University:
Massachusetts Institute of Technology
Course:
8.022 | Physics II: Electricity and Magnetism
Academic year:

2023
Views:

176

Pages:

11
Author:

Ismael Haas

Massachusetts Institute of Technology Department of Physics 8.022 Spring 2004 Solution to Quiz #2 PROBLEM 1 (9 pts) Answer the following questions, justifying your answers with one short sentence. (a) Assume that lightbulbs act as resistors and examine the following two circuits: V V C L Both switches are closed at t = 0. Qualitatively describe the behavior of the lightbulbs as time passes. In the left circuit, current ﬂows until the charge builds on the capacitor, opposing the battery. In the right circuit, the inductor initially opposes current ﬂow — Lenz’s law! — and only gradually allows the current to ﬂow. Hence, the left bulb will glow brightly and then fade away. The right bulb will start out dark and gradually get brighter. (b) A metal ring is dropped (feels uniform gravitational acceleration down) into a region of constant magnetic ﬁeld, pointing out of the page: �� Region of constant magnetic field Describe the motion of the ring as it falls. Be sure to contrast the behavior when it is out of the ﬁeld, when it is partially in the ﬁeld, and when it is completely in the ﬁeld. Lenz’s law tells us that an induced current will ﬂow when there is a changing magnetic ﬂux in the loop; the direction of this ﬂow will be such that the magnetic force on the current will oppose the falling motion. The loop will fall freely until it starts to enter the magnetic ﬁeld. It will then slow down very suddenly — falling but not freely. When it is completely inside the region, it will again fall freely since the ﬂux through the loop no longer changes. As it starts to leave the region, it will again fall slowly; it will then fall freely once it completely leaves the region. ∂ = B0 zẑ a possible magnetic ﬁeld? If so, justify your statement. If not, ﬁx it. (c) Is B ∂ ·B ∂ ≥= 0! There are lots of ways to ﬁx it. It is NOT a valid ﬁeld: Notice that ∞ The simplest is probably to include a component along the x or y directions — ∂ = B0 (zẑ − xˆ e.g., put B x). PROBLEM 2 (35 pts) In this problem, you will look at a circuit that contains a constant current source: the total current which comes out of this device is always I0 no matter what EMF is required. The switch S is closed at t = 0. First, this source is hooked up to a pair of resistors in parallel: I0 I1 I2 R1 R2 (a) Find the EMF E produced by the constant current source, as well as the currents I1 (ﬂowing through resistor R1 ) and I2 (ﬂowing through R2 ). The resistors in parallel have an equivalent resistance given by 1/R eq = 1/R1 +1/R2 , or Req = R1 R2 /(R1 + R2 ). The EMF produced by the current source is thus E= I0 R 1 R 2 . R1 + R 2 The currents through the two resistors are given by E I0 R 2 = R1 R1 + R 2 E I0 R 1 = = . R2 R1 + R 2 I1 = I2 The circuit is now modiﬁed: a capacitor and a switch are added. The capacitor is initially uncharged. S C I0 I1 I2 R1 R2 The switch is closed at t = 0. (b) Find the initial currents I1 (t = 0), I2 (t = 0), and the late time currents, I1 (t ≤ �), I2 (t ≤ �). You should be able to do this with very little calculation. (Express your answers in terms of I0 and the parameters of the circuit.) At t = 0, there is no charge on the capacitor and it behaves identically to the circuit in part (a). At late times, the capacitor is fully charged and no current ﬂows through R2 . I0 R 1 I0 R 2 , I2 (t = 0) = R1 + R 2 R1 + R 2 I1 (t ≤ �) = I0 , I2 (t ≤ �) = 0 . I1 (t = 0) = (c) Use Kirchhoﬀ’s laws to write down two equations relating I1 (t), I2 (t), I0 , and the charge on the capacitor Q. Write down a third equation relating Q to I2 . I0 = I 1 + I 2 , I 1 R1 − I 2 R2 − Q =0, C I2 = dQ . dt (d) Using the results of (c), ﬁnd the late time charge, Q(t ≤ �). (Express your answer in terms of I0 and the parameters of the circuit.) For t ≤ � we have I2 = 0, I1 = I0 . Plugging into Kirchhoﬀ and solving for Q, we ﬁnd Q =0 C −≤ Q = I0 R1 C . I0 R 1 − (e) Assume I1 (t) = A + Be−t/T1 , I2 (t) = D + Ee−t/T2 , Q = F + Ge−t/T3 . Using your results from parts (b) and (c), ﬁnd A, B, D, E, F , and G. I1 (0) = A + B = I0 R 2 ; R1 + R 2 I1 (�) = A = I0 −≤ A = I0 , I2 (0) = D + E = I0 R 1 ; R1 + R 2 −≤ D = 0 , B=− I0 R 1 R1 + R 2 I2 (�) = D = 0 E= I0 R 1 R1 + R 2 Q(0) = F + G = 0 ; Q(�) = F = CI0 R1 −≤ F = CI0 R1 , G = −CI0 R1 (f) Using your answers to parts (c) and (e), ﬁnd T1 , T2 , and T3 . Let’s begin with I2 = dQ/dt: I2 = I0 R1 −t/T2 e ; R1 + R 2 dQ CI0 R1 −t/T3 = e dt T3 These are equal at t = 0 if T3 = C(R1 + R2 ) These are equal for ALL t if T2 = T3 . Plugging into the remaining Kirchhoﬀ equation, we ﬁnd that everything works IF we also choose T1 = T3 . So, T1 = T2 = T3 = C(R1 + R2 ) . (g) After a long time, the switch is opened. Find the total energy dissipated in the resistors after the switch is opened. Hint: this can be done without a single integral! It’s the energy stored in the capacitor: Udiss = 1 Q(�)2 = CI02 R12 . 2C 2 PROBLEM 3 (20 pts) The segment of wire shown below is oriented along the x axis, has a total length L and a cross-sectional area A. It is made of material whose conductivity varies along the wire as �c (x) = �0 (L/x). This segment of wire is connected to a battery using zero-resistance cables. The battery puts out an EMF V . A x=L x=0 V In what follows, assume that everything is in a steady state (φ�/φt = 0), and that the wires bring current uniformly to and from the ends of the segment. x̂ points from x = 0 to x = L. (a) What is the resistance R of the wire segment? Treat the segment as a sequence of resistive slabs of length dx. Each slab has resistance dR = 1 dx 1 = x dx �c (x) A �0 LA To get the total resistance, we add the slabs in series — i.e., we integrate: R= � L 0 dR(x) = L 1 �L x dx = . �0 LA 0 2�0 A (b) What is the current density J∂ that ﬂows in the wire? Include direction. The current is found by Ohm’s law, V = IR, and ﬂows in the positive x direction. The magnitude of the current density is J = I/A. Putting all this this together, we have 2�0 V V xˆ . J∂ = xˆ = RA L ∂ in the wire? Include direction. (c) What is the electric ﬁeld E ∂ so The other form of Ohm’s law tells us that J∂ = �c E, ∂ = 1 J∂ = 2V x xˆ . E �c L2 (d) What is the charge density � in the wire? �= 1 ∂ ∂ 1 φEx V ∞·E = = 4γ 4γ φx 2γL2 PROBLEM 4 (25 pts) Consider a solid, conducting inﬁnitely long cylinder, of radius R, oriented parallel to the zaxis. The unit vector ẑ points up; the unit vector �ˆ (cylindrical coordinates) points “around”. ẑ and �ˆ are related by right-hand rule: point your right-hand thumb along ẑ, and your ﬁngers curl in the direction of �ˆ. r̂ (not shown) points radially out from the z-axis. ^ z R ^ φ ∂ = A0 (r/R)ˆ There is a vector potential A z inside this cylinder (for r < R). Note that A0 > 0. ∂ inside the cylinder (r < R). (The formula sheet may help (a) Compute the magnetic ﬁeld B you here.) ∂ = curl A ∂ = − φAz �ˆ = − A0 �ˆ B φr R (b) Compute the total current I ﬂowing in the cylinder. Does it ﬂow up or down? Use Faraday’s law on a contour just at the surface of the cylinder: ∂ · d∂s = 4γI B c 4γI A0 c A0 2γR = −≤ I = . R c 2 � This current ﬂows DOWN (in order to produce a magnetic ﬁeld pointing in the −�ˆ direction). (c) Compute the magnetic ﬁeld for r > R. Use Faraday’s law for a circular contour at r > R: ∂ · d∂s = 4γI B c 4γI 2γrB(r) = c � Now, plug in I = A0 c/2, and use right-hand rule to get direction: ∂ = − A0 �ˆ . B r Now, replace this cylinder with one of radius 2R. z^ ^ φ 2R A current density J∂ = J0 (r/R)2 ẑ ﬂows in this cylinder. J0 may be positive or negative. ∂ for all r that arises from this current. (d) Write down the magnetic ﬁeld B More fun with Faraday’s law. For r � 2R, we have 2γrB(r) = = = ∂ = −≤ B 4γ � r ∂ J(r) · d∂a c 0 8γ 2 J0 � r 3 r dr c R2 0 2γ 2 J0 r4 c R2 γ J0 r 3 ˆ �. c R2 For r � 2R, we have 4γ � 2R ∂ J(r) · d∂a c 0 8γ 2 J0 � 2R 3 r dr = c R2 0 2γ 2 J0 (2R)4 32γ 2 = J0 R 2 = c R2 c 2 ∂ = 16γ J0 R �ˆ . −≤ B c r 2γrB(r) = Both cylinders are now placed next to one another, separated by a distance D > 4R (so that D/2 > 2R is outside both cylinders). 2R R D (e) Find the value of J0 such that the total magnetic ﬁeld vanishes midway between them (i.e., at a distance r = D/2 from the centers of both cylinders). Let out of the page be the positive direction. Then, ∂ left + B ∂ right = 0 B 16γ J0 R2 A0 + =0 D/2 c D/2 cA0 −≤ J0 = − 16γR2 PROBLEM 5 (11 pts) Velocity vector v Separation d Each line has charge/length λ0 in its rest frame Two inﬁnite lines of charge with charge per unit length σ0 in their rest frame are separated by a distance d. These charges are moving in a direction parallel to their length with speed v. This speed could be close to the speed of light! (a) In the rest frame, what is the electric force per unit length that the top line feels due to the bottom line? Give direction (up or down) and magnitude. The bottom line generates an electric ﬁeld of magnitude E= 2σ0 d at the top line. The force per unit length is thus E Frest /L = 2σ20 . d The force is repulsive, so the direction is up. (b) In the lab frame, what is the electric force per unit length that the top line feels due to the bottom line? Give direction (up or down) and magnitude. Same answer ... but now all lengths contract, so the density is increased by a factor of ρ: E Flab /L = 2ρ 2 σ20 2(ρσ0 )2 = . d d The direction is up. (c) In the lab frame, what is the magnetic force per unit length that the top line feels due to the bottom line? Give direction (up or down) and magnitude. The moving line charge generates a current of magnitude I = ρσ0 v, which gener ates a magnetic ﬁeld B= 2ρσ0 v cd The top line charge generates the same current, so there is a force per unit ∂ length I∂ × B/c: B Flab /L = 2(ρσ0 v)2 2ρ 2 v 2 σ20 = . c2 d c2 d The direction is down. (d) What is the total force per unit length in the lab frame? TOT E B Flab /L = Flab /L + Flab /L = 2σ2 2ρ 2 σ20 � 1 − v 2 /c2 = 0 d d The net direction is up. This is identical to the force per unit length in the rest frame!

8.022, Solution to Quiz #2

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