P115_Sp23_E2_Ver_A_Solution

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University of Washington *

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116

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Physics

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Apr 3, 2024

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Physics 115 Spring 2023 Midterm Exam 2 1. [5 pts] Consider the two circuits at right. Is the brightness of bulb A greater than, less than , or equal to the brightness of bulb B? A) Greater than B) Less than C) Equal to D) Not enough information Bulb A is connected in series to a 5 ࠵? resistor, while bulb B is connected in series to a parallel network of two 10 ࠵? resistors. The equivalent resistance of the two 10 ࠵? resistors is 5 ࠵? as shown below. 1 ࠵? !"! = 1 10 Ω + 1 10 Ω = 2 10 Ω ࠵? !"! = 10 Ω 2 = 5 Ω The total resistance of each circuit is thus the same, and since both batteries are identical, the same current flows through each circuit. Bulb brightness is proportional to current, so both bulbs have equal brightness. 2. [5 pts] In the circuit at right, the current through the battery, I bat , is: A) I 1 /2 B) 2 I 1 /3 C) I 1 D) 3 I 1 /2 E) 2 I 1 For resistors connected in parallel, the voltage across each resistor is the same. The current I 1 can be defined as: ࠵? # = ࠵? $%! ࠵? The current through the 2R resistor is & !"# ’( , which is # ’ ࠵? # . Applying Kirchhoff’s junction law, the current through both resistors combines to flow through the battery. The current through the battery is therefore ) ’ ࠵? # (࠵? # + # ’ ࠵? # ) . A B

Physics 115 Spring 2023 Midterm Exam 2 3. [5 pts] The switch in the RC circuit shown at right is closed at t = 0. The capacitor is initially uncharged. What is the current through the resistor at the time t = 0.56 s? A) 0.13 A B) 0.29 A C) 0.48 A D) 0.62 A E) 0.97 A We can use the following equation to determine the current through the resistor: ࠵?(࠵?) = ࠵? * ࠵? +!/(- = ࠵? $%! ࠵? ࠵? +!/(- ࠵?(0.56 s) = 12 V 16 Ω ࠵? +*./0 2/(#0 4)6’*×#* $% 89 ࠵?(0.56 s) = 0.13 A

Physics 115 Spring 2023 Midterm Exam 2 4. [5 pts] The capacitor in the figure at right is initially uncharged and the switch, in position c, is not connected to either side of the circuit. The switch is now flipped to position a for 10 ms, then to position b for 10 ms. Which of the graphs below best represents the current through the capacitor over this 20 ms period? When the switch is in position a, the capacitor is charging, and when the switch is in position b, the capacitor is discharging. The current through the capacitor for both charging and discharging circuits is an exponential decay function, which is best represented by graph B.

Physics 115 Spring 2023 Midterm Exam 2 5. [5 pts] A negative charge moves with a velocity in the positive y -direction through a region of uniform magnetic field. The negative charge experiences a magnetic force that points out of the page. What is the direction of the magnetic field? A) Positive x -direction (to the right) B) Negative x -direction (to the left) C) Negative y -direction (toward bottom of the page) D) Into the page E) Out of the page By the right-hand rule, a force out of the page is consistent with a velocity in the positive y-direction and a magnetic field that points in the negative x-direction. However, since the charge in this question is negative, the magnetic field must point in the positive x-direction. 6. [5 pts] Four charges of equal mass are initially moving to the right at equal speeds and enter a magnetic field. The magnetic field exerts a force on each of these charges resulting in the paths shown in the diagram. Which of the following statements are true? A) All charges are positive, and the magnetic field is into the page. B) All charges are positive, and the magnetic field is out of the page. C) All charges are negative, and the magnetic field is out of the page. D) Charges A and B are positive, C and D are negative, and the magnetic field is into the page. E) More than one of the above is correct. At point P, the velocity of the charges is to the right, and to turn clockwise, the force on the charges must be downward (toward the center of the circular path). By the right-hand rule, a downward force, with a rightward velocity, would be consistent with an out of the page magnetic field for positive charges, or an into the page magnetic field for negative charges. Since there is no choice that has an into the page magnetic field with negative charges, only choice B is correct. P ࠵? ⃗ ࠵? ⃗

Physics 115 Spring 2023 Midterm Exam 2 7. [5 pts] The top wire in the figure at right produces a magnetic field at point P with a magnitude B top . The magnitude of the net magnetic field at point P is 5 B top and it points into the page. What is the magnitude and the direction of current in the bottom wire? A) 50 A to the right B) 50 A to the left C) 80 A to the right D) 80 A to the left E) 120 A to the left Using the right-hand rule, the magnetic field at point P due to the current in the top wire points into the page, and we know it has a magnitude B top . The magnitude of the net magnetic field at point P is 5B top , which means that the bottom wire could produce a magnetic field of 4B top into the page at point P or a 6B top field that points out of the page. Let us first look at the case where the magnetic field needs to be 4B top into the page. For the bottom wire to produce a magnetic field into the page, the current must move to the left through the bottom wire, and the current must be 80 A as shown below. |࠵? $"!!": | = 4=࠵? !"; = ࠵? * ࠵? $"!!": 2࠵?(2 cm) = 4 ࠵? * (10 A) 2࠵?(1 cm) ࠵? $"!!": = 4(10 A) 2 cm 1 cm = 80 A This is consistent with choice D. For the case where the magnetic field needs to be 6B top out of the page, the bottom wire to needs to carry a current to the right, and the current must be 120 A as shown below. |࠵? $"!!": | = 6=࠵? !"; = ࠵? * ࠵? $"!!": 2࠵?(2 cm) = 6 ࠵? * (10 A) 2࠵?(1 cm) ࠵? $"!!": = 6(10 A) 2 cm 1 cm = 120 A Choice E has the correct current magnitude but not the correct current direction.

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