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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 ...
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This note was uploaded on 12/25/2009 for the course PHYS 213 at Cornell University (Engineering School).
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