, if two object are in ytem with known B...16 Find B at center of quare loop, with I. i ditance from center...17 Find B if the figure had n-ide...17 Find B if the figure wa a circle (n -> )...17 Find B if the figure i a hell...17 Find B if the figure i a um of a half circle and a wire...17 Find B at P, olenoid...17 Find B at arbitrary point, two circle...18 Find B outide and inide, I, cylindrical wire, with K or J...18 Find B, inide and outide of two lab...18 Find B, in 3 region of 2 olenoid...19 Find B at Origo of rectangular loop...19 Find B at z of rectangular loop...19 If having to find it at large z Find B at arbitrary point of rectangular loop...2 Show that B ha the right limit at urface of a cylinder...2 Find B, exact, for,,z of a rectangular loop (2a,2b) ek 8/9-prob Find K ϕ,z, B inner, outer, A in,out in a piral cylindrical figure with two urface current...2 Show that B i right at it boundary condition...21 Section 6: Magnetization...22 Find m, rectangular loop...22 Find B due to M, inide and outide, an -cylinder, uniform M...22 Find B due to M, inide and outide, an -cylinder, known M...22 Find B in region between tube, two coaxial cable ( )...23 Find H inide and outide, cylinder with known I

4 Show that H fill the limit, Cylinder with urface current...23 Section 7: Electromagnetic force, emf Find the flux of B, quare loop of wire at ditance from long wire, I Find emf, if loop of wire i pulled away at v...24 Find flux in a big and mall loop in ame ytem, fig Unit lit

5 Section 1: The mathematic 5

6 Section 2: Electrotatic Here we work with the E-field, and what it i when there i a charge nearby. emember, that if you have a point Q, then the E-field will decreae the further you move from thi point. emember that the E-field alway point away from poitive charge. emember: E = V x potential ditance Find Ez, a circle; λ Problem 2.5 E z = 1 λdl 4πε P P = λz 2ε r P 3 z Where, P i the ditance, P = r 2 + z 2 and dl = rdθ, θ =..2π Find Ez, a charged circle; σ Problem 2.6 E z = 1 σdl 4πε P P = zσ 2ε ( 1 z 1 z 2 + 2)z Where, P i the ditance, P = r 2 + z 2 and da = rdrdθ, θ =..2π SMAT TICK: σda = λdl σdl dr = λdl σdr = λ - therefore ue ame formula a before. Find Eoutide, infinite wire Problem 2.13, Symmetry tell u E da = Q ε E = Find Eregion, two wire; ρ, σ Problem 2.16, Symmetry give u Gau law Q ε 2πL = λl ε 2πL = λ 2πε E da = Q ε = 1 ε 2π dφ d L dz E = ρ 2ε E = ρ a2 2ε Q tot =, < a a < < b emember that at < a =, but at > a, where the charge denity end at a, = a. Find Vinide and outide, Sphere, uniform charge denity, ref. at Problem 2.21, Firt find E then potential E da = Q ε = 1 ε 2π dφ π in θ dθ r r 2 dr E = ρr 3ε = Q 4πε r Q, ρ = 3 vol. = Q 4 3 π3 6

7 The potential can be found by the following formula, but keep your tongue in your mouth at the value. V = r E dl Inide, we are inide the phere, o we got to take outide + inner phere : V = E outide dl r E inide dl r V = Q 1 4πε r V = Q 1 4πε 2 Outide, we jut need the outer part, o we get; V = E outide dl 3 r2 2 3 = Q 1 4πε r 2 Find V at ditance P above line, L Problem 2.25: We have to tart determining the ditance from a point dl. V z = 2λ 1 4πε x 2 + z dx 2 L = Q 4πε 1 r = 2λ 4πε inh 1 L z Find V at ditance P above urface charged circle Problem 2.25c: We have to ue the ame formula above and look at what da i. V r = ς 4πε 2π dφ r r 2 + z 2 dr = ς 2ε 2 + z 2 z If we imagine having two circle with radiu a and b with urface charge, with ref point V r = V b V a = ς 2π b r dφ dr = ςz b 4πε r 2 + z 2 ε 2 + z 2 a 2 + z 2 Find V at ditance P above circle Problem 2.5, find potential intead. a V r = 1 4πε λ r dl = λ 2ε r r 2 + z 2 Find V, ytem of 2 phere, +Q,-Q, with known dipole moment of ytem If you have a ytem containing two object at a ditance d/2 from origo, then you can find the dipole moment of the ytem. p = n i=1 q i r i 7

8 Then the potential of the ytem, which look like a dipole can be found by Find W, uniform phere Problem 2.32a) The other method, W = 1 2 ρv dτ, V r< = W = ρ 2 q 1 4πε 2 2π d φ V = 1 p r 4πε r 2 q 1 4πε 2 π in θ dθ 3 r2 2 r 2 3 r2 2 dr = 1 3q2 4πε 5 W = ε 2 all pace E 2 dτ = 4πε 2 r 2 E 2 dr = 4πε 2 r 2 2 E r< dr + r 2 2 E r> dr = 1 4πε 3 5 E out and E in for uniform phere E r< = 1 q 4πε r 2 r, E r> = q r 4πε 3 r Find C, 2 cylinder Problem 2.39, find firt E then V and at lat ue the formula E = 1 2λ 4πε ; = a a < b a < < b emember when calculating potential that you go from a reference point outide of the cylinder therefore b come firt. V = a E tot dl b = 1 4πε 2Q L ln b ln a C = Q V = 2πε L ln b ln a 8

9 Section 3: Harder Mathematic Find the force on the charge q (xy plane conducter) We know that the formula for finding a force on a tet charge i, 1 q F = QE = Q 4πε r 2 But it i crucial to remember the image problem, and that i that we take the ytem we have and mirror it to the according axe. Then you make an imaginary plate at the charge which you want to find, and then meaure the ditance from thi plate. Example, look up problem 3.6. Find V in region, λ- wire, above conducting plane To find the potential in region above the conducting plane, we ue the formula a normal, by V = E dl, E = 1 2λ 4πε d cond to wire But remember, ince we work with a conducting ytem, we will have to take an image of the wire below the conducting plane, o, V r = V +λ, + + V λ, Jut remember that the other image potential from the wire i the ame a the + λ, but jut negative with a negative line charge, but the ditance i till poitive. Find σ, λ- wire on conducting plane Finding the induced urface charge on the conducting plane can be done imply by differentiating, V. ς = ε dv dn, were n i te normal to te conductor plane But remember that to find the induced urface charge, we have to et z to, ince the conducting plane i at z =. Find p, urface charged pherical hell Example 3.9 tell u what the potential i inide and outide a pherical hell, and we can et it to equal the dipole contribution to the potential eq. 3.99, then we hould get the dipole moment V r, θ = k r co θ r [3.86] V 3ε dip = 1 p r 4πε r 2 [3.99] And now you can jut iolate the dipole moment, otherwie the dipole moment could be found by, p = r ρ r dτ 9

10 Find Vapprox at point far from phere Problem 3.28 check with olution manual Find V at arbitrary point r>, ref at, phere, Q We can find V r> by olving Laplace equation in pherical coordinate. Laplace equation: 1 d r 2 dr r2 dv dr + 1 r 2 in θ d dθ inθ dv dθ + 1 r 2 in 2 θ d 2 V dφ 2 = emember that there i ymmetry in the following problem, o only the radial direction give a reult. To ee more, check problem 6) at 14/5-21 1

11 Section 4: D-field / Polarization / Surface and denity charge When we are working with dielectric there are ome golden rule that we hould remember. Polarization can ONLY occur in a dielectric material, unle it i frozen-in. If there i a dielectric material than it mean that you know the order of the charge, ince they are in pair. Calculate the bound charge, phere with P The bound urface charge can be found by, ς = P n Okay, the problem here i that it i hard to remember which way the normal vector point, but remember the following thumb rule: The E-field alway point away from minu, and ince we work with ome that i polarized, all charge are in a dipole, meaning that the minu end of the dipole i at center, and the plu at the urface, but remember, if the phere i charged then jut ee thi charge a one point of the dipole. And remember to tranform the ditance to the ditance of the charge! ρ = P Calculate E inide and outide, phere with P Here we can ue gau law, but have to remember what Q enc i. Where, E da = Q enc ε Q enc r < = r ρ b dτ and Q enc r > = ρ b dτ + ςda And remember that I only included the ditance value, remember the other one, a [..2π+ and *..π+. Normally the lat equation would go out, ince the field outide a polarized material i zero. Find E at arbitrary, cylinder with P Since, the cylinder i polarized; all charge are bound o let u find them. ς = P n And here, it i crucial to find which way the polarization i point, typically in x-direction, meaning that at the back of the cylinder we find minu, and at the front poitive, o at x=, negative, x=l, poitive. Next up the bound denity charge, and if the polarization i contant the bound denity charge i zero. ρ = P Now we have to imagine that cylinder i jut like a long line of urface charged circle, which we calculated to have an E field of, 11

12 E x = ς bx 2ε 1 x 1 x emember that when working with polarization, you can ee the whole ytem a one big dipole, with two different urface charge, one negative at x=, and one poitive at x=l, o the total field would be, E = E x=x + E x=x L Find E inide and outide, cylinder with P We know what the E-field i at an arbitrary point, o let take a point within the cylinder. We would run into a problem, ince the poitive urface charge would go oppoite the x-axe, while the E-field from the negative urface charge would point along the x-axe. E x = ς bx 2ε 1 x 1 x x ς b x L 1 x 2ε x L ς b x 1 +x 2ε x And then jut take the um of thoe two electric field and you hould get, E inide = ς b 2ε x x x L x L x And if it i outide, we jut know that both e-field point in the ame direction, o we loe the lat term (-2). E outide = ς b 2ε x x x L x L x Find D inide and outide, cylinder with P Ue the following formula, and inert the calculated E. D = ε E + P emember, the polarization outide an object (=vacuum) i zero. Calculate D inide and outide, phere with P Thi reult hould alway give, becaue of ymmetry there would be no free charge. Or ele ue the normal formula. Find D,E,P,Vdiff, σb and C in each lab, parallel-plate capacitor with εr Problem 4.18 The electric diplacement can be found by, D da = Q enc = ςa D = ς emember, which way it point within the lab. 12

13 The electric field can be found by, ince we know ε r Polarization can be found by, The potential difference can be found by E = D ε ε r P = ε χ e E = ε (ε r 1)E V = b E dl ΔV = V b V a a emember, the direction of dl to a certain middle point (look up problem). Find all the bound charge ς = P n emember, that negative goe to the poitive charge urface and the poitive goe to the negative in a dielectric material due to the dipole. ρ = P Find the capacitance, C = Q V = ςa ΔV Find E, σ inner, outer, charged phere, Q, Shell of conducting material We have a problem, where we have a phere with radiu a, and between the phere and conducting material there i vacuum, b c, and after a ditance c, we aume vacuum again. E da = Q enc ε E(4πr 2 ) = Q ε r < a: E 4πr 2 = Q enc ε = 1 ε ρ r dt = 1 ε ρ r = a 4 3 πr3 E = 1 4πε Q V r=a 4 3 πr3 r 2 = Q 4πε r a 3 a < r < b: E 4πr 2 = Q enc ε = Q ε E = Q 4πε r 2 E = for b < r < c 13

14 c < r: E 4πr 2 = Q enc ε = Q ε E = Q 4πε r 2 To find the charged urface, we need to remember that it jut a charge divided by the urface area ς inner = Q 4πb 2, ς outer = Q 4πr 2 Find D, E, P, charged phere with Q (a), dielectric layer (a b), conducting layer (b c), vacuum (other) We can find D by uing the Gau law, D da = Q fenc or D = ε E (if no free carge are canged) So here we get the following value D = for b < r < c D = D = Qr 4πa 3 for r < a Q fora < r < b & for c < r 4πr2 To find E now, with different value than before, we can t not expect the ame value in the range from a to b. The new value can be found by D = ε 1 + χ e E E = 1 ε 1 + χ e D The other E-field are the ame, ince they are not affected by the dielectric material. The polarization can only be found in material which i dielectric, therefore it i important to remember that thi can only be true in a to b. P = ε χ e E EMEMBE: It i important to remember that the reaon for not being able to ue the electric diplacement (4.21) i that the equation (4.23), which we ued before, i developed from (4.21)! The formula ued for polarization above include all of E, E(total). Find F, parallel-plate capacitor The force which act on the capacitor i, F = dw dx 14

15 Where, the energy can be found by, W = 1 2 CV2, V = Q C So if Q=Q F = dw dx = Q 2 2C 1 λ 1 λx 2, C x = 1 λx C 1 If V=V F = dw dx = V λ 2 C 1 Verify that the calculated ρb and σb are right in a dielectric material Thi can eaily be calculated, ince we remember that in a dielectric material there are 3 golden rule, een in the entry of thi ection. Here we ue that the total charge i in a dielectric material. Q tot = b ρ b dτ a a b + ς b da + ς b da + Q oter = A lat dave-thumb rule Thi i probably wrong, but it give me control over a ituation that uually give me panic. It i the ituation where you have a figure coniting of a charge, then an area with di-electric and a conducting layer and vacuum otherwie. Here it i important to remember; 1. If having a uniform charge denity, and finding E, remember to et the charge denity (r=a), radiu of inner phere, to obtain the right reult 2. Sign of the urface charge are alway hard to remember, but I jut picture + - and vica vera, if inner phere i poitively charged, then the conducting layer will tart with a minu to neutralize the E-field and therefore have E= (thi i probably a very wrong picture to give!) 3. If having a problem aking quetion in following direction; Find E, Find D, Find E with D, Find P, then don t panic, thi jut mean that: a. To find E, ue gau law E da = Q enc ε b. To find D, ue gau law D da = Q fenc and remember, that you don t know if the Q(enc) are free or bound, o jut ay they are free. So when calculating inner phere, ue the charge denity a if It wa free, with r=a. c. To find E with D, then the only thing that will change i the layer with di-electricum, everything ele i ame a in a. E = D = D ε ε ε r d. To find the Polarization, you have to be calm, and not ue equation 4.21! Why? Maybe becaue you ued it in term of gau law to find P, and why you can do that? I don t know, but ue equation 4.3. P = ε Χ e E Which will be the E found in c., and remember POLAIZATION IS ONLY IN DIELECTICUM. 1 15

16 Section 5: B-field In thi ection it i crucial to have a good feeling of the different right-hand rule. 1. FBI: Thi rule i called FBI, which wa hown to me by Mitch Campbell. It i the imple principle of F(force), B(magnetic field) and I(Inertia/movement velocity), and jut form your hand a a 3D-ytem. 2. Thumb-up: Thi rule I call Thumb up, ince the thumb ymbolize the current (poitive if up), and the B- field a the path the finger would take if you tighten your fit. So if your thumb i up, the B-field will go in a counter clockwie direction. Problem 5.1 Find K, when I i uniform, wire K = I = I l 2πa Find J, wire J = I I = J a a = K ddθ = 2πKa Find B in region, if two object are in ytem with known B Thi problem occur ometime, and might be a mind-tricker/bender, but it i relatively eay. You have to remember that uperpoition i valid, o you can jut add up within the ytem. The two B-field that we know from the book, i, the B-field of an inf. Plane with a uniform urface current, and an inf. Straight wire. B = µ K 2 y 1 for z > +1 for z <, B = µ I 2π φ So if we imagine having two inf. wire with oppoite directional current, and need to find the B-field outide the firt wire, between them, and outide the econd, then it would look omething like thi: So we have a ytem, where we have 3 pot, where we need to find the B-field. The ytem would look, by the help of the econd thumb-rule, like thi: Starting from right (1). In, out, out, in. So, we take our B-field and calculate: 1 B = µ I 2π 1 µ I 2π 2 2 B = µ I 2π 1 + µ I 2π 2 3 B = µ I 2π 1 + µ I 2π 2 So the point i, that you have to look at the direction of the B-field. At (2) they have the ame direction, therefore ame ign, while at (1) 1 i going into the paper, while 2 i going out, therefore oppoite ign. 16

17 Find B at center of quare loop, with I. i ditance from center Here it i crucial to look at the given example 5.5. Here we ee that the magnetic field for one egment line i, B = μ I 4π in θ 2 inθ 1 Where, in a line egment of a quare the line are ¼ of the maximum angle, 2π. Jut remember the ign of the angle. So in concluion the magnetic field in middle of a quare loop i, B = μ I π 2, remember in θ = π 2 = 2 2 Find B if the figure had n-ide A above decribed, we jut know that the um of all angle in a figure i: 2π. So each ide i 2π/n but remember that there are two angle, o you get: θ 2 = θ 1 = π n B = μ I 4π 2 in π n Find B if the figure wa a circle (n -> ) If thi wa the cae, then we have to remember what inu reache when n. B = μ I 4π 2 π n = μ I 2n, in θ ~θ Find B if the figure i a hell You could eaily ue the reult above, and take to be the ditance from inner phere to outer, but let try to do it another way, by the ue of example 5.6 B z = µ I z 2 3 2, were z =, ince we are in plan i of coure, the ditance of the outer phere (from center) ubtracted with the inner ditance. Find B if the figure i a um of a half circle and a wire Then you jut take the figure one by one. So you ue the formula, For circular wire: µ I 2 For a line of wire: EMEMBE: if the wire i infinite B = µ I 2π B = µ I 2π inθ 2 inθ 1 Find B at P, olenoid To undertand the reult, pleae take a look at problem 5.11, which i a hard to do. The reult i, 17

18 B Z = µ I 2 coθ 2 co θ 1 Find B at arbitrary point, two circle A olenoid i jut a piece of wire going in clockwie motion, o here we will jut ue the formula, where we are working with a circle from example 5.6 B z = µ I 2 2 z Now we have to remember to take a point which i in between the circle. The ditance between them can alway be et to L. Then it i jut important to remember which direction the B-field i by the econd right hand rule. And another thing i to remember, if the ditance to the center i negative or poitive. Find B outide and inide, I, cylindrical wire, with K or J Let take a firt example, where all the current i placed at the urface, uniformly K i uniform. K = I l I = K 2π dθ = K2π z emember: If there i only a urface current, the B-field will alway be zero inide an object due to the encloed current. To find the magnetic field, we can ue ampere loop due to ymmetry. B da = µ I enc B2π = µ K2π If inner: B2π =, ince te inner value do not include urface If outer: B2π = µ K2πa µ K2πa 2π, K = I 2πa B = µ I 2π φ If we look at the charge denity, we can ay that it i i.e. J = k EMEMBE! What i k? I = J = I = I I = k a π2 k2π a 3 3 k = 3I 2πa 3 a k2πa3 ddθ = z 3 If inner: ( < a) B da = µ I enc B2π = µ k 2π3 B = µ k = Iµ 2 2πa 3 φ If outer: (>a) B da = µ I enc B2π = µ k 2πa3 B = µ k 3 a 3 3 = Iµ 2π φ Find B, inide and outide of two lab So the problem i that you have two lab, one at z=-a, and other one z=+a. Thee can be charged with ome denity, J. The method would be a the one een above, find the current. Then ue ampere loop, but 18

19 thi time, remember what direction the B-field goe, econd thumb rule. So if it goe counterclockwie, then you get the following, B dl = μ I enc 2Bl = μ J da = μ J2 z dz l dy B = µ Jz y Okay, to undertand the thing we jut made, we will have to take a look at example 5.8, and jut remember that here we have a ytem where the upper and button contribution add up, o we get anwer above. So now we have the function of z in between the lab, and to find the value outide, we jut inert the z-value. Find B, in 3 region of 2 olenoid We have to remember the golden rule of olenoid: Inide there i a B-field of, µ ni and outide it zero. So the problem get fairly eay. Jut remember the econd thumb rule, when working with olenoid. So keep your mind on the ign. Inner olenoid: In between (remember golden rule) B dl = μ I enc B = µ I n 1 + n 2 z B dl = μ I enc B = µ In 2 z Outer region: (EMEMBE GOLDEN ULE) B dl = μ I enc = Find B at Origo of rectangular loop Becaue we are working with a rectangular figure, we can ue the traight line egment. So we ue equation 5.35 B = μ I 4π in θ 2 inθ 1 It i jut important to remember that the figure i exact on each egment to ue it. It i important the you make the rectangular figure into an exact amount of triangle. Sometime, it can be quicker to ue trigonometry. Imagine a 2a, 2b rectangular figure. in θ = oppoite ypotenue B = μ I 4π 2a b + 2b a Find B at z of rectangular loop If w e need to find the magnetic field of a magnetic dipole, it i eential to ue eq B dip r = A = µ m 2coθ r + inθ θ 4πr3 Now it i jut inerting the value you normally know, and be calm with what angle are in which direction. The angle I ue in the rectangular cae wa, π/4. If having to find it at large z1 Then you have to remember that you have to change the pherical coordinate to Carteian 2coθ r + inθ θ r =... + co θ z, θ =... inθz 2 co 2 θ in 2 θ = 2 1 z 19

20 And r in the firt link of the equation i, r = x 2 + y 2 + z 2 So if you have to find B(,,z), you will get; B dip,, z = 2µ m 4πz 3 Find B at arbitrary point of rectangular loop We ue the ame method a at origo, but remember that the magnetic field varie from y-direction and x- direction. So jut take each egment, one by one, and remember what angle you re working with. emember, max i 2π. Show that B ha the right limit at urface of a cylinder To how thi, we need an example, let ay the following value are given for a cylinder: B inner =, B outer = µ K φ, K = Kz To how omething we need to how that the magnetic field i in the right direction, clockwie. B outer B inner = µ K φ φ And remember a tated in the problem, we are at the urface of the cylinder o, =. µ Kφ = µ K z = µ K n And thi lat tatement tate that the limit are correct. Find B, exact, for,,z of a rectangular loop (2a,2b) ek 8/9-prob. 5 Here you ue the equation for rectangular loop, and in ome complicated way, you end up with. B,, z = µ abi π 1 1 a 2 + b 2 + z 2 z 2 + b z 2 + a 2 z Find Kϕ,z, Binner, outer, Ain,out in a piral cylindrical figure with two urface current Thi problem i an exam quetion from 27/6/28, and i evil. So we have to thing to think about, and I will write down how I olved them. So let u tart with the firt quetion, finding the urface current. It i important to remember that the current going in ϕ-direction ha n-turn and i L-long. K φ = di = nli φ dl L = ni φ, K z = di = I dl 2π Now we have the urface current, and the problem i olved. Let not think about them for the next problem, or we will get confued trying to incl. the urface current in the magnetic field. B z dl = μ I enc B z L = µ nli B z,in = µ ni, B zout = 2

21 B φ dl = μ I enc B φ 2π = µ I z B φ,in =, B φout = µ I z 2π It i in the moment, where we are placing the value of inner and outer, that we need to remember where the different urface are. To find the vector potential A, we need to take advance of the ymmetry, meaning that A =, which lead u to having the following equation: A = A φ φ + A z z And we know from the dipole magnetic field that B = A da z d Thi mean that we get the follow equation: φ + 1 d A φ d z A z = B φ d ( < ) µ I z 2π d = µ I z 2π log ( > ) A φ = 1 B z d 1 1 B z d = 1 2 µ ni B z d = µ ni 2 2 ( < ) ( > ) The lat reult for B z cannot be explained, but I refer to your local teacher, I m jut writing it down, ince it i written in the olution to the exam. Show that B i right at it boundary condition Thi i decribed in chapter 5.4.2, the requirement are imple, the B-field you have found mut have ame direction a on the boundary, o. B above B below = µ K n It i crucial to remember that the normal vector i the one going out of the object, perpendicular to the urface of the object. Sphere, r, Cylinder, and Carteian object, z. 21

22 Section 6: Magnetization A good teacher once told me, like dipole work in a pecial way under polarization, o doe magnetic dipole. Magnet Paramagnet Diamagnet Ferrromagnet Attribute Parallel with B-field Oppoite to B-field etain magnetization after external field i removed It can ometime be a good idea when finding the magnetic field, to find the H-field firt. Thi can be done if the magnetization i frozen-in. Then there are no free electron, and eq. 6.2 will yield H=. Then by the ue of eq we can find the magnetic field H = 1 µ B M µ M = B And remember M i zero outide the material. 3. Thi H rule i one, I m almot hundred percent ure of. It tate that if a current i going along the thumb, then the H-field will alo be decribed by your finger clockwie direction, and it i alway the ame direction of B, while M i alway the oppoite direction. Find m, rectangular loop Thi can eaily be found by eq m = I da = I A loop Find B due to M, inide and outide, an -cylinder, uniform M Firt we will look for a urface current, and ince M i uniform then we get K b = M n = M φ Meaning that n i orthogonal on the magnetization. Now we know that there i ome kind of urface current. It ha the direction, which remind of a olenoid. We can already ay that the magnetic field would be outide (golden rule!). And inide we get the following. Inide: B dl = μ I enc A B = µ I enc B = µ I enc A = µ K b = µ M Find B due to M, inide and outide, an -cylinder, known M If M i not uniform but changing, then we have to tackle the problem by finding the current denity a well K b = M n, J = M When thee value are found, then we need to EMEMBE that the current are not the ame at urface and within the cylinder. 22

23 I urf = K dl, I vol = J da And ince there i an active magnetization in the cylinder, then we expect the magnetic field to be zero outide. Inide: B dl = μ I enc B2π = µ I vol < a B = µ k 2 EMEMBE: that you alo have to remember what k i, and in problem 6.8, M = k 2 So the anwer will be, B = µ M Find B in region between tube, two coaxial cable ( ) They are eparated by a magnetic uceptibility, χ,m. We could now ue the method decribed in the introduction to thi chapter. Hdl = I fenc H = I 2π a < < b We can eaily find the magnetic field in an area with magnetic uceptibility. B = µh = µ 1 + χ m H = µ I 1 + χ m 2π φ Find H inide and outide, cylinder with known I The H-field can be found by a practical formula; Hdl = I fenc Then you jut put the current inide and the right value by the ue of the emember: If you don t know what dl i uppoe to be then ue the thumb rule 2 from ection 5. It mean that dl and the current will alway be perpendicular, o if you have a current running along a urface, z, then dl will long along the φ-path. Show that H fill the limit, Cylinder with urface current Thee method are all different compared to the ytem we work in. In thi ytem, we have to remember that there i no magnetization therefore we can write the following expreion up H outer H inner = K f n, n point out of cyl. We found K to be a certain value, let u ay, K = I L φ H outer H inner = I L φ = I L z And thi i true, ince the H-field i in z-direction, which can be found by ampere law: the only free current i at urface. f H dl = I enc, where 23

24 Section 7: Electromagnetic force, emf. Here we are introduced to an electromotive force, which can induce a electric field. Find the flux of B, quare loop of wire at ditance from long wire, I. To find the flux, we ue equation 7.12, Φ = Bda Where, da = dzd and the value of the magnetic field i the one for a long traight wire, B = µ I 2π Φ = dz Bdzd b +a, were b, a are dim. of loop Find emf, if loop of wire i pulled away at v The emf can be found by the flux rule, ε = dφ dt emember: then when a peed i involved will turn into a time dependent function t = + vt. If we imagine pulling it horizontal with cable, we would get a contant flux, therefore no electromotive force. Find flux in a big and mall loop in ame ytem, fig 7.36 Small loop: Here we can find the magnetic field by ue of ex. 5.6, where we found B(z). B z = µ I z Φ = B( = b)da, da = rdrdθ Where, b i the radiu of the big loop. Big loop: Here we need to find the magnetic field by the dipole, which fintate: B = µ m 4πr 3 2coθ r + inθ θ Φ = Bda, da = r2 in θ dθdφ To calculate i much more complex than before, o it i important to know two rule; co θ in θ = inθ 2, in θ = oppoite ypotenue 24

25 Unit lit Electric Field, E = N C Magnetization, Polarization: M = A m, P = C m 2 Surface current, current denity: K = A m, J = A m 2 = C m 2 25