Evaluating this approximately uniform field at the little loop s center which happens to lie on the big loop s axis we find

Transcription

1 PHY 35 K. Solutions for problem set #1. Problem 7.: a) We assume the small loop is so much smaller than the big loop or the distance between the loops that the magnetic field of the big loop is approximately uniform over the little-loop, B big r littleloop) const. 1) Evaluating this approximately uniform field at the little loop s center which happens to lie on the big loop s axis we find B big r littleloop) B big,,z) = µ I big b b +z ) 3/,ẑ, ) where the second equality here is the textbook equation 5.41) for the magnetic field of a circular loop. Consequently, the magnetic flux of the big loop field s) through the little loop is Φ little a little B big,,z) = πa B z big,,z) = µ I big πa b b +z. 3) ) 3/ b) Now consider the magnetic flux of the little loop through the big loop. Again, we assume that the little loop is very small compared to the distance to the big loop, and this allows us to use the dipole approximation, where B little r ) = µ 3m ˆr )ˆr m r 3 4) m = πa )I littleˆẑ 5) is the magnetic moment of the little loop, and the radius vector r is taken relative to the little loop s center. In cylindrical coordinates, the magnetic field 4) becomes where z = z z littleloop. B little s,φ,z) = µ πa )I little z s z +s ) 5/ ẑ + 3z ) s z +s ) 5/ ŝ 6) 1

2 Now let s calculate the flux of this magnetic field through the big loop. Spanning the big loop with a flat disk of radius b in the xy plane hence at constant z = while z = z littleloop, we have Φ big = disk B z little s,φ)d A = = µ πa )I little b b πsds µ πa )I little z s )sds z +s ) 5/ changing variables from s to ν = s +z = µ πa )I little = µ πa )I little = µ πa )I little = µ πa )I little b +z z 3z ν) 1 dν ν 5/ b +z d z ν 3/ + 1 ) ν 1/ z 1 b +z z ) b +z ) 3/ b b +z ) 3/ z s z +s ) 5/ )) 1 z z z 3 7) Altogether, Φ big = µ I little b πa ) b +zlittle. 8) )3/ c) In terms of mutual inductances, eq. 3) from parta) becomes Similarly, eq. 8) from part b) becomes Φ little = M l,b I big 9) where M l,b = µ b πa b + z. 1) ) 3/ Φ big = M b,l I little 11)

3 where M b.l = µ b πa b + z. 1) ) 3/ By inspection of eqs 1) and 1), the two mutual inductances are equal, M l,b = M b,l. 13) Quod erat demonstrandum. Problem 7.5: Let s find the magnetic flux between the wires when the current flows one way through the top wire and back through the other wire. When the distance d between the wires is much smaller than the length l of the whole hairpin configuration, we may approximate each wire as infinitely long for the purpose of calculating the field between the wires. Thus, in the coordinate system where the wires run along the x axis with the bottom wire at z =, y = Y b while the top wire is at z =, y = Y t, the magnetic field in the z = plane the plane of the whole hairpin) is Bx,y,z) µ I π ẑ µ I y Y b π ẑ y Y t. 14) The flux of this field strictly between the wires that is, from the top edge of the bottom wire at y = Y b +ǫ to the bottom edge of the top wire at t = Y t ǫ is Φ = hairpin B d a = l dx Y b ǫ Y b +ǫ dyb z x,y,z = ) = l Y t ǫ Y b +ǫ µ I π 1 1 ) dy y Y b y Y t = l µ I π ln Y t Y b ǫ ǫ = lµ I π ln d ǫ ǫ. ) ǫ ln Y b +ǫ Y t 15) Finally, the self-inductance L of the hairpin-shaped wire is the ratio of this flux through the 3

4 hairpin to the current I in the wire which creates the flux in the first place. Thus L = Φ I = l µ π ln d ǫ ǫ. 16) Note that this self-inductance is proportional to the hairpin s length l. Problem 7.3: By symmetries of the coaxial cable, the magnetic field has form Bs,φ,z) = Bs only) ˆφ 17) where Bs) obtains from the Ampere s law: πs Bs) = µ I[through disk of radius s] = I s /a ) inside the inner wire, I between the wires, outside the whole cable, 18) In the limit of the outer wire and the insulation being very thin compared to the inner wire s radius a, this gives us Bs) = µ Is πa for s < a, for s > a. 19) The self-inductance of the wire is the ratio of the magnetic field s flux to the current, where the flux should be take through the loop formed by the current. For the coaxial cable, the loop surrounds the longitudinal section of the insulator as well as the adjacent part s of the wire. Specifically, the loop goes along the wire s axis in the +z direction the direction of the current in the inner wire) from one end of the wire to the other end, then radially to the outer rim of the outer wire, then in the z direction back to the first end of the wire along the outer rim, and finally radially back to the axis. The plane of this loop is to the 4

5 magnetic field s direction ˆφ, hence the magnetic flux through this loop is Φ = B d a = dz a dsb φ s) = l a ds µ Is πa = l µ I ) where l is the length of the cable. In terms of the self-inductance L = Φ/I of the cable, this formula means the self-inductance per unit length is µ /) = H/m. L = µ l, 1) Problem 7.9: The geometry of the coil is shown on textbook figure 7.34 page 37): rectangular crosssection b a) h, where a is the inner radius of the toroid and b is the outer radius. Assuming the toroid is uniformly and densely wound, the magnetic field vanishes anywhere outside the coil, while inside the coil B inside = µ NI πs ˆφ. ) According to the textbook equation 7.35), the net energy stored in this magnetic field is U = 1 µ whole = 1 µ space toroid B d 3 Vol = 1 µ µ N I = µ N I 8π ) µ NI d 3 Vol πs h h πln b a = I µ N hlnb/a). π b dz a πsds s 3) 5

6 On the other hand, treating the toroid as an inductor with self-inductance L = µ N hlnb/a) π 7.8) see the textbook example 7.11 page 35) for the calculation), we expect the net magnetic energy stored in this inductor to be U = I L = I µ N hlnb/a). 4) π By inspection, this is precisely the same energy as in eq. 3). Problem 7.31: a) Once the switch is in position B, the current flows through an LR circuit which contains nothing but the inductor L and the resistor R. Consequently, the current obeys I = V onr R = V onl R = L di dt 1 R, 5) or equivalently di dt = R L Solving this differential equation, we arrive at It). 6) It) = I exp t ), 7) τ where the time constant τ is τ = L R, 8) and I is the initial current at time t = when the switch was thrown from A to B). This is the same current as flowed through the inductor back when the switch was in position A, thus I = E R, 9) assuming negligible Ohmic resistances of the battery and of the inductor compared to R 6

7 b) The total electric work dissipated inthe resistor once the switch is thrown into position B is W = Pt)dt 3) where the power Pt) is Pt) = R I t) = R I exp t ). 31) τ Therefore, W = RI e t τ = RI τ = I Rτ = I L 3) where the last equality follows from eq. 8) for the time constant. c) The net energy stored in the inductor at the moment the switch was thrown was U = LI. 33) Comparing this formula to eq. 3), we immediately see that this is precisely the net electric work dissipated by the resistor. Problem 7.33: a) When the cylinder rotates with angular velocity ω, the charges on its surface move at linear velocity v = Rωˆφ, which creates a surface current K = σv = σrω ˆφ 34) This current is similar to what we have in a solenoid, so the magnetic field it creates is similar to the solenoid s field: B = { µ σrωẑ inside the cylinder, outside the cylinder. 35) Strictly speaking, this formula applies only for the steady current around the cylinder, which for us means ω = const. However, if the rotation speed ω changes slowly, we may use the 7

8 quasi-static approximation in which Br,t) = { µ σrωt)ẑ for r inside the cylinder, f orroutsidethecylinder. 36) The electric field induced by this time-dependent solenoid-like magnetic field is explained in detail in my notes on magnetic inductionpages 11 1). By symmetry, the induced electric field points in the circular direction ˆφ while its magnitude depends only on the cylindrical radius s, E induces = Es) ˆφ 37) where Es) follows from the Ampere-like law E d l = πs Es) = d Φ[through loop of radius s]. 38) dt Specifically, for the solenoid-like cylinder at hand, { } πs Φt) = B inside for s < R, t) πr for s > R, { } πs for s < R, = µ σr ωt) πr for s > R, 39) hence E induced s) = µ σr dω dt s R s for s < R, for s > R. 4) In particular, at the cylinder s surface itself E = R) = µ σr dω dt. 41) By the way, besides the induced electric field, there is also a static electric field due to the 8

9 charged cylinder, rotation or not, so altogether E static = inside the cylinder, σ ŝ ǫ s outside the cylinder, 4) Er,t) = E static r) + E induced r,t). 43) Fortunately, thestaticfieldisradial,sotheforceitexertsonthecylinder itselfhasnottorque. On the other hand, the force of the induced electric field is in the circular φ direction, which leads to net torque τ = R F net φ = R πrlσ Eφ = R) where l is the cylinder s length = πr lσ E = R) since Eφ static = = πr lσ µ σr ) dω dt = πµ lr 4 σ dω dt. 44) When we spin up the cylinder starting with ω = ), this torque performs negative mechanical work W = t end t τ dφt) = πµ lr 4 σ t end This work depends only on the finite rotation frequency; indeed hence t end t dω dt ωdt = t end t dω dt ωt)dt 45) t ωt) dωt) = 1 ω t end ) 1 ω = 1 ω final, 46) W = πµ lr 4 σ 1 ω final. 47) Consequently, to spin the cylinder from rest to the rotational frequency ω, we need to perform 9

10 positive mechanical work +W = +πµ lr 4 σ 1 ω. 48) b) The work 48) we had to perform in order to spin up the charges cylinder is stored in the magnetic field inside the rotating cylinder. Indeed, the net energy of the magnetic field 36) is U = 1 B d 3 Vol µ whole space = 1 µ cylinder µ σrω ) d 3 Vol = 1 µ µ σrω ) πr l = πµ lr 4 σ 1 ω. 49) By inspection, this is precisely the work 48) need to spin-up the cylinder and create the magnetic field. Quod erat demonstrandum. Problem 7.34: Inside the gap, there is no real electric current, J =, but there is Maxwell s displacement current J d = D t = ǫ r t. 5) Indeed, as the current flows through the wire, electric charges accumulate at the wire ends in the gap which acts as capacitor plates at the rate dq dt = I. 51) The time-dependent charges ±Q on the gap s sides create time-dependent displacement field 1

11 inside the gap Dt) = σt)ẑ = Qt) A ẑ 5) where A is the wire s cross-sectional area. Consequently, the displacement current density through the gap is J d = D t = dq dt ẑ A = I ẑ. 53) A Note: the net displacement current through the gap is the same as the net current I through the wire, I d = J z d A = I. 54) The Maxwell Ampere equation for the magnetic field is B = µ J+J d ) 55) where on the RHS we add the displacement current J d to the electric current J. For the problem at hand, there is electric current J in the wire and the displacement current J d in the gaps, and they seamlessly turn into each other at the gap s borders. Consequently, the magnetic field inside the gap is the same as the field inside the wire at the same distance from the axis: B = µ Is πa ˆφ. 56) Problem 7.4: A parallel-plates capacitor filled with sea water has capacitance C = ǫǫ A d 57) where A is the plates area and d is the distance between the plates. But the capacitor is leaky, it has a rather low DC resistance R = dρ A 58) where ρ is the sea water s resistivity. 11

12 Wren this capacitor is connected to an AC outlet, so the voltage on the capacitor oscillates with time as Vt) = V cosωt), 59) the capacitor s charge also oscillates as Qt) = C Vt) = CV cosωt). 6) Consequently, the displacement field D inside the capacitor oscillates as Dt) = Qt) A = C A V cosωt) = ǫǫ d V cosωt). 61) Therefore, there is Maxwell s displacement current inside the capacitor, of density J d t) = D t = ǫǫ d V ω)sinωt), 6) So the net displacement current through the capacitor is I d t) = A J d t) = ǫǫ A d ωv sinωt). 63) At the same time, due to see water being a conductor, there is a conduction current It) = Vt) R = A dρ V cosωt) 64) Comparing the displacement and conductance currents 63) and 64), we see that they oscillate at phases differing by 9. As to their amplitudes, Idisp max = C ω V = ǫǫ A ω V,, 65) d I max cond = V R = A dρ V, 66) 1

13 they are in ratio I max disp I max cond = R C ω = ǫǫ ρ ω. 67) Specifically, for the sea water with ǫ = 81 and ρ =.3 Ωm, we have ǫǫ ρ F Ω = s, 68) hence at frequency ω = π 4 MHz, I max disp I max cond ) Problem 7.6: a) When the current density Jr) is time-independent but may have non-zero divergence, the continuity equation Jr,t) + ρr,t) = 7) t require ρ/ t to be time-independent. Hence, at any particular point r, the charge density ρr,t) must be a linear function of time, ρr,t) = ρ r) + ρ r) t. 71) b) The problem claims that the quasi-static Coulomb formula Er,t) = 1 ρr,t)gr r )d 3 Vol 7) ǫ where ) Gr r ) = r r 1 r r 3 = r r r = + r ) 1 r r 73) gives the exact electric field for this problem, while the un-modified Biot Savart Laplace formula Br) = µ Jr ) Gr r )d 3 Vol 74) gives the exact magnetic field. 13

14 Our task is to verify that the fields 7) and 74) do obey all the Maxwell equations E = ρ ǫ, 75) E = B t, 76) B =, 77) E B = µ J + µ ǫ t. 78) The Gauss Law for the electric field7) is completely straightforward it works exactly similar to the electrostatic case while the Induction Law is trivial: the electric field 7) has zero curl, while the magnetic field 74) is time independent. The non-trivial calculations involve the divergence and the curl of the magnetic field74). Fortunately, the calculations in my notes on the divergence and the curl of a static magnetic field carry over almost verbatim, with just one extra term in B due to J. We shall see that it is this extra term which accounts for the displacement current term in the Maxwell Ampere equation. For the divergence, eqs. 11) and 13 18) from my notes carry over without any changes: µ Br) = ) Jr ) Gr r ) = µ r Jr ) Gr r ) )) d 3 Vol 79) where r Jr ) Gr r ) ) = Gr r ) r Jr ) ) Jr ) r Gr r ) ). 8) But the current Jr ) here depends onthe r rather thanr, so itscurl WRTrisautomatically zero, while r Gr r ) vanishes since G is a gradient WRT to r. Consequently, r Jr ) Gr r ) ) = 81) and therefore B =. 8) 14

15 For the curl, let me start by repeating eqs. 1) and ) from my notes: µ Br) = ) Jr ) Gr r ) where according to textbook 1..6, rule vi), = µ r Jr ) Gr r ) )) d 3 Vol, 83) r Jr ) Gr r ) ) = G r ) J J r ) G + J r G ) G r J ). 84) Again, since J depends on the integration variable r rather than r, it has zero derivatives with respect to r and its components, thus G r ) J = and G r J ) =, 85) hence r Jr ) Gr r ) ) = Jr ) r Gr r ) ) Jr ) r ) Gr r ), 86) and therefore Br) = µ Jr ) r Gr r ) ) d 3 Vol µ Jr ) ) r Gr r )d 3 Vol. 87) The first term on the RHS here is simply the Ampere s µ Jr). Indeed, r Gr r ) = r ) 1 r r = +δ 3) r r ), 88) hence ) first term = µ Jr ) r Gr r ) ) d 3 Vol = µ Jr )δ 3) r r )d 3 Vol = µ Jr). 89) For the second term in eq. 87), we use G i r r ) r j = G ir r ) r j 9) 15

16 hence Jr ) ) r Gi r r ) = Jr ) ) r Gi r r ) integrating by parts 91) = r Jr )G i r r ) ) + r Jr ) ) G i r r ). Consequently, applying the Gauss theorem to integral of the total divergence here, we obtain ) second = µ Jr ) ) r Gi r r )d 3 Vol term i V = + µ G i r r )J r ) d a µ r Jr ) ) G i r r )d 3 Vol S V 9) The integration volume V here can be any volume enclosing all the currents, but a larger volume up to the whole space is OK too. Let s take a larger volume, so that there are no currents on its surface S; this kills the surface integral in eq. 9) and leaves us with just the volume integral ) second term = µ V r Jr ) ) Er r )d 3 Vol. 93) Back in my notes on the divergence and the curl of a static magnetic field this second term vanished because of divergence-less current, J =. But in the present case, the divergence of the current density does not vanish; instead J = ρ t. 94) Consequently, ) second r) = + µ term V = + t µ ρr,t) t = µ ǫ Er,t) t V. Gr r )d 3 Vol ρr,t)gr r )d 3 Vol where the last equality follows from eq. 7) for the electric field. 95) 16

17 Altogether, we arrive at B = µ J + µ ǫ E t, 96) in perfect agreement with the Maxwell Ampere equation 78). This completes our verification of the Maxwell equations for the fields 7) and 74). Quod erat demonstrandum. 17