Unformatted text preview: AP® Physics C: Electricity and Magnetism 2009 Scoring Guidelines The College Board
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General Notes About 2009 AP Physics Scoring Guidelines
1. The solutions contain the most common method of solving the freeresponse questions and the allocation of points for this solution. Some also contain a common alternate solution. Other methods of solution also receive appropriate credit for correct work. 2. Generally, double penalty for errors is avoided. For example, if an incorrect answer to part (a) is correctly substituted into an otherwise correct solution to part (b), full credit will usually be awarded. One exception to this may be cases when the numerical answer to a later part should be easily recognized as wrong, e.g., a speed faster than the speed of light in vacuum. 3. Implicit statements of concepts normally receive credit. For example, if use of the equation expressing a particular concept is worth one point and a student’s solution contains the application of that equation to the problem, but the student does not write the basic equation, the point is still awarded. However, when students are asked to derive an expression it is normally expected that they will begin by writing one or more fundamental equations, such as those given on the AP Physics Exam equation sheet. For a description of the use of such terms as “derive” and “calculate” on the exams, and what is expected for each, see “The FreeResponse Sections⎯Student Presentation” in the AP Physics Course Description. 4. The scoring guidelines typically show numerical results using the value g = 9.8 m s 2 , but use of 10 m s 2 is of course also acceptable. Solutions usually show numerical answers using both values when they are significantly different.
5. Strict rules regarding significant digits are usually not applied to numerical answers. However, in some cases answers containing too many digits may be penalized. In general, two to four significant digits are acceptable. Numerical answers that differ from the published answer due to differences in rounding throughout the question typically receive full credit. Exceptions to these guidelines usually occur when rounding makes a difference in obtaining a reasonable answer. For example, suppose a solution requires subtracting two numbers that should have five significant figures and that differ starting with the fourth digit (e.g., 20.295 and 20.278). Rounding to three digits will lose the accuracy required to determine the difference in the numbers, and some credit may be lost. © 2009 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. AP® PHYSICS C: ELECTRICITY AND MAGNETISM 2009 SCORING GUIDELINES
Question 1
15 points total (a) (i) 3 points For indicating the field points radially inward dV Er = dr For correctly substituting the electric potential for r < R into the equation for the electric field 2 dV d Ê Q0 È r ˘ˆ Einside =  inside =  Á 2 + 3 dr dr Ë 4π 0 R Í R ˙˜ Î ˚¯ 1 point Distribution of points 1 point () For correctly taking the derivative of the potential function to determine the magnitude of the electric field Q0 È r1 Einside = (3)(2) R R ˘ Í ˙ 4π 0 R Î ˚ 6Q0r Einside = 4π 0 R 3 1 point ( )( ) (ii) 2 points 1 point For indicating the field points radially outward dV Er = dr dV dÊ Q ˆ Eoutside =  outside =  Á 0 ˜ dr dr Ë 4π 0r ¯ For the correct answer Q0 Eoutside = 4π 0r 2 1 point © 2009 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. AP® PHYSICS C: ELECTRICITY AND MAGNETISM 2009 SCORING GUIDELINES
Question 1 (continued)
Distribution of points (b) (i) 3 points For using Gauss’s law Q E i d A = enclosed 1 point Ú 0 For substituting the correct expression for the flux through a Gaussian sphere Q E 4π r 2 = enclosed , r < R
0 1 point 4π 0r 2 For substituting the appropriate expression for the electric field, consistent with the answer to part (a), including a minus sign Qenclosed , r < R 6Q0r = 4π 0 R 3 4π 0r 2
Qenclosed , r < R =  Einside = Qenclosed , r < R 1 point 6Q0r 3
R3 (ii) 2 points From Gauss’s law as shown above Q E = enclosed 4π 0r 2 For correctly substituting the electric field from part (a) into the expression above Qenclosed , r > R Q0 Eoutside = = 4π 0r 2 4π 0r 2 Qenclosed , r > R = Q0 The field points outward, so the charge is positive. For the correct enclosed charge for r > R 1 point 1 point Qenclosed, r > R = +Q0 © 2009 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. AP® PHYSICS C: ELECTRICITY AND MAGNETISM 2009 SCORING GUIDELINES
Question 1 (continued)
Distribution of points (c) 2 points There is charge residing on the surface of the sphere. For indicating that the total enclosed charge equals the charge at the surface plus all the charge inside the sphere Qenclosed , r > R = Qsurface + Qenclosed , r < R when r = R 1 point Qsurface = Qenclosed , r > R  Qenclosed , r < R when r = R
For substituting the correct value of Qenclosed , r < R when r = R 1 point Ê 6Q R 3 ˆ Qsurface = Q0  Á  03 ˜ Ë R¯ Qsurface = Q0  ( 6Q0 ) Qsurface = 7Q0
Note: The student could earn 1 point for a correct qualitative description of the charge configuration, even if the surface charge was not calculated. (d) 3 points For having the graph for r < R consistent with the answer to part a (i) For having the graph for r > R consistent with the answer to part a (ii), with a finite value at r = R For having a step discontinuity at r = R indicating the presence of charge at the surface (Having both graphs asymptotic to r = R does not constitute the correct discontinuity.) 1 point 1 point 1 point © 2009 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. AP® PHYSICS C: ELECTRICITY AND MAGNETISM 2009 SCORING GUIDELINES
Question 2
15 points total Distribution of points (a) 3 points For correctly substituting the given values into the expression for resistance R=ρ A 1 point R = 4.5 ¥ 104 Wim (0.080 m ) 5.0 ¥ 106 m 2 = 7.2 W For correctly combining V = IR and P = IV to get an expression for power in terms of voltage and resistance P = V2 R 1 point ( ) ( ) P = ( 9.0 V ) 7.2 W For the correct answer P = 11 W
2 1 point Alternate solution: For correctly substituting the given values into the expression for resistance R=ρ A Alternate Points 1 point R = 4.5 ¥ 104 Wim (0.080 m ) 5.0 ¥ 106 m 2 = 7.2 W
For correctly combining V = IR and P = IV to get an expression for power in terms of current and resistance P = I 2R I = 9.0 V 7.2 W = 1.25 A 1 point ( ) ( ) P = (1.25 A)2 (7.2 Ω) For the correct answer P = 11 W 1 point (b) 3 points For an arrow directed from a to b For correctly indicating that the conventional current is from a to b OR that the electron current is from b to a For correctly indicating that the electric field is in the same direction as the conventional current OR in the opposite direction to the electron current
Alternate solution: For an arrow directed from a to b For stating that point a is at a higher potential than point b For stating that electric field points from higher to lower potential 1 point 1 point 1 point Alternate Points 1 point 1 point 1 point Note: The third point could be earned for an incorrect field direction consistent with an incorrect current or potential drop direction.
© 2009 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. AP® PHYSICS C: ELECTRICITY AND MAGNETISM 2009 SCORING GUIDELINES
Question 2 (continued)
Distribution of points (c) 2 points For recognizing that the electric field is uniform in the bar and stating the relationship between V and E for a uniform field E=V E = (9.0 V ) (0.080 m ) For the correct answer E = 110 V m
Alternate solution: E = ρJ For correctly calculating the current density J = I A = 1.25 A 5.0 ¥ 106 m 2 1 point 1 point Alternate Points 1 point J = 2.5 ¥ 105 A m 2 E = J = 4.5 ¥ 104 Wim 2.5 ¥ 105 A m 2
For the correct answer E = 110 V m ( )( )
1 point (d) 2 points For applying the equation for the magnetic force on a wire with current F=IB For substituting for current, in terms of either the given quantities or numerical values consistent with previous calculations, for example, the current determined in part (a) Examples: F = (V R ) B = V B R OR F = I B F = (9.0 V ) (0.08 m ) (0.25 T ) ( 7.2 W) OR F = (1.25 A ) (0.080 m ) (0.25 T ) F = 0.025 N 1 point 1 point (e) 1 point For arrows indicating an electric field directed toward the top of the bar 1 point © 2009 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. AP® PHYSICS C: ELECTRICITY AND MAGNETISM 2009 SCORING GUIDELINES
Question 2 (continued)
Distribution of points (f) 3 points For recognizing that when there is no longer deflection, the electric force is equal and opposite to the magnetic force FE = FB For the correct expressions for the electric force and the magnetic force FE = qE
FB = qu B qE = qu B E = uB For correctly substituting values 1 point 1 point E = 3.5 ¥ 10 ( 3 m s (0.25 T ) ) 1 point E = 8.8 ¥ 104 V m Units point For the correct units on at least two numerical answers 1 point © 2009 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. AP® PHYSICS C: ELECTRICITY AND MAGNETISM 2009 SCORING GUIDELINES
Question 3
15 points total Distribution of points (a) 4 points Using the expression for the magnitude of the induced emf f e =  ddtm For recognizing that the area is constant in time
dB dt For correctly taking the time derivative of the magnetic field dB d = (at + b ) = a dt dt For a correct expression for the emf For a positive value for the magnitude of the emf e = aL2 =A 1 point e 1 point 1 point 1 point (b) (i) 2 points The resistors are in series, so the total current flows through both of them.
Rtot For correctly calculating the resistance of the two resistors in series Rtot = Â Ri = 2 R0 I= ε 1 point 1 point For correctly substituting the value of the emf from part (a) aL2 I= 2 R0 Alternate solution: R2 The resistors have equal resistance, so the emf across each is half the total emf. For correctly substituting the emf across bulb 2, consistent with the answer to part (a) aL2 2 I2 = R2 For correctly substituting the resistance of bulb 2 aL2 2 aL2 = I2 = R0 2 R0 I2 ε =2 Alternate Points ( ( ) ) 1 point 1 point © 2009 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. AP® PHYSICS C: ELECTRICITY AND MAGNETISM 2009 SCORING GUIDELINES
Question 3 (continued)
Distribution of points (b) (ii) (continued) 1 point For indicating that current flows upward in bulb 2 or counterclockwise around the loop (c) 2 points
P = I1V1 = I12 R1 = V12 R1 1 For substituting one correct value into one of the above expressions For substituting a second correct value in the chosen expression 1 point 1 point 1 point
R0 Ê aL2 ˆ Ê aL2 ˆ P=Á Ë 2 R0 ˜ Á 2 ˜ ¯Ë ¯
P = a 2 L4 4 R0
(d) 4 points OR Ê aL2 ˆ P=Á R Ë 2 R0 ˜ 0 ¯ 2 OR Ê aL2 ˆ P=Á Ë2˜ ¯ 2 For indicating that bulb 1 is brighter For any indication that the emf is the same as in the original circuit (since there is no flux through the added loop) For stating that adding a bulb in parallel with bulb 2 decreases the overall resistance of the circuit For stating that decreasing the overall resistance increases the overall current, which is equal to the current in bulb 1 (e) 2 points For indicating that bulb 1 is the same brightness as in the original circuit For a complete and correct justification Examples: Since each loop has half the area, each has half the original emf. But each also has half the resistance. This means the current and thus the power in bulb 1 is the same. The two loops are essentially identical. Separately, the flux through each loop would create an emf that is counterclockwise. Since these emfs are in opposite directions in the central wire, the net effect is that there is no emf in that wire. Therefore the situation is equivalent to the original one. 1 point 1 point 1 point 1 point 1 point 1 point © 2009 The College Board. All rights reserved. Visit the College Board on the Web: www.collegeboard.com. ...
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 Spring '11
 MR.Smith
 Magnetic Field, The College Board

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