Electromagnetic Induction, Solution Manual

Electromagnetic induction Induced voltage from the tides Assume that the speed of the tidal current is 1 m/s. The force per unit charge in the water is vB = (1 m/s)(5 · 10−5 T) = 5 · 10−5 V/m. This is the effective E field, so the induced voltage across the length of the wire is (using 960 ft ≈ 300m) Eℓ = (5 · 10−5 V/m)(300 m) = 0.015 V, or 15 millivolts. Maximum emf The maximum emf equals the maximum value of dΦ/dt. The flux is given by Φ = N AB cos(ωt + ϕ), where N is the number of turns and A = πr2 is the area. (We’ll assume that the coil is oriented optimally, with its axis of rotation lying perpendicular to the field.) The maximum value of dΦ/dt = −ωN AB sin(ωt + ϕ) is then ωN AB, so ) ( (488) Emax = ωN AB = (2π · 30 s−1)(4000) π(0.12 m) · 2 (5 10−5 T) = 1.7 V. Oscillating E and B Consider a circle of radius r centered on the axis. The flux through this circle is πr2 B . Since B takes the form of B0 cos(ωt+ϕ), the amplitude of dB/dt = −ωB0 sin(ωt+ϕ) is ωB0 . So the amplitude of the emf around the circle is Emax = (dΦ/dt)max = πr2 (ωB0 ). The electric field along the circle is related to E by 2πrE = E. Therefore, the amplitude of E is Emax = (2π · 2.5 · 106 s−1 )(0.03 m)(4 · 10−4 T) Emax ωrB0 = = = 94 V/m. 2 2 2πr (489) Vibrating wire The amplitude of the vibration is x0 = 3 · 10−4 m, the frequency is ν = 2000 Hz, the length of wire within the gap is ℓ = 0.018 m, and the magnetic field is B = 0.5 T. The position of the wire takes the general form of x0 cos(ωt+ϕ), so taking the derivative tells us that the maximum speed is vmax = ωx0 = 2πνx0 . Imagine connecting the ends of the wire with another wire to form a complete loop, which is in fact what you would be doing if you measured the voltage between the ends by connecting them to a voltmeter. Then the movement of the wire implies that an area is being swept out; the area enclosed by the loop is changing. The maximum rate of swept area equals the maximum speed times ℓ. So the maximum induced voltage is ( ) dΦ Emax = = Bℓvmax = 2πBℓνx0 dt max = 2π(0.5 T)(0.018 m)(2000 s−1 )(3 · 10−4 m) = 0.034 V. (490) This result depends linearly on all four of the given quantities, which makes intuitive sense. We can also solve this exercise by looking at the qvB magnetic force on the charges in the wire. Multiplying this force by the distance ℓ along the wire over which it acts, and dividing by q to obtain the work per charge, gives a voltage difference of vBℓ, as above. This is maximum when v is maximum, since B and ℓ are constants. Pulling a frame We could solve this exercise piecemeal, but let’s instead derive a single expression for the force, which will take care of all the questions. Let ℓ be the total perimeter of the rectangle, and let b be the length of the side that sweeps through the field. The current in the frame is I = E/R, where E = dΦ/dt = Bbv, and where R = ρℓ/A = ρℓ/πr 2 . So I = Bbv/(ρℓ/πr2 ) = Bbvπr2 /ρℓ. The force on the trailing side of the frame is F = IBb, and you can show with Lenz’s law and the right-hand rule that this force is directed to the left; that is, it is a drag force. The force required to balance the magnetic drag force therefore equals F = B 2 b2 vπr2 . ρℓ (491) (You can check that this does indeed have units of force.) Since we are ignoring the inertia of the frame, the applied ∑ force must be exactly∑equal to the magnetic force, in F = ma implies that F = 0.) For any particular F magnitude. (If m = 0, then that you pick, Eq. (491) can be solved for the velocity v that the frame will have. We can now answer the various questions. Eq. (491) implies that twice the force means twice the velocity. So a force of 2 N will pull the frame out in half the time, or 0.5 sec. Keeping everything else the same, doubling ρ means halving F (there is half as much current). So a brass frame will be pulled out in 1 sec by a force of 0.5 N. z (side view) current out of page h b/2 B Sliding loop B b/2 x v Figure 130 Doubling the radius increases F by a factor 22 = 4 (there is four times as much current). So a 1 cm aluminum frame will be pulled out in 1 sec by a force of 4 N. (We effectively have four of the original frames stacked on top of each other, each of which requires 1 N.) In Fig. 130 the y axis points into the page. We’ve arbitrarily chosen the current in the wire to flow in the negative y direction (out of the page), but the sign doesn’t matter since all we care about is the magnitude of the emf. √ At the leading edge of the square loop, the magnitude of B is µ0 I/2πr, where r = h2 + (b/2)2 . Only the z component matters in the flux, and this brings in a factor of (b/2)/r. So Bz = µ0 I b/2 µ0 Ib = . 2πr r 4π(h2 + b2 /4) (492) At the trailing edge, Bz has the opposite sign. If the loop moves a small distance v dt, there is additional positive flux through a thin rectangle with area b(v dt) at the leading edge, and also less negative flux through a similar rectangle at the trailing edge. Both of these effects cause the upward flux to increase. Therefore, E= µ0 Ib2 v dΦ b(v dt)Bz = 2bvBz = =2 . dt 2π(h2 + b2 /4) dt (493) The flux is increasing upward. So for our choice of direction of the current in the wire, the induced emf is clockwise when viewed from above, because that creates a downward field inside the loop which opposes the change in flux. For h = 0 (or in general for h ≪ b) E reduces to 2µ0 Iv/π. This is independent of b because the field at the leading and trailing edges decreases with b, while the length of the thin rectangles at these edges increases with b. You can show that our result for E has the correct units, either by working them out explicitly, or by noting that E has the units of B (which are the same as µ0 I/2πr) times length squared divided by time, which correctly gives flux per time. Sliding bar (a) Let v be the instantaneous velocity of the bar. The area of the circuit increases at a rate b(v dt)/dt = bv, so the induced emf is E = dΦ/dt = Bbv. The current is therefore I = E/R = Bbv/R. The general expression for the force on piece of wire (the bar in our setup) is F = IBb, which yields B 2 b2 v/R here. So F = ma gives (including the minus sign because the force opposes the motion, as you can check with Lenz’s law and the right-hand rule) ∫ t 2 2 ∫ v ′ B 2 b2 v dv B b dv dv =⇒ − =⇒ − dt′ = =m ′ dt dt R 0 mR v0 v ( ) v mR B 2 b2 t = ln =⇒ v = v0 e−t/T , where T ≡ 2 2 . (494) − B b mR v0 −F = m =⇒ (You can check that T has units of time.) We see that the velocity decreases exponentially, so technically the rod never stops moving (in an ideal world). This exponential decay of v is a familiar result for forces that are proportional to (the negative of) v . (b) The total distance traveled in the limit t → ∞ is ∞ ∫ ∞ ∫ ∞  v0 mR v0 e−t/T dt = −v0 T e−t/T  = v0 T = 2 2 . x= v dt = B b 0 0 0 (495) So the rod travels a finite distance in an infinite time. (c) The initial kinetic energy of the rod is mv02/2. This must eventually show up as heat in the resistor, so let’s check this. The instantaneous power dissipated in the resistor is P = I 2 R, where I is given above as I = Bbv/R = (Bbv0 /R)e−t/T . The total energy loss in the resistor is therefore ∞ ∫ ∞ ∫ B 2 b2 v02 T −2t/T  B 2 b2 v02 ∞ −2t/T dt = − e e I 2 R dt =  R R 2 0 0 0 B 2 b2 v02 T 1 2 B 2 b2 v 20 mR (496) = = mv0 . = 2B 2 b2 2 2 R R Ring in a solenoid (a) The magnetic field inside the solenoid is B(t) = µ0 nI(t) = µ0 nI0 cos ωt. Faraday’s law applied to the given ring yields E =− dB dΦ = −πr2 = πr2 µ0 nI0 ω sin ωt. dt dt (497) With the given positive direction of I, the right-hand rule gives the positive direction of B as upward, and then also gives the positive direction of E as counterclockwise when viewed from above (as for I). The current in the loop is Iloop (t) = E/R = (πr2 µ0 nI0 ω/R) sin ωt. (b) The force on a little piece of the ring is F (t) = Iloop (t) dl × B. With positive I counterclockwise and positive B upward, this force is radial and equals F (t) = πr2 µ0 nI0 ω πr2 µ20 n2 I 02 ω dl sin ωt cos ωt. (498) sin ωt · dl · µ0 nI0 cos ωt = R R The force is radially outward if this quantity is positive, inward if it is negative. Since sin ωt cos ωt = (1/2) sin(2ωt) we see that the force is maximum outward when ωt = π/4 (plus multiplies of π), and maximum inward when ωt = 3π/4 (plus multiplies of π). (c) Since the force lies in the horizontal plane, it serves only to stretch/shrink the ring (negligibly, if the ring is rigid). A loop with two surfaces Surface (a) has two sides, whereas surface (b) has only one side; it is a Mobius strip. The two-sided surface is the one we must use to calculate the flux through the loop. The Mobius strip can’t be used, because the direction of the area vector da isn’t well defined; if you travel around the strip and return to your starting point, da will point in the opposite direction. (a) If surface (a) is stretched vertically at the twist and then viewed from the side, it looks like the surface shown in Fig. 131(a). So it can be considered a two-turn coil. The three-turn coil in shown in Fig. 131(b), and so on for N turns. If you trace along the wire as it spirals upward, it takes the same shape as the windings of a solenoid. All of the “pancakes” have the same orientation, so the total flux through all of them is simply N times the flux through one. The pancakes are truly all part of a single surface. We therefore obtain the result that a coil of N turns has an emf that is N times that of a single loop. (b) Induced emf in a loop Figure 131 The magnetic field due to an infinite current-carrying wire is µ0 I/2πr, so the fields at the near and far sides of the rectangle are B1 = µ0 (100 A) = 1.33 · 10−4 T, 2π (0.15 m) and B2 = µ0 (100 A) = 0.8·10−4 T. (499) 2π(0.25 m) The induced emf is therefore E= dΦ = wv(B1 − B2 ) = (0.08 m)(5 m/s)(0.53 · 10−4 T) = 2.1 · 10−5 V. dt (500) The flux is downward and decreasing, so E will be in the direction to drive a current that would make more flux downward. The current is therefore clockwise when viewed from above. Let’s now estimate roughly how large the resistance must be to make the effect of the current in the loop negligible. The current in the loop at any instant is I ′ = E/R. This causes a field B ′ and a flux Φ′ through the loop. Because E is changing with time as the loop moves away from the wire, Φ′ is changing too, resulting in an extra emf E ′ , which we have so far ignored. The question is, how large must R be so that E ′ is negligible compared with E ? As a very rough estimate, we have B ′ ≈ µ0 I ′ /2πd, where d is a typical dimension of the loop, say, d = 5 cm. The flux1 is then Φ′ ≈ B ′ A = (µ0 I ′ /2πd)wℓ, where ℓ is the length of the loop (we could set ℓ ≈ w ≈ d here since we’re being rough, but we’ll keep them separate). The (very rough) time characteristic of the change in Φ′ is τ = h/v, where h is the mean distance from the loop to the wire (20 cm), and v is the speed of the loop. So in order of magnitude, we have (using I ′ = E/R) E′ = dΦ′ µ0 Ewℓv Φ′ µ0 I ′ wℓ v · = . ≈ = 2πRhd 2πd h dt′ τ (501) ′ Our goal is to have E ≪ E , which is equivalent to µ0 wℓv 2πRhd =⇒ R µ0 wℓv 1 =⇒ R ≫ 2πhd ) ( m 4π · 10−7 kg C2 (0.08 m)(0.1 m)(5 m/s) ≫ = 8 · 10−7 Ω. 2π (0.2 m)(0.05 m) ≪ (502) This is roughly equal to 10−6 Ω, so the condition that the current in the loop is negligible (more precisely, the condition that E ′ ≪ E) can be written as R ≫ 10−6 Ω. This is a rather small resistance, so this bound is easily satisfied by a typical copper wire. We can alternatively write the condition in Eq. (502) as µ0 wℓ 1 h · ≪ . v 2πd R (503) ′ But from the definition of the self-inductance L, the above expression for Φ yields L = µ0 wℓ/2πd. So the condition can be written as L/R ≪ τ . In other words, the inductive time constant of the loop itself, L/R, should be short compared with the time scale of the change of the externally induced emf. Note that in the case where all of the above lengths (w, ℓ, h, d) are of the same order of magnitude (which is the case here), the condition reduces to the simple expression (ignoring the 2π): R ≫ µ0 v. We therefore see that the smallness of the above lower bound on R (in SI units) comes from the smallness of µ0 (in SI units). Work and dissipated energy The induced emf is E = wv(B1 − B2 ), so the current is I = E/R = wv(B1 − B2 )/R. From Lenz’s law it is counterclockwise when viewed from above. From the righthand rule, the forward-directed force that must be applied to the loop to balance the retarding magnetic force and keep the loop moving at constant speed is F = IB1 w − IB2 w = Iw(B1 − B2 ). Using (B1 − B2 ) = IR/wv, the rate at which work is done is therefore ( ) IR F v = Iw(B1 − B2 )v = Iw v = I 2 R, (504) wv 1 Strictly speaking, we should expect the flux of B ′ to involve a factor like ln(rloop /rwire ); see the comments at the end of Section 7.8. But unless the wire is extremely thin, this logarithm won’t be a very large number. which is the power dissipated in the resistance, as we wanted to show. In Fig. 7.14 the energy that is dissipated in the stationary loop has to be supplied by whatever agency is moving the coil. A force is indeed required to move the coil because of the magnetic field arising from the induced current in the loop. To see how this works out in a simple case, let the coil have the same rectangular shape as the loop, but with N turns. And let the coil have current I0 . Then the difference in the B fields (due to the loop) at the leading and trailing edges of the coil is smaller than the difference in the B fields (due to the coil) at the leading and trailing edges of the loop by a factor of I/I0 (because these are the currents that produce the B fields) and also by a factor of N . So the F v = Iw(B1 − B2 )v relation in Eq. (504) for the rate at which work is done in moving the N -loop coil becomes ( ) I 1 F v = N · I0 w (B1 − B2 ) · · v = Iw(B1 − B2 )v, (505) I0 N which agrees with the expression in Eq. (504), and hence equals I 2 R. Sinusoidal emf If the field is uniform, the emf will be sinusoidal, regardless of the shape of the planar loop (assuming a constant rate of rotation). To see why, imagine slicing the loop into strips oriented perpendicular to the axis of rotation. Each strip is essentially a thin rectangle, so the flux through each strip varies sinusoidally. The fluxes through all the different strips have the same frequency and phase, so the total flux also varies sinusoidally. nonzero B 3 4 2 ω 1 5 loop B Figure 132 3 2 1 Actually, the emf will be sinusoidal even if the loop isn’t planar. This is true for the following reason. Consider a nonplanar surface bounded by the loop, and divide it into many tiny planar patches. The flux through each planar patch varies sinusoidally, so it takes the form of Φi sin(ωt + ϕi ). You can quickly verify, by using the trig sum formula for sine, that the sum of two such fluxes (with arbitrary Φi and ϕi values) again takes the same form. Therefore, by mathematical induction, the sum of an arbitrary number of such fluxes takes this form. The point is that the frequency of all the individual sinusoidal fluxes is the same, which means that there is no way for any other frequency to creep into the total flux. Alternatively, imagine looking at the loop along the line of the B field. If you close one eye, so that you don’t have any depth perception, then for all you know the loop could be planar. However, if the field isn’t uniform, then the emf need not be sinusoidal. This is easily seen in an extreme case where B is zero except in a given region; see Fig. 132. The emf will be zero except when the loop is sweeping through that region, so the emf will look something like the curve in Fig. 133. The nonzero parts of the curve belong to a sine curve. The numbers shown correspond to the orientations in Fig. 132. In the case where the field is created by a ring (instead of a solenoid which creates a uniform field), the field will weaker in the middle and stronger near the ring. The former fact will lessen the peak in the standard sinusoidal curve, and the latter fact will increase E near its zeros. The emf will therefore look more like a “square wave.” Emfs and voltmeters 4 5 Figure 133 The three cases are: (a) The electric field caused by the changing flux is directed clockwise around the ∫b solenoid, so the line integral a E · ds along path 1 equals +E0 . The voltage ∫b difference Vb − Va ≡ − a E · ds therefore equals −E0 . Path 2 encloses zero changing flux, so Va − Vb equals zero along path 2. We are assuming that a and b are infinitesimally close to each other. (b) We now have a simple electrostatic setup, so the voltage differences are path independent. Point a at the higher potential, so we have Vb − Va = −E0 , and Va − Vb = E0 (along any paths). (c) The answers here are the same as in part (b), except with −E0 replaced with −E0 /N , and E0 replaced with E0 /N . So in the N → ∞ limit, the potential differences are zero. The results are compiled in this table: Case (a) Case (b) Case (c) Path 1 Vb − Va −E0 −E0 0 Path 2 V a − Vb 0 E0 0 Cases (a) and (b) are equivalent for path 1, but not for path 2. Cases (a) and (c) are equivalent for path 2, but not for path 1. Cases (b) and (c) are different for both paths. The total voltage drop around a closed path is zero in cases (b) and (c) (consistent with the fact that these are electrostatic setups), but nonzero in case (a). Getting a ring to spin The flux through the ring at any given time is Φ = πa2 B. From Faraday’s law, the induced emf is E ∫= dΦ/dt = πa2 (dB/dt) (ignoring the sign). But by definition, the emf is also E = E · ds = 2πaE. The tangential field is therefore E = E/2πa = (a/2)(dB/dt). A little element of charge dq feels a tangential force E dq, and hence a torque Ea dq. So the total torque is τ = Eaq = (qa2 /2)(dB/dt). The final angular momentum acquired by the ring is therefore ∫ ∞ ∫ ∞ 2 ∫ qa dB qa2 B0 qa2 0 L= τ dt = dB = − . (506) dt = 2 2 2 dt B0 0 0 We haven’t been keeping track of signs, so the minus sign here doesn’t mean much. The correct statement to make is that Lenz’s law implies that the ring will spin in the direction that has a positive right-hand-rule relation to the direction of the initial magnetic flux. So you can quickly show that if q is positive, the direction of L is the same as the direction of B0 . The angular momentum can be written as L = Iω = (ma2 )ω, so we have ω= L qB0 , = 2m ma2 (507) as desired. We see that L depends only on the net change in B, and not on the rate at which this change comes about. Intuitively, if B changes more slowly, then E (and hence the torque) is smaller, so the angular momentum increases at a lesser rate. But the process takes longer, so the increase goes on for a longer time. These two competing effects exactly cancel. Note that ω is independent of a. This is due to the fact that E is proportional to a, which means that the linear acceleration in the tangential direction, and hence the tangential speed v at a given time, is proportional to a. The angular velocity, ω = v/a, is therefore independent of a. (The E ∝ a fact quickly follows from Faraday’s law, although it isn’t terribly intuitive. In the end, it comes down to whether or not you find ∇ × E = −∂B/∂t to be intuitive.) Faraday’s experiment We have enough information to calculate the resistance of each coil, if we know the resistivity of copper. From Table 4.1 let’s use the rough value of 2·10−8 ohm-m at room temperature. The length ( of each) wire is ℓ = (203 ft)(12 in/ft)(0.0254 m/in) = 62 m. And the radius is r = (1/40) in (0.0254 m/in) = 6.4 · 10−4 m, which gives a crosssectional area of A = πr2 = 1.3 · 10−6 m2 . The resistance of each wire is then R= twine coil #1 coil #2 1/20 inch ρℓ (2 · 10−8 ohm-m)(62 m) ≈ 1 ohm. = 1.3 · 10−6 m2 A (508) We don’t know the coil size or the number of turns. Suppose the coil is cylindrical, with length h and radius a. Then if N is the number of turns in each wire, the length of each wire (which we know is 62 m) is ℓ = 2πaN . We have another clue: the two coils are wound closely together separated only by twine. The winding must look something like what is shown in Fig. 134. If we assume that the twine is about as thick as the wire, then the turns in one coil are spaced at intervals of 4/20 inch, or about 0.005 m. The number of turns in each coil is then N = h/(0.005 m) = 200h/(1 m). Figure 134 It seems likely that Faraday’s “block of wood” would have been roughly “squarish” in proportions, so let’s assume h = 2a. Then the three relations, ℓ = 2πaN , N = 200h/(1 m), and h = 2a, quickly give ℓ(1 m) = 200πh2 . From ℓ = 62 m we find h ≈ 0.3 m. And then a = 0.15 m, and N ≈ 60 turns. So the coil is about a foot long and a foot in diameter – very reasonable. internal resistance 100-plate battery Because end effects were neglected, this somewhat overestimates the inductance (see Exercise 7.40). But it would be silly to worry about this error, given all the other approximations we’ve made. This L is also the mutual inductance of the two coils, because they link the same flux. The reconstructed circuit is shown in Fig. 135. Rb L =10-3 H R =1 Ω L =10-3 H R =1 Ω coil #1 M =10-3 H coil #2 Rg G internal resistance An approximate formula for the inductance L of a coil with N turns is easily derived. Using the long-solenoid field of B = µ0 nI = µ0 (N/h)I, we have Φ = Nπa2 B = πa2 µ0 N 2 I/h. The inductance is obtained by erasing the I, so we have ( ) 4π · 10−7 kgC2m π(0.15 m)2 (60)2 µ0 πa2 N 2 = ≈ 1 · 10−3 H. (509) L= h 0.3 m galvanometer Figure 135 50 A 10-3 s Figure 136 We’ll assume that the 100-plate battery has an emf of roughly 100 volts. Nothing would be gained by using so large a battery if its internal resistance were much greater than 1 ohm, but it probably wasn’t much less. So let’s assume Rb ≈ 1 Ω. Then with the switch closed, the steady current through coil #1 is 50 amps. When the switch is opened (causing the current in coil #1 to drop to zero), the current must rise instantaneously to 50 amps in coil #2, because the flux through the coil cannot decrease discontinuously.2 Thereafter, the current in circuit #2 decays with the time constant L/(R + Rg ). Assuming Rg ≪ R (but maybe it wasn’t!) we have L/R = (10−3 H)/(1 Ω) = 10−3 seconds. The current pulse through the galvanometer therefore looks something like the curve in Fig. 136. 2 If coil #2 weren’t present, then the current in coil #1 would not stop abruptly; charge would build up on either side of the switch, and a spark might jump across. But the existence of coil #2 allows the current in coil #1 to stop. M for two rings From Eq. (6.53), the magnetic field along the axis of a ring of radius a, a distance b from the center, is B = µ0 Ia2 /2(a2 + b2 )3/2 . For b ≫ a this can be approximated as B = µ0 Ia2 /2b3 . In this limit we can also neglect the variation of B over the interior of the other ring. The flux through the other ring is therefore Φ = πa2 B = µ0 πIa4 /2b3 . The mutual inductance is then Φ/I = µ0 πa4 /2b3 . Connecting two circuits (a) In Fig. 7.40(a), if I2 is increasing we have an increasing upward flux through the top circuit. This will induce an E1 in a direction to drive current that makes a downward flux; this direction is opposite to the positive direction assigned to E1 . Hence the sign in the first equation must be a “−.” A similar argument shows that the sign in the second equation is also a “−.” Had we assigned the opposite positive directions for I2 and E2 , the sign in front of M in both equations would be a “+.” (b) In Fig. 7.40(b), the current I is the same in both coils, so I = I1 = I2 . And the emfs add, so the total emf is E = E1 + E2 . If we add the two given equations, we obtain E = −L1 dI dI dI dI dI −M −M = −(L1 + L2 + 2M ) . − L2 dt dt dt dt dt (510) The circuit is therefore equivalent to a single coil with L′ = L1 + L2 + 2M . In Fig. 7.40(c), the current is now I = I1 = −I2 , and the emf is E = E1 − E2 . If we subtract the two given equations, we obtain E = −L1 dI d(−I) dI dI d(−I) +M = −(L1 + L2 − 2M ) . + L2 −M dt dt dt dt dt (511) ′′ The circuit is therefore equivalent to a single coil with L = L1 + L2 − 2M . Since M is positive, we see that L′ is larger than L′′ . (c) A circuit with L < 0 would violate Lenz’s law; it would be unstable. That is, an increasing current would cause an emf that would increase the current even more. This would violate conservation of energy. Therefore, we must have L′′ ≥ 0, which implies that M ≤ (L1 + L2 )/2 for any pair of circuits. Equality is achieved if we have two identical coils wound right on top of each other. (An even stronger inequality, M 2 ≤ L1 L2 , can be derived by considering coils connected in parallel. This is indeed stronger, due to the arithmetic-geometric-mean inequality.) Another (less enlightening) way of obtaining the M ≤ (L1 + L2 )/2 result is the following. The energy stored in a system of two inductors is U = L1 I12 /2 + L2 I 22 /2+M I1 I2 (see Problem 7.10). As you can check, this energy can be written as ) ( L1 + L2 L (512) − M I 21 + M (I 12 + I1 I2 ) − 2 (I 12 − I22). U= 2 2 This holds for any I1 and I2 . In particular, if I2 = −I1 , then only the first of the three terms is nonzero. Since the energy must be positive, we must therefore have M ≤ (L1 + L2)/2. Flux through two rings Figure 137 shows a side view of the field due to the inner ring. (The dots are the intersections of the rings with the plane of the paper.) The key point here is that the flux through the outer ring comes not only from the field lines pointing upward in the interior of the inner ring, but also from the field lines pointing downward in the region between the rings. The latter flux partially cancels the former flux. The larger the outer ring is, the larger this canceling effect is, and so the smaller the net flux is. The field lines within the dotted curves yield a net flux of zero through the outer ring, so it is only the lines in the central region that contribute to the net flux. The larger the outer ring is, the smaller this central region is. Net flux through outer ring comes from this central region ring 1 ring 2 Figure 137 Using the mutual inductance for two rings With current I1 in the outer ring, Eq. (6.54) tells us that the field at the location of the (much smaller) inner ring is B1 = µ0 I1 /2R1 . The flux through the inner ring is therefore Φ21 = πR22B1 = µ0 πR22 I1 /2R1 . If we increase R1 by ∆R1 , then this flux decreases by an amount ∆Φ21 = µ0 πR22I1 ∂Φ21 ∆R1 = − ∆R1 . ∂R1 2R12 (513) Now consider a current I2 in the inner ring. Let B2 be the desired field due to the inner ring at the location of the outer ring at radius R1 . If we expand the outer ring by ∆R1 , the flux Φ12 through this ring decreases by the amount of flux in the annular region between radii R1 and R1 + ∆R1 (Exercise 7.37 explains why it is a decrease). The area of this region is 2πR1 ∆R1 , so the decrease in flux is ∆Φ12 = −B2 2πR1 ∆R1 . Now, if our theorem Φ21 /I1 = Φ12 /I2 always holds, in particular if it holds for any value of R1 , then it must also hold if the Φ’s are replaced with ∆Φ’s. That is, we must also have B2 2πR1 ∆R1 ∆Φ12 µ0 R22 I2 µ0 πR22 ∆Φ21 ∆R1 = − = . (514) =⇒ B2 = =⇒ − 2R21 4R31 I2 I1 I2 We can pick any radius for the outer ring, so in more general notation we can write B = µ0 R2 I/4r3 at any point at radius r in the plane of a ring with radius R and current I, if r ≫ R. Note that this result can be written as B = µ0 (πR2 )I/4πr3 = (µ0 /4π )(IA/r3 ), where A is the area of the ring. We will have more to say about this form of B in Chapter 11. Figure 138 Small L One way to wind resistance wire into a “non-inductive” coil is shown in Fig. 138. Of course, the inductance is not exactly zero. The residual inductance is approximately that of the long, narrow “hair-pin” configuration shown in Fig. 139. Technically, if the wire is infinitely thin, then the self-inductance is actually infinite, due to the issue discussed at the end of Section 7.8. But real wires have thickness, so the self-inductance of the hair pin will indeed be small. Figure 139 Note that the configuration in Fig. 138 effectively consists of two solenoids with currents in opposite directions. So there is essentially no B field inside the cylinder. However, this is actually irrelevant, because the area relevant to the flux is not the cross-sectional areas of all the circular loops. Rather, the area spanned by the wire in Fig. 138 is the hair-pin area in Fig. 139, which wraps around the surface of the cylinder. This area has nothing to do with the inside of the cylinder. L for a cylindrical solenoid The field inside a long solenoid is B = µ0 nI = µ0 (N/ℓ)I. The flux is Φ = N πr2 B = µ0 πr2 N 2 I/ℓ. The self-inductance is obtained by erasing the “I,” so we have ( ) 4π · 10−7 kgC2m π(0.05 m)2 (1200)2 µ0 πr2 N 2 = = 7.1 · 10−3 H. (515) L= ℓ 2m We have neglected the fact that the field inside the solenoid is not constant. It decreases near each end, to the point where the flux through the last turn is only about half the flux through a turn in the middle (see Exercise 6.63). This means that we have overestimated the inductance; the true value is smaller than the above approximate result. We might expect the error to be on the order of the diameter divided by the length, because the diameter should determine the length scale of the region near the end where the field differs appreciably from the idealized B = µ0 nI value. This is 5% in the present example. The actual error is only about 2% in this case (which is consistent with 5%, as far as order of magnitude goes), as you can discover by referring to tables that give exact values for the inductance of cylindrical coils. Opening a switch After the switch has been closed a while, the currents are steady and the inductor is irrelevant. So 10 V is the initial voltage across each of the branches of the circuit. The initial currents across the 150 Ω and 50 Ω resistors are therefore 0.067 A and 0.2 A, respectively. They are both directed downward. Initially both A and B are at 10 V with respect to ground. Right after the switch is opened, we have the circuit shown in Fig. 140. The current through the inductor cannot change abruptly (otherwise there would be an infinite dΦ/dt and hence infinite E, which would cause the current to not change abruptly after all). Therefore, the current through the circuit is 0.2 A in the clockwise direction. The current does change abruptly in the 150 Ω resistor; it goes from 0.067 A downward to 0.2 A upward. The potential at B with respect to ground is still VB = (0.2 A)(50 Ω) = 10 V, but the potential of A is now VA = −(0.2 A)(150 Ω) = −30 V. -30V A 150 Ω B 10V 0.2 A Figure 140 50 Ω VA VB 10 V VB 1 ms VA -30 V Figure 141 2 ms The circuit in Fig. 140 is a simple RL circuit, so as time goes on, the current equals I(t) = I0 e−(R/L)t , where I0 = 0.2 A and where the time constant L/R equals (0.1 H)/(200 Ω) = 5 · 10−4 s = 0.5 millisec. The potentials at A and B are proportional to I, so they decrease like e−(R/L)t . After 0.5 millisec they have decreased by a factor 1/e = 0.37, and after 1 millisec by 1/e2 = 0.14. After 5 millisec the factor is 1/e10 = 4.5 · 10−5 , which is negligible. The plots are shown in Fig. 141. We have only plotted up to t = 2 millisec, because the curves are essentially zero after that. Note the discontinuity in VA . RL circuit ( ) From Eq. (7.69) the current is I(t) = I0 1 − e−(R/L)t , where I0 = E0 /R. In the problem at hand, I0 = E0 12 V = = 1200 A R 0.01 Ω and R 0.01 Ω = 20 s−1 . = 0.5 · 10−3 H L (516) So the time scale is L/R = 0.05 s. The current reaches a value of (0.9)I0 when e−(R/L)t = 0.1 =⇒ (20 s−1 )t = ln 10 =⇒ t = 0.115 s. (517) At this time, the current is I = (0.9)(1200) = 1080 A, so the energy stored in the magnetic field is 1 1 2 LI = (0.5 · 10−3 H)(1080 A)2 = 292 J. (518) 2 2 The instantaneous power delivered by the battery is E0 I, but since I is changing we must perform an integral to find the energy delivered by the battery between t = 0 and t = 0.115 s: ∫ t=0.115 s 0 E0 I(t′ ) dt′ ∫ t ′) 1 − e−(R/L)t dt′ ( ) t L −(R/L)t′  = E 0 I 0 t′ + e  R 0 ( ) L −(R/L)t L = E0 I0 t + e − R R ( ) = (12 V)(1200 A) 0.115 s + (0.05 s)(0.1) − (0.05 s) = E0 I0 0 ( = 1008 J. (519) From conservation of energy, apparently 1008 J − 292 J = 716 J has been dissipated in the resistor. The task of Problem 7.15 is to show that the energy delivered by the battery does indeed equal the energy stored in the magnetic field plus the energy dissipated in the resistor, at any general time t. Energy in an RL circuit The energy delivered by the battery is E0 Q = E0 the resistor is I 2 R. So our task is to show that E0 ∫ 0 t 1 I dt = LI 2 + 2 ∫ ∫t 0 t 0 I dt, and the power dissipated in I 2 R dt. (520) Using the expression for I given in Eq. (7.69), this becomes ∫ t ) ) 1 E02 ( E0 ( E0 1 − e−(R/L)t dt = 1 − 2e−(R/L)t + e−2(R/L)t L R 2 R2 0 ∫ t 2 ) E0 ( 1 − 2e−(R/L)t + e−2(R/L)t R dt + 2 0 R ) ) L( L ( −(R/L)t e −1 = ⇐⇒ t + 1 − 2e−(R/L)t + e−2(R/L)t (521) R 2R ) ( ) ) 2L ( −(R/L)t L ( −2(R/L)t e −1 , + t+ −1 − e 2R R which is indeed true, as you can check. Magnetic energy in the galaxy The energy density is B2 (3 · 10−10 T)2 = ( ) = 3.6 · 10−14 J/m3 . m 2µ0 2 4π · 10−7 kg C2 (522) The volume of the galaxy is πr2 h = π(5 · 1020 m)2 (1019 m) ≈ 8 · 1060 m3 , so the total energy contained in the magnetic field is (3.6 · 10−14 J/m3 )(8 · 1060 m3 ) ≈ 3 · 1047 J. The radiation power of starlight in the galaxy is 1037 J/s, so the magnetic energy is worth (3 · 1047 J)(1037 J/s) = 3 · 1010 s ≈ 1000 years. Magnetic energy near a neutron star The energy density is (1010 T)2 B2 = ( ) = 4 · 1025 J/m3 . m 2µ0 2 4π · 10−7 kg C2 (523) Using E = mc2 , the amount of energy contained in 1 kg is 9 · 1016 J. So the above energy density is equivalent to a mass density of ρ= 4 · 1025 J/m3 = 4.4 · 108 kg/m3 = 4.4 · 105 g/cm3 . 9 · 1016 J/kg (524) 3 This is very large. By comparison, the mass density of water is 1 g/cm . Decay time for current in the earth From Eq. (6.54) the B field at the center of a ring is µ0 I/2r, which gives B = µ0 I/2(a/2) = µ0 I/a here. The stored energy is then )2 ( ) B2 ( µ0 πaI 2 1 µ0 I U= volume = (πa2 · a) = . (525) 2µ0 a 2 2µ0 Since R = π/aσ, the ohmic energy dissipation is I 2 R = πI 2 /aσ. The decay time is therefore µ0 πaI 2 /2 µ 0 a2 σ U (526) ∼ µ0 a2 σ, τ≈ 2 = = 2 I R πI 2 /aσ up to numerical factors. With a = 3000 km and σ = 106 (ohm-m)−1 , we have ( )2 ( ) kg m )( τ ∼ 4π · 10−7 3 · 106 m 106 (ohm-m)−1 ≈ 1 · 1013 s, (527) C2 which is about 3000 centuries. A dynamo The device on the bottom is the dynamo. To see why, suppose a current I is flowing in the coil, in the clockwise direction in the upper circular loop. Such a current produces a magnetic field B pointing downward through the disk. With v the velocity of any part of the rotating disk, v × B is a vector pointing radially outward. Positive charges in the disk will be pushed outward, negative charges pushed inward. Either effect causes current to flow in the direction postulated. If we assume the opposite direction for the coil current I , the field B and hence the cross product v × B will be reversed, and the force will again be in the direction to sustain or increase the current. This conclusion is independent of the sign of the mobile charges. You can quickly verify that in the device on the top, the effect of the qv × B force on the charges in the disk is to decrease whatever current happened to be flowing in the coil. See if you can formulate an unambiguous rule to distinguish the potential dynamo on the bottom from the non-dynamo on the top, a rule that refers only to the relation of disk rotation to coil configuration. Would a mirror image of the figure on the top represent a dynamo? A dynamo of this kind runs equally well with current in either direction. The current can also be zero. However, in any circuit not at absolute zero there are slight random motions of charge, or randomly fluctuating currents. Some fluctuation, tremendously amplified by the “positive feedback” of the dynamo action, becomes the steady dynamo current. It retains the direction of its initial excitation. (In a conventional dc generator there is some residual magnetic field in the iron poles, even at zero current, which suffices to determine the eventual polarity.) The magnitude of the current in this purely ohmic dynamo is determined by the input mechanical power. The current will be such that the ohmic power loss in the coil and disk is precisely equal to the applied torque times the shaft’s angular speed.