Electromagnetic Theory PHYS 402. Electrodynamics. Ohm s law Electromotive Force Electromagnetic Induction Maxwell s Equations

Transcription

1 Electromagnetic Theory PHYS 4 Electrodynamics Ohm s law Electromotive Force Electromagnetic Induction Maxwell s Equations 1

2 7.1.1 Ohms Law For the EM force Usually v is small so J = J = σ Current density conductivity force per unit charge 1 σ = ρ resistivity σ E the Ohm s law Empirical, approximate and not applicable to all media. J E = σ For good conductors σ>>1 so E for steady state currents and uniform conductivity 1 E = J = ρ = ε E = σ Just like in the static case charge density is zero in a conductor V = f Laplace equation So can apply

3 In case of a conductor electric field is proportional to the potential difference I=JA= σea= σa(v/l) V=IR Ohm s Law I=JA V l potential current resistance [ in ohms (Ω) ] 3

4 Example: σ uniform uniform I=? R=? V A resistor has across sectional areaa and conductivity σ. if the potential difference between ends is V what is the current. Electric field is uniform within the wire V σ A L I = JA = σ EA = σ A = V R = L L σ A in series L1, L R = R1 + R with the same σ and A you are adding L 1 to L in parallel A, A = + R R R 1 1 = = 1 ( R par R 1 R ρl A + A 1 ) I = V R 1 + V R same V with the same σ and L you are adding A 1 to A 4

5 Example: To prove the field E is uniform A=const V= V=V σ =const J = J nˆ = on the cylindrical surface E nˆ = nˆ V = V = i.e., n V = V ( ) z V z = L V E = V = zˆ L has only z dependance 5

6 Microscopic aspects of the Ohms law In an electric field E, charges accelerate. So why doesn t the current is increasing with time? charges (electrons) undergo collisions and lose speed so their speed won t increase indefinitely, have an average constant velocity. If the average distance a charge travel in between collisions is λ, acceleration a and the average time it takes is t, = average velocity = = not proportional to E! Charges have random thermal velocities which are much larger than the change in speed due to field. Time between collisions are determined by the thermal speed λ 1 aλ t = vave = at = v v thermal thermal 6

7 molecule density nfqλ F nf λq J = n f q vave = = E v thermal m vthermal charge (e) free electrons per molecule Mass (of e) J = nf λq σ mvthermal Thermal velocity increases with temperature, so according to this simple model conductivity decreases with temperature which agree with observations. Collisions convert work done by the field to heat. Work done per unit charge = V ; charge following per unit time =I =! =! Joule s law E 7

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9 7.1. Electromotive Force The current is the same all the way around the loop. If the current is not the same, fields produced by accumulating charges acts in such a way to even out the flow. There are two forces driving the current around the circuit. driving force from the source (ie. From battery, generator.) electrostatic forces (that communicate the influence of the force and smooth out the flow) "#=" $ % electromotive force (emf) = ε = f dl = f dl E dl = s ( ) 9

10 Within an ideal source net force on charges is zero (i.e. &='" '>>1 f=) E = f s The potential difference between terminals a,b V E d f = = d = f l l d l = ε b b ab a a s s a b equal to the electromotive force of the source. So the function of the source (ie. a battery) is to maintain a voltage difference equal to the emf. The resulting electrostatic field drives the current around the rest of the circuit. The battery operates like a pump that moves charges from a lower to a higher electrical potential. 1

11 7.1.3motional emf B Suppose a conductive loop is moved at a velocity v in a magnetic field B as shown in the figure. Magnetic force on charges in the conductor ab: magnetic force per unit charge This produces an emf f F mag, v mag, v = qvb Fmag, v = = vb q ε = f, mag v dl = vbh Which moves charges up along the conductor at a velocity u 11

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13 B So the motion of charges is the sum of v andu. u motion of charges produce a Lorenz force toward left =-. to move the conducting loop to right this force has to be overcome by a similar force " /11 =-. The amount of work done on the loop (note: work is done by the pull force, not by B) h = f d l = (ub)( )sin θ pull cosθ = ubh tan θ = vbh = ε ε = vbh = Bh( dx ) = d ( Bhx) = d ( BA) = d Φ dt dt dt dt ( Φ magnetic flux) 13

14 General proof: electromotive force in the loop 3= dφ = Φ ( t + dt) Φ ( t) = Φ = B da ribbon ribbon da = v dldt = ( v + u) dldt = ( w dl) dt dφ = B ( w dl) dt = ( w B) dl = f dl = ε f mag mag ε = dφ dt 14

15 Example: A metal disk of radius a is in a uniform magnetic field B and rotates with angular velocity ω as shown in the figure. Find the current in the resister. =? ε = a fmag ds f = v B = ωs( w B) = ωsb sˆ mag I = a wsb ds = wba ε wba = = R R 15

16 7. Electromagnetic Induction Faraday experiments and Faraday s law: In the first experiment a loop of wire in a magnetic field was moved, and a current in the loop was observed. This is due to motional emf 3= In the second experiment the magnet was moved holding the loop still, a current in the loop was observed. Here charges are not moving, so there isn t a magnetic force to create an emf. So Faraday suggested that The changing magnetic field must be creating an electric field that produced the observed current (emf). 16

17 A changing magnetic field induces an electric field. Experimentally Faraday found that the emf is equal to the rate of change of flux: 3= ;<= B E dl = da Faraday Law t B E dl = ( E) da = da E = t B t 17

18 In experiment 3 When the strength of the magnetic field was changed (using an electromagnet) a current in the loop was observed. Again the changing magnetic field induces an electric field producing an emf according to the Faraday law. So whenever the magnetic flus through a loop changes, an emf will be induced in the loop

19 Example 7.5: The emf produced when a long cylindrical magnet is passed through a circular wire ring of slightly larger diameter. Lens Law: The induced current flow in such a direction that the flux it produces tend to cancel the change. Nature abhors a change in flux 19

20 7..The Induced Electric Field In a time varying magnetic field If there are no charges (pure Faraday field) These are mathematically equivalent to magnetostatic equations E B = µ J B = t and dφ E d l = similar to Ampere law dt E = ( ρ = ) So similar techniques can be applied as the symmetry permits. B = B dl = µ I enc

21 Example: If the uniform magnetic field B is changing find the induced electric field. d d db s db Φ = π = π = π = dt dt dt dt E dl E. s [ s B( t)] s E If B is increasing E runs clockwise as viewed from above 1

22 Example: A charge ring of linear density λ and radius b is suspended horizontally so that it is free to rotate. In the center there is a magnetic field B up to radius a. What happens when the field is turned off. dφ db E d l = = πa dt dt Torque on dl dn = b F = zb ˆ λedl) db N = dn = zb ˆ λ E dl = zb ˆ λ[ π a ] = zb ˆ λπ a dt Change in the angular momentum on the wheel db dt N dt = bλπ a db = λπ a bb B

23 B B I π s µ ˆ ϕ similar to a solenoid induced E field should be parallel to axis (wire) Example: A infinity straight wire carries a slowly varying current I(t). Find the induced electric field. d d µ I E dl = B da = l ds' dt dt π s ' µ l di s 1 µ ( l di E s) l E( s) l = ds ' == (ln s ln s) π dt s s ' π dt µ ( ) [ l di E s = ln s + K] zˆ π dt K is independent of s but could depend on t. This diverge with s,: at large distances this technique is not valid sine EM disturbances propagate at a finite speed, not instantly. 3

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25 7..3 Inductance Consider two conductive loops, some of the fields lines produced by the current I 1 in loop 1 pass through loop µ dl rˆ B I I 4π r 1 1 = 1 1 Φ = ( A ) da = A dl µ I1 dl 1 = dl 4π R µ dl1 dl M 1 = 4π R 1 1 Φ = B da I Flux through loop 1 1 M 1 mutual inductance µ I1 d A 1 1 = 4π l A 1 is given by R The mutual inductance is a purely geometrical quantity = M I 1 1 M 1 = M 1 = M Φ 1 = M 1 I Φ 1 = Φ if I 1 = I 5

26 Example A short solenoid (length l, radius a n 1 turns per unit length) lies in the axis of a long solenoid ( radius b n turns per unit length). If current I flows through the small solenoid what is the flux through the long solenoid. Calculation flux due to the field of inner solenoid is complicated, but Φ 1 = Φ if I 1 = I, and the field of the long solenoid is uniform. B Φ = n l Φ 1 1 1, per turn 1l π µ π 1 l a n1 nl I = n a B = a n n I = µ π = Φ M a n1 n = µ π l = µ n I I

27 If the current in the loop 1 is changed since that will change the flux through the loop that induces current in loop d Φ di ε = = M 1 dt dt In fact when the current through loop 1 is changed flux in loop 1 also changes, so according to Faraday's law that should induces an emf in loop 1 itself. Since again flux is proportional to current Φ = LI di ε = L dt L self-inductance depends on the geometry of the loop unit: Hendry (H) unit: volt/ampere/second volt.second/ampere This emf in a direction to counter the change in current (back emf) 7

28 Example: Find the self-inductance of a toroidal coil with rectangular cross section (inner radius a, outer radius b and total N turns) Φ = N B da µ B = NI b 1 = N h ds π a s µ N Ih b µ N h b = ln ( ) L = ln ( ) π a π a µ NI πs 8

29 Example: find the current in the circuit I( t ) =? (particular solution) di ε L = dt ε I ( t) = + R IR ke R t L (general solution) ε if I() =, k = R R t τ ε t ε I ( t) = (1 e L ) = R R (1 e If the circuit is disconnected (cut the wire) the current drops instantaneously, producing a large emf (reason for spark when tyrn off the circuit) 9 ε ) R τ = L R time constant

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31 7..4 Energy in a Magnetic Field In electrostatics, the energy in an electric field is given by: 1 ε ε ( ) We = ρvdτ = E Vdτ = E dτ magnetic fields do no work on charges, but to drive (start) a current back emf has to be overcome. This requires some work to be dome and that energy is stored in the magnetic field produced. dwm di d = ε I = L I = ( LI ) Wm = LI = IΦ dt dt dt Φ = B da = ( A) da = A dl s s loop W 1 1 m ( ) I loop A d = = loop A I l d l 1 Wm = ( A J ) dτ V Similarly for a volume current density J

32 1 1 Wm = A Jdτ = A Bdτ V µ V 1 1 = B dτ.( A B) dτ µ V µ V ( A B) da s s 1 W = m B dτ µ V since ( A B) = B ( A) A ( B) = 1 ε ( Vρ) dτ = E d τ = 1 1 ( A J ) dτ = B dτ µ W elec W mag 3

33 Example: Find the magnetic energy stored in a coaxial cable of length l carrying a current I. µ I B = ˆ φ a< s<b B s < a = πs s > b µ I WB = l dwb = l sds µ π µ I b ds = l 4π a s = µ I l b ln( ) 4π a 1 ( ) ( π ) s 1 µ l WB = L I L = π b ln ( ) a 33

34 So far: E = B = ρ ε 7.3 Maxwell's Equations (Gauss Law) B E = t B = µ J (Faraday s Law) (Ampere s Law) B B E = = = = t t as expected B µ = = J = µ ρ Which is zero for steady t currents but not always zero. 34

35 Consider a charging capacitor, here the current passing through the loop is ill defined. I enc is the current passing through a surface that has loop as the boundary. For the surface in between plates, which has I enc = For surface on the other side I enc =I ρ E J = = ε t t ε E ε E = + = t t J So if the Ampere s law is modified such that ε E B = = µ J + = t The extra term is called the displacement current the modified Ampere s law implies that a changing electric field induces a magnetic field. B = µ J + µ ε E t Which would be true all situations J d E = ε t 35

36 for the capacitor problem The E field between capacitors E 1 1 Q = σ = E 1 dq 1 = = I ε σ A t ε A dt ε A E B dl = µ Ienc + µ ε da t Therefore in between plates I enc = and ε ε = = E da I. A I t εa Same as any surface intersecting the wire. 36

37 Maxwell s Equations ρ E = B ε = B E = t B = µ J + µ ε t E Gauss s law Faraday s law Ampere s law with Maxwell s correction Force law continuity equation F = q( E + v B) ρ J = t ( the continuity equation can be obtained from Maxwell s equations ) 37

38 -Q(t) Q(t) Example: Two concentric spherical shells carry charges Q(t) and Q(t), space in between them is filled with a material with conductivity σ. Show that the magnetic field inside is zero. How is it expanded in terms of displacement currents. Due to radial symmetry, magnetic field only has a radial component B. B = B. da = B(4 π r ) = B = 1 Q Q J σ E σ σ = = rˆ I = Qɺ = J. da = 4 πε ε r ohmic currant alone cannot explain B= E σ d Q σq Jd = ε = rˆ = rˆ J + Jd = t 4π dt r 4πε r 38

39 7.3.5 Maxwell s Equations in Matter In matter there are: bound charges ρ = P Change in polarization results in a flow of a current which must be included in the total current σb P nˆ P di = da = da = da t t t P JP = t J p J p σ J b σ b p = = = J p + = t t t t b bound currents = M J b P increased Continuity equation satisfied. charge density current density ρ = ρ f + ρb = ρ f J = J + J + J = f b P P J + M f + t P 39

40 1 Gauss's law E = ( ρ f P) D = ρ f ( D = εe + P) ε Ampere s law ( with Maxwell s term ) B = µ ( J f + M + P) + µ ε t t ( B µ M ) = µ J f + µ ( εe + P) t H = J f + D t In terms of free charges and currents, Maxwell s equations become 1 H = B M µ D B = ρ f E = B = H = J f + D t t D, H and E, B are mixed. P = x E D = εe ε = ε (1 + x ) In these equations M = ε x m e H 1 H = B µ e µ = µ (1 + x m ) E 4

41 7.3.6 Boundary conditions Maxwell s equations in integral form can be used to deduce boundary conditions = D da Q s f, enc B da = s L L d E dl = B da dt s d H dl = I fenc + D da dt s Over any closed surface S for any surface S bounded by the closed loop L 41

42 D, B 1 1 applying above to thin Gaussian pillbox D1 a D a = σ f a D D 1 = σ f 1 = B B D, B E d l E l = B d dt S E E 1 a = = 1 = H1 l H l = I free = K f ( nˆ l ) = l ( K f nˆ ) H H = K f nˆ = = B1 B = K f nˆ µ µ 1 = applying to a thin Ampereian loop = 4

43 (i) 1 = f 1E1 E = f D D σ ε ε σ (iii) E 1 E = (ii) B 1 B = (iv) 1 1 B B = K f nˆ µ µ

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