Key # listed as VERSION # on bottom of each page Physics 240
Third Exam
1. Last Name:____________________ Name:____________________
Fall 2012
First Name:_____________________ Name:_____________________
2. Please circle your discussion instructor and section below: Instructor
Time
Section
David Gidley
9-10
100
David Winn
10-11
200
11-noon
300
David Winn
3. Special Scantron Instructions: a) Bubble in UNIQ- NAME in “last “last name” field b) Bubble in 8 digit UM student ID number — this this is critical c) Bubble in Form #=version# (critical) and section number d) Print your name in shaded upper right signature box 4. There are 20 multiple choice problems problems worth 1 points each for a total of 20 points. Choose the one best answer and transfer it to the scantron sheet. You must mark the correct answer on the scantron sheet to get credit for the problem. There is no penalty for guessing, so be sure to enter an answer for every question even if it is just a best guess. 5. At the end end of the exam hand your your scantron to one of the proctors. You may keep this exam booklet for your records. 6. This is a closed book, 90 minute exam. Turn off and put away all electronic devices. You may use a calculator and three 3”x5” note card or equivalent.
k = 1/4πε0 =9x109 Nm2 /C2 -19 e = 1.6x10 C μ = micro =10-6
ε0 = 8.85x10-12 C2 /Nm2 -19 1 eV= 1.6x10 J 2 g = 9.8 m/s
me = 9.1 x 10
MP=1.67 x 10
-31
kg
-7
µ 0 = 4π x 10 Tm/A
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-27
kg
1. The ferromagnetic core of a transformer is made of laminated iron sheets to: A) facilitate easy assembly B) reduce i R losses in the coils C) increase the magnetic flux D) save weight E) prevent eddy currents
2. An AC generator operating at a frequency of f = 60 cycles/second has an RMS emf of 80 -4 V. It is connected to the leads of a 2.0 x 10 F capacitor with negligible resistance. What is the RMS current supplied by the generator (in A)? A) 27 B) 13 C) 19 D) 6.0 E) 42
3. The inductor is a solenoid with an internal resistance of R = 0.15 What is the current flowing through the battery 8 seconds after closing the switch?
6 V
L
A) B) C) D) E)
30 A 3A 40 A 18 A 12 A
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2H
4. A circular coil of wire has 25 turns and has a radius of 0.075 m. The coil is located in a variable magnetic field whose behavior is shown on the graph. At all times, the magnetic field is directed at an angle of 75° relative to the normal to the plane of a loop. What is the average emf induced in the coil in the time interval from t = 5.00 s to 7.50 s?
A) B) C) D) E)
180 mV 140 mV 18 mV 92 mV 49 mV
5. A cylindrical region of radius R contains a uniform conduction current density, parallel to t
its axis, with magnitude that is exponentially decaying with time, J (t ) J 0e . The
The time dependence of J is irrelevant because no time derivatives appear in the conduction current part of amperes law.
magnitude of the magnetic field at radius r at a particular moment in time inside the cylindrical region (r
6. A toroidal solenoid with 400 turns of wire and a mean radius of 6.0 cm carries a current of 0.25 Amps. The wires are wrapped around a core with a relative permeability of 60. What percentage of the magnetic field within the core of the toroid is produced by atomic currents? A) 37% B) 98% C) 63% D) 1.3% E) 0%
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7. Two of Maxwell's equations contain a path integral on the left side and an area integral on the right. These equations require that: A) the area must be well-separated from the path B) the path must be along a field line and the area must be perpendicular to the field line C) the area must be flat and contain the path See figure 29.21 (page 975) for a diagram D) the path must pierce through the area illustrating a path around the boundary of a E) the path must be the boundary of the area non-flat surface. Also, recall the giant "windsock" demo.
8. Find the magnitude and direction of the magnetic field produced at point P by the current element shown in the diagram. The lengths in the diagram are x1=40 cm and z1=30 cm. The current is 5 amps and the current element is 1 cm long. (Note that in the coordinate system pictured in the diagram the y axis points into the page).
A) B) C) D) E)
-
5.3 x 10 3.4 x 10 9.1 x 10 0 Tesla 1.2 x 10
T in negative z direction T in negative y direction T in negative x direction T in negative y direction
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9. The resistance R in the rail gun pictured below is initially 100 Ohms. Consider what happens to the terminal velocity v t and the initial acceleration a 0 at time t=0 if we replace this resistor with one that is 200 Ohms. (t=0 is the moment you release the rod from rest). (Note, the bar is sliding on frictionless rails). Initially, there is no motional emf because there is no motion, so acceleration depends on battery voltage, resistance, magnetic field, and mass of the rod.
Doubling R cuts initial acceleration in half.
A) B) C) D) E)
vt is half and a0 half vt is twice and a 0 twice vt is half and a0 is the same vt is the same and a 0 half vt is the same and a 0 is the same
Terminal velocity occurs when the motional emf cancels the battery emf and reduces the current (and hence the magnetic force) to zero. Since current is zero at terminal velocity the resistance is irrelevant.
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10. The solid circle of radius R in the diagram shows the cross section of a long solenoid with its axis pointing straight into the page. The current in the solenoid is increasing steadily. The dashed concentric circles are four integration paths in a plane perpendicular to the solenoid axis that have radii of r 1 = R /4, r 2 = R /2, r 3 =2 R, and r 4 = 3 R. Rank the paths according to the magnitude of the circulation of the electric field,
E dl
along each
path, least to greatest.
R
1
A) B) C) D) E)
2
3
4
The circulation is the integral on the left side of Faraday's law (as defined in the problem). Basically, you need to compare the circulations by looking at the right side of Faraday's law. Path 3 and 4 tie because there is no flux outside the solenoid and they both surround all of the flux inside. Path 2 is larger than path 1 because it surrounds more flux than path 1, but smaller than 3 and 4 because it does not surround all of the flux.
1, then 2 and 4 tie, then 3 all tie 4, 1, 3, 2 1, 4, then 2 and 3 tie 1,2, then 3 and 4 tie
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11. A solid conductor of radius a is supported by insulating disks on the axis of a conducting tube with inner radius b and outer radius c as shown in the figure. The central conductor and the tube carry equal currents in opposite directions. The currents are distributed uniformly over the cross sections of each conductor. Which one of the following statements about the magnetic field created by this arrangement is FALSE.
A) B) C) D) E)
B is a maximum at r=a. B=0 for r>c B decreases to zero as you go from r=b to r=c. B is proportional to 1/ r for r
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12. The AC generator in a hydroelectric power plant supplies a small town with an RMS current of 4000 A at 120 V rms. At the power plant a 50-to-1 step up transformer is used prior to high voltage transmission. Near the town a substation has a 100-to-1 step down transformer and then each house has a nearby 20-to-1 step down transformer to provide a safe, low voltage. Assuming ideal transformers and no transmission losses what RMS current must be produced in the AC generator? 50-1 Step up
A) B) C) D) E)
HV transmission
100-1
20-1
Step down
Step down
141 A 40,000 A 100 A 25 A 1000 A
4000 A rms
Remember, N_s < N_p in a step down transformer. We have to work backwards through the three transformers.
13. An electric field E E0 cos(2 ft ) is applied along the axis of a cylindrical 2 semiconductor of cross-sectional area A=20 cm and resistivitym At what frequency, f, are the amplitudes of the conduction and displacement currents equal? (1 Hz = 1 cycle/s) A) 5.0 x 10 Hz B) 4.0 x 10 Hz C) 2.0 x 10 Hz D) 8.0 x 10 Hz E) 9.0 x 10 Hz
Set the amplitudes equal to each other and solve for f.
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14. Two concentric circular loops of wire lie in a plane. The inner loop radius, r = 0.1 m, is much smaller than the outer loop radius of R = 2 m. What is the approximate mutual inductance of this configuration of two loops?
r
A) B) C) D) E)
R
Strategy: estimate the flux through the smaller loop by assuming that the magnetic field at the center of the larger ring is uniform over the area of the smaller one. Use this flux in the definition of inductance to calculate inductance.
-
5.2 x 10 H 8.4 x 10 H cannot determine without knowing the current in the outer loop 9.9 x 10 H cannot determine without knowing the rate change of current in the outer loop
Note: the mutual inductance is the part of the flu calculation that does NOT depend on current. You do not need to know the current to calculate it. We gave half credit to the two incorrect NUMERIC answers (A and B on this version).
15. The switch in the circuit below has been closed for a long time. Immediately after opening the switch what is the rate (in Joules/s = Watts) at which magnetic energy stored in the inductor, L, is decreasing?
12 V
L 2 mH
60 Ω
20 Ω A) B) C) D) E) 1. calculate current in inductor immediately before the switch is opened.
29 W 0W 48 W 7W 15 W
This is the current that must go through the two resistors when the switch is opened. Use it to calculate power.
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The rate that energy in the inductor decreases is equal to the power dissipated in the two resistors because that is where the energy is going.
16. Current I is charging a circular parallel plate capacitor with plate radius of R. The circulation of the magnetic field,
B dl , around a coaxial circular path between the
plates with r = R/3 has some definite value. Consider two surfaces bounded by this circular path; the round flat surface (shaded); and the open soup-can (dashed). What can we conclude from the Ampere-Maxwell Law? The total current of all kinds flowing in to the cylinder must be exactly equal to the total current of all kinds flowing out. (this is a paraphrase of the Kirchhoff junction rule). I
A) B) C) D) E)
The cylinder only covers 1/9 of the area of the capacitor plate. This means only 1/9 of the displacement current of the capacitor goes through the shaded side. To make up the difference 8/9 of current must go out the sides in the form of conduction current (remember, the charge has to spread out across the entire plate...).
The "current" (conduction + displacement ) through the soup-can is not equal to the "current" ( conduction + displacement ) through the shaded surface. The conduction current through the soup-can is equal to the conduction current through the shaded surface. There must be conduction current (8/9) I flowing radially outward in the capacitor plate through the soup-can sides There must be conduction current of (3/4) I flowing radially outward in the capacitor plate through the soup-can sides There must be conduction current equal to (1/9) I flowing to the right through the shaded surface.
17. In a simple, purely inductive AC circuit a sinusoidally oscillating emf of angular frequency ω forces AC current to flow through a solenoid of inductance L. This AC current generates an oscillating magnetic field along the axis of the solenoid. The induced electric field that circulates around the axis inside the volume of the solenoid is: A) sinusoidal in time and its amplitude does not depend on ω B) sinusoidal in time and its amplitude is inversely proportional to ω C) constant D) 0, since the inductor/solenoid voltage is 90° out of phase with the current E) sinusoidal in time and its amplitude is proportional to ω The emf of the generator must always equal the emf of the inductor because of the loop law. But... the emf of the inductor is the integral of the electric field in the windings of the solenoid. That means that E depends only on the amplitude of th generator emf, not on its frequency. Version 1
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18. In the circuit below the three resistors are the same at 10 Ω each and the inductor has L = 35 mH. What is the potential difference across the inductor immediately after closing the switch? We do know that the current through the battery is 4.0 A after the switch has been closed for a very long time.
L
A) B) C) D) E)
Two options. 1) calculate the 19. B field of the wire in the middle and multiply by 18. This is NOT an exact answer (it underestimates B by about 20%) but it is close enough to the correct answer to allow you to choose it. 2) integrate the current. This gives a very precise answer, but is more diffecult than option 1.
10 50 V 15 V 30 V 0
In the instant after the switch is closed the curren through the inductor is zero. All current goes through loop 1 only. The voltage across the inductor must be the same as the voltage across the second resistor in loop 1. Since the two resistors in loop 1 are identical the voltage across each must be half the battery voltage. That means the inductor also has half of the battery voltage across it, so V_L=30 volts.
Eighteen very long, straight conductors with square cross sections carrying current of 5 A straight out of the page are laid side by side to form a current sheet as shown in the diagram. What is the magnitude and direction of the magnetic field at point P, 0.03 m from one edge of the sheet as shown? 0.03 m
P
0.12 m A) B) C) D) E)
-4
2.8 x 10 1.5 x 10 6.9 x 10 9.5 x 10 3.2 x 10
T, T, T, T, T,
down up up left down
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20. An LC circuit is composed of an ideal solenoid and a capacitor (no resistance in the circuit). It oscillates at some natural frequency f 0. If we now fill the volume of the solenoid with a ferromagnetic material that has permeability K m = 3000 what happens to the natural frequency? A) it is 55 times higher B) it is 3000 times higher C) it is unchanged D) it is 3000 times lower E) it is 55 times lower
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Answer Key 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20.
E D D C B B E E D E D C E D A C A D A E
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