{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

exam2sol - Adam S Bolton [email protected] MIT 8.02 Spring...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Adam S. Bolton April 10, 2002 [email protected] MIT 8.02 Spring 2002 Exam #2 Solutions Problem 1 (Homework problem 6.6; simplified version) (a) Before the switch is closed, all currents are zero. Immediately after the switch is closed, [3 will still be zero, for the inductor prevents instantaneous changes in [3. That leaves us V 2 R111 + R212, and [1 = [2 because [3 = SO [1 = [2 = + (b) The ohmic resistance of the self—inductor is zero. So if we wait a long time, until the self—inductor is no longer opposing changes in [3, we have a wire with zero resistance in paral— lel with R2. Thus [2 will be zero, and [1 = [3. Applying Ohm’s law, we find [1 = V/R1 (2 [3). Problem 2 One way: The Lorentz force on the left segment is I ()3 to the left (with uniform force per unit length), the force on the right segment is I ()3 to the right, the force on the top segment is I (13 upward, and the force on the bottom segment is I (13 downward. This combination of forces produces no torque on the loop. Another way: The loop’s magnetic moment [1, has direction perpendicular to the plane surface bounded by the loop. B is also perpendicular to that surface. So the torque is 7' = [.LXB = 0. Problem 3 (Homework problem 4.6.) The kinetic energy acquired during acceleration is M122 / 2 = (2e)V, where v is the ion’s final speed. In the B—field region, (Lorentz force) 2 (centripetal force) gives (2e)vB 2 M122 /d. Eliminating v and solving for M gives M = (E3B2d2 / V. Problem 4 (Homework problem 5. 7.) Assuming an abrupt drop to zero, integrating around the dashed path shown gives 39 E - dl 75 0, for there is an E—field between the plates. Since this is a static situation, we have dQB/dt = 0 for the rate of change of magnetic flux through the open surface bounded by our loop, and thus Faraday’s law asserts 39 E - dl = 0. Therefore the E—field cannot drop abruptly to zero outside the capacitor. Problem 5 Tiny elements dl along the AC wire segment are parallel to the unit vector f' directed from the element to point P, giving I dl X f' = 0. So, by the Biot—Savart law, the AC wire segment makes no contribution to the magnetic field at P. MIT 8.02 Spring 2002 7 Exam #2 Solutions 2 For tiny elements dl along the CD wire segment, I dl X f' is purely in the “out—of—the—paper” direction, so the contribution of the CD wire segment to the B—field at P is directed purely out of the paper. By the principle of linear superposition, this contribution will be exactly one—half the field at P due to a complete infinite straight wire which extends beyond C. The familiar application of Ampere’s law for an “infinitely” long wire gives B = MOI/27rd, so our answer is half that value I Bp ML out of the paper. 2 47rd Problem 6 (See the lecture supplement of March 15.) A voltmeter will read a negative value if the (small) current that flows through it enters through the negative terminal, so the current in our circuit must be counterclockwise. (a) Both voltmeters have equal internal resistances and they are wired in series, so both will have the same 0.1 Volt drop across themselves. The induced EMF driving the current must therefore be 8 = 0.2 Volts (counterclockwise). (b) The counterclockwise current enters the positive terminal of the left—hand voltmeter, so it will read Vleft = +0.1 Volts. Problem 7 If 0 is the angle between the resistor wire and the moving bar, the area of the plane surface bounded by the closed circuit is 9 1 2 =_ 2 (7rD)—27F 2D9 . The magnetic flux through this surface is then <I>B = éBD20, and the magnitude of the induced EMF is dQB 1 d0 1 6' = — = —BD2— 2 —BD2 . I I dt 2 dt 2 w The induced current will be I _ fl _ BD2w _ R _ 2R (By Lenz’s law, this current will flow in a counterclockwise sense, so as to counter the chang— ing flux due to the motion of the bar.) END ...
View Full Document

{[ snackBarMessage ]}