Problem Set #6: 7.4, 7.6, 7.8, 7.12, 7.18, 7.26, 7.30, 7.38 (Due Tuesday, April 22nd)

Transcription

1 Chapter 7 Electrodynamics Problem Set #6: 7.4, 7.6, 7.8, 7.12, 7.18, 7.26, 7.30, 7.38 (Due Tuesday, April 22nd) 7.1 Ohm s law If the electric field is generated by stationary charges, the magnetic fields are generated by moving charges. If F/ is the force per unit charge, then one can define current density as J = σ F (7.1) where σ is the conductivity with as an inverse of resistivity ρ = 1 σ. (7.2) For a perfect conductor, σ conductor = ρ conductor = 0 (7.3) and for a perfect insulator, σ insulator = 0 ρ insulator =. (7.4) In the case when driving force is mainly due to electric field F E (7.5) 69

2 CHAPTER 7. ELECTRODYNAMICS 70 the current density is J = σ E = σe. (7.6) This implies that in a perfect conductor E conductor = J σ conductor =0 (7.7) even with finite current density. Consider a wire of a cross sectional area A, lengthl and conductivity σ, with potential difference V 1 between the ends. Inside the wire the potential must obey Laplace euation 2 V =0 (7.8) with boundary conditions at the ends of the wire and V (0) = V 0 V (z) = V 0 + V 1 (7.9) E ˆn =0 (7.10) or V =0 (7.11) n on the sides of the wire. Since the problem is essentially one-dimensional the solution is linear V (z) = V 1 L z + V 0 (7.12) and thus, E = V 1 L ẑ. (7.13) It is now easy to calculate the current through the wire I J da =(σe) (Aẑ )= σa L V 1. (7.14) Another example is a system of two coaxial cylinders of length L at potential difference V separated by a material with conductivity σ. In this case the electric field inside the material is E = λ 2πϵ 0 sŝ (7.15)

3 CHAPTER 7. ELECTRODYNAMICS 71 and the current I = J da = σ ( E da = σ λ ) 2πϵ 0 sŝ (L2πs) = σλl ϵ 0. (7.16) Moreover is the radii of cylinders are a and b then V = b a E ds = b a λ 2πϵ 0 s ds = λ 2πϵ 0 ln ( ) b a (7.17) and I = 2πσL V. (7.18) ln(b/a) Once again we see the the current is linearly proportional to potential difference which is just a statement of the Ohm s law V = IR. (7.19) Unlike Coulomb s or Ampere s laws, the Ohms law is not exact and may fail in many extreme circumstances such as very low temperatures. However, under normal conditions it is a very good approximation to the observed behavior. It is mainly due to the random collisions that charge particle experience as they move in the wire. If we assume that the randomly oriented thermal velocity of the charged particles is much larger than the drift velocity caused by the potentialdif- ference in the wire v thermal v drift. (7.20) Then the average time between collisions t λ v thermal (7.21) where λ is the mean free path of the charged particles, the average velocity in the direction of the drift v at 2 = Fλ 2v thermal m = λ E. (7.22) 2v thermal m where m is the mass and is the charge of the charged particle (e.g. electron). If we also assume that there are n molecules per unit volume and there aref free electrons per molecule then the current density J =(nf) v = nf2 λ E. (7.23) 2v thermal m

4 CHAPTER 7. ELECTRODYNAMICS 72 There are two forces that drive a charge around a circuit: source force in the battery F source and electrostatic force in the wire Then the combined force is F electrostatic = E. (7.24) F total = F electrostatic + F source. (7.25) But since 0= E dl = F total dl F source dl (7.26) we can define the integral of the total force per unit charge E Ftotal dl = Fsource dl (7.27) known as electromotive force. In the case of ideal battery (σ = ) thetotalforceinthebatteryis F total and the source force in the wire is = J σ =0 (7.28) F source =0. (7.29) Therefore, the potential difference between terminals (a and b) V = b a E dl = b a F source dl = Fsource dl = E. (7.30) Clearly, what we called electromotive force is actually work done by the source to move a unit charge across the battery! The following analogy is useful: for a closed system of pipes the gravitational force plays the role of electric force and force of the pump plays the role of the source force. Most of the batteries use chemical energy to move the charged particle (electrons) across the battery and thus, to generate electric current. However one can also generate current bysimply moving a wire throughmagnetic field. Consider a piece of wire (for example, connected to a light bulb) h = hˆx (7.31) moving with velocity v = vŷ (7.32)

5 CHAPTER 7. ELECTRODYNAMICS 73 in a uniform magnetic field B = Bẑ. (7.33) Then the free charges in the wire will experience magnetic force Fsource E = dl = (v B) dl = vbh. (7.34) This is work, but it is not done by magnetic field, but by whatever is pulling the wire. Once charges started to move in the x direction with velocity uˆx the Lorentz force gets a component in the y direction and thus the work per unit charge F pull = (u B) = ubŷ (7.35) W = Fpull is exactly the same as the electromotive force. dl = ubh (7.36) 7.2 Faraday s Law The first observation of the effect of the time-dependent magnetic fields were made by Faraday who observed that the change in time of the total magnetic flux through some surface creates an electromotive force (and therefore electric current) on the loop bounding the surface, i.e. E E dl = k d B ˆn da (7.37) where E is the electric filed at dl in the coordinate system where dl is at rest. To determine k we can make an assumption of Galilean invariance (i.e. physical laws are invariant under Galilean transformation (x,t) (x = x vt, t = t)). Consider a loop moving with some velocity v (whose spatial derivatives vanish), then d B B ˆn da = ˆn da + (v ) Bda B = ˆn da + (B v) da + v ( B) da B = ˆn da + (B v) dl (7.38)

6 CHAPTER 7. ELECTRODYNAMICS 74 By inserting (7.38) into(7.37) weobtain B (E k (v B)) dl = k ˆn da. (7.39) Alternatively one can consider the same loop but in a coordinate system moving with the loop and then v =0and B E dl = k ˆn da (7.40) where E is the electric field in the new frame. The Galilean invariance implies that E = E k (v B) (7.41) or E = E + k (v B). (7.42) However, is a test charge at rest experiences an electric force F = E (7.43) then in a moving reference frame the same charge represents a current J = vδ(x x 0 ) and experiences both electric and magnetic forces, F = E + J(x) B(x)d 3 x = (E + v B). (7.44) Therefore the Galleon invariance (i.e. F = F )impliesthat E = E +(v B) (7.45) and thus k =1in (7.42). Then Faraday s Law (7.37) takesthefollowing form E dl = d B ˆn da. (7.46) If we choose the reference frame where the current is held fixed then E dl = B ˆn da (7.47) and ( E + ) B ˆn da =0 (7.48) must be satisfied for an arbitrary integration volume. Therefore we can write the Faraday s law in a differential form E + B =0 (7.49) which is a time-dependent generalization of the electrostatic euation E =0.

7 CHAPTER 7. ELECTRODYNAMICS Maxwell euations So far we have derived two electrostatic euations and two magnetostatics euations E = ρ ϵ 0 (7.50) E = 0 (7.51) B = 0 (7.52) B = µ 0 J (7.53) which are to be modified due to Faraday s observation, E = ρ ϵ 0 (7.54) E + B = 0 (7.55) B = 0 (7.56) B = µ 0 J. (7.57) However, this is not a self consistent set of euation for time varyingconfigurations. Clearly, ( B) =µ 0 J =0 (7.58) and the continuity euation (5.36) allowonlysteadystatesconfigurationsof charges ρ =0 (7.59) which cannot be true in general. The problem was fixed by Maxwell who found a small modification to (7.57) whichmakesthenewsetof(maxwell)euationsself-consistent. By starting from the continuity euation (5.36) weget ρ + J = 0 (7.60) (ϵ 0 E) 1 ( B) (ϵ 0 E) µ 0 B 1 c 2 E = J (7.61) = J (7.62) = µ 0 J + C (7.63)

8 CHAPTER 7. ELECTRODYNAMICS 76 where c = (µ 0 ϵ 0 ) 1/2 is the speed of light and C is an arbitrary timeindependent integration constant. If we set C =0,thenweobtain aself-consistentsetofmicroscopicmaxwelleuations 7.4 Induction E = ρ ϵ 0 (7.64) E + B = 0 (7.65) B = 0 (7.66) B 1 E c 2 = µ 0 J. (7.67) The Faraday s law (in the integral form) expresses the induced electromotive force E = E dl = dφ (7.68) in terms of the change of flux of magnetic field Φ= B ˆn da. (7.69) Roughly speaking Faraday s law says that when the magnetic flux through a given loop is changing the induced current tends to produce a magnetic flux to cancel the change (known as Lenz s law). This due to the negative sign in (7.68). This means, for example, that if you place two loops of wire one on top of the other and suddenly turn on the current in the bottom loop, the top loop will jump off in the air. Consider a time-dependent uniform magnetic field B(t) =B(t)ẑ. Then a circular loop of radius s would feel an induced electric field such that E dl = d B ˆn da E (2πs) = πs 2 db E = s db 2

9 CHAPTER 7. ELECTRODYNAMICS 77 or in vector from E = s db 2 ˆφ. The induced electric field can be calculate even if there is no actual loop of wire where the current would be induced. For example if an infinitely long straight wire carries a slowly varying current I(t), thenbyfaraday slaw E dl = d B ˆn da E dl = µ 0l di s 1 2π s 0 s ds or in a vector form E(s 0 )l E(s)l = µ 0l 2π E(s) = [ ] µ0 l di 2π + K ẑ. di (ln s ln s 0) This is the induced electric field that would induce a current if the loop of wire would be present in the system. For a pair of loops of currents (described by l 1 and l 2 )thefluxofthe magnetic field B 1 (generated by current I 1 )throughl 2 is given by Φ 2 = B 1 ˆn da = ( A 1 ) ˆn da 2 = A 1 dl 2. (7.70) where the vector potential is given by A 1 = µ 0 4π J1 (l 1 ) l 2 l 1 d3 r 1 = µ 0I 1 4π dl 1 l 2 l 1. (7.71) and, thus, Φ 2 = µ 0I 1 4π dl1 dl 2 l 2 l 1. (7.72) Then it is convenient to define a (purely geometrical uantity) known as mutual inductance M = µ 0 dl1 dl 2 (7.73) 4π l 2 l 1 so that Φ 1 = I 2 M (7.74) Φ 2 = I 1 M. (7.75)

10 CHAPTER 7. ELECTRODYNAMICS 78 Then by varying a current I 1 one changes flux Φ 2 and thus generates an electromotive force E 2 = dφ 2 = M di 1 (7.76) and vise versa E 1 = dφ 1 = M di 2. (7.77) Another important point is that the changing magnetic flux also generates and electromotive force on the original wire. This can be expressed through (a purely geometrical) self inductance Φ=IL (7.78) where L is calculated using (7.73) byintegratingtwiceofthesameloopto obtain E = L di. (7.79) Now consider a circuit with resistance R and (self) inductance L connected to a battery with electromotive force E 0.Thenthetotalelectromotive force is E 0 + E = E 0 L di (7.80) and according to Ohm s law I(t) = E 0 R E 0 L di = RI. (7.81) The solution of the differential euation is given by an exponent ( 1 exp ( RL t )). (7.82) We have already discussed how the Ohm s law can be derived whenthe collisions of electrons with nuclei are taken into account. Such collision convert the electric potential energy into work corresponding to delivered power E = V (7.83) W = E = V = VIt (7.84) dw = VI = I 2 R. (7.85)

11 CHAPTER 7. ELECTRODYNAMICS 79 Similarly the self-induction creates an electromotive force and thus delivers power dw = IE = LI di (7.86) This can be integrated to obtain an expression of the total work needed to generate current I, W = dw = More generally, one can show that I 0 W magnetic = 1 2µ 0 LI di = 1 2 LI2. (7.87) B 2 d 3 x (7.88) similarly to how we showed that W electric = ϵ 0 2 E 2 d 3 x. (7.89) Then the sum of the magnetic and electric potential energy densities is the energy density of electromagnetic filed given by u = 1 ) (ϵ 0 E 2 + 1µ0 B 2. (7.90) 2