Phys115_Practice_3A_solutions (1)

PHYSICS 115 PRACTICE 3A Seat No_________ Last Name (Print): ______________________ First Name (Print): _____________________ Honor Pledge: All work presented here is my own. Signature: ______________________________ Student ID: _________________________ READ THIS ENTIRE PAGE NOW Do not open the exam until told to do so. You will have 110 minutes to complete the examination. NO CELL PHONES, TEXT MSG, etc. ALLOWED AT ANY TIME. Before the exam begins: • Print and sign your name, and write your student ID number in the spaces above. During the exam • When the exam begins, print your name and student ID number on the top of each page. Do this first when you are told to open your exam. • If you are confused about a question, raise your hand and ask for an explanation. • If you cannot do one part of a problem, move on to the next part. • This is a closed book examination. All equations and constants are provided. • You may use a calculator, but not a computer, or other internet connected devices (smart-phones, iPads, etc.). End of exam: • Out of respect to other students, please remain seated for the last 20 minutes of the exam. At the end of the exam, please remain seated until all exams have been collected

Name: _______________________________________ Student ID _____________ Score _________ last first Physics 115 – Practice Exam 3A 2 1. [7 pts] A heavily loaded boat is floating in a pond. The boat sinks because of a leak. What happens to the surface level of the pond? a. It stays the same. b. It goes up. c. It goes down. d. More information is needed to reach a conclusion. Solution: The reason why the boat is floating in the first place despite its heavy load is because of the way it is shaped, which spreads out the buoyant force over a larger area, and that lowers the effective density and its load. When the boat sinks the shape of the boat loses its function, and if we neglect the mass of the boat (in the sense that it is much less than the weight of the load), then we see it is only the load now with its high density that is sinking. The mass is the same but volume displaced is less than before, and that’s why the level of the water will go down. 2. [7 pts] For a certain patient, the build-up of fatty tissue on the wall of an artery has decreased the arterial radius by 10%. By how much would the pressure provided by the heart have to be increased to maintain a constant volume of blood flow? Assume that the average flow velocity in the unconstricted region is 50 cm/s and that the density of blood is 1060 kg/m 3 . Model the blood as an ideal incompressible fluid. a. 42 Pa b. 71 Pa c. 10 5 Pa d. 91 Pa Solution: Information was added to simplify the original version of the question. It should be noted that blood is not an ideal fluid, it is a viscous fluid. If it were ideal, then we should expect the pressure before and after the constriction to be the same, which of course is not true, so what this problem is asking us to do is to find the difference in pressure before and after the fatty tissue and not the pressure difference supplied by the blood (unless we make the assumption that blood is viscous). With those caveats in mind, let’s find the change in pressure. Using the continuity equation, we see that: 𝑄𝑄 1 = 𝑣𝑣 1 𝐴𝐴 1 = 𝑣𝑣 2 𝐴𝐴 2 = 𝑄𝑄 2 = const. Or in terms of radii : 𝑣𝑣 1 𝜋𝜋𝑟𝑟 1 2 = 𝑣𝑣 2 𝜋𝜋𝑟𝑟 2 2 ⇒ 𝑣𝑣 2 = 𝑣𝑣 1 � 𝑟𝑟 1 𝑟𝑟 2 � 2 = 0.50 � 1 1 − 0.1 � 2 = 0.62 m/s Now, we use Bernoulli’s equation to find the change in pressure: 𝑝𝑝 2 + 1 2 𝜌𝜌𝑣𝑣 2 2 + 𝜌𝜌𝜌𝜌𝑦𝑦 2 = 𝑝𝑝 1 + 1 2 𝜌𝜌𝑣𝑣 1 2 + 𝜌𝜌𝜌𝜌𝑦𝑦 1 We can assume ∆𝑦𝑦 ≈ 0 , since the fatty tissue within the artery would be very small, therefore: ∆𝑝𝑝 = 𝑝𝑝 2 − 𝑝𝑝 1 = 1 2 𝜌𝜌 ( 𝑣𝑣 1 2 − 𝑣𝑣 2 2 ) = 1 2 (1060)(0.50 2 − 0.62 2 ) = − 71.2 Pa This is negative because it is a drop in pressure, the heart would need to supply 71.2 Pa to maintain the same flow rate (assuming the drop is maintained after the constriction due to the fatty tissue, which is not realistic in ideal flow, as explained earlier).

Name: _______________________________________ Student ID _____________ Score _________ last first Physics 115 – Practice Exam 3A 4 6. [7 pts] I place a 500-g ice cube (initially at 0°C) in a Styrofoam box with wall thickness 1.0 cm and total surface area 600 cm 2 . If the air surrounding the box is at 20°C and after 4 hours the ice is completely melted, what is the conductivity of the Styrofoam material? ( L f = 80 cal/g) a. 9.6 × 10 − 5 cal/s ⋅ cm ⋅ °C b. 2.8 × 10 − 6 cal/s ⋅ cm ⋅ °C c. 1.15 × 10 − 2 cal/s ⋅ cm ⋅ °C d. 2.3 × 10 − 4 cal/s ⋅ cm ⋅ °C Solution: 𝑄𝑄 = 𝑘𝑘𝑘𝑘 � Δ𝑇𝑇 𝐿𝐿 � 𝑡𝑡 Since the has melted fully we know that 𝑄𝑄 = 𝑚𝑚𝐿𝐿 𝑓𝑓 = 500 g × 80 cal g = 40 kcal 𝑘𝑘 = 𝑄𝑄𝐿𝐿 𝑡𝑡𝑘𝑘∆𝑇𝑇 = (40 × 10 3 cal)(1.0 cm) (4 × 3600 s)(600 cm 2 )(20 ∘ C) = 2.3 × 10 −4 cal/s ⋅ cm ⋅ ∘ C Consider the graph at right for questions 7 and 8. The pV diagram at right illustrates four different processes which takes a gas from state 1 to state 5. The process between state 3 and 4 is an isothermal process. 7. [6 pts] Is the net work done on the gas positive, negative, or zero? Explain. a. Positive b. Negative c. Zero d. Not possible to answer Solution: Between 1 and 2 the gas is being compressed, so work on it is positive. Between 2 and 5 the gas is expanding, so work done on it is negative, and since work is area under the pV curve, and the positive portion has more area under, we see the net work on the gas is positive. 8. [6 pts] Rank the magnitude of the heat transferred with the gas in each of the four processes. a. | 𝑄𝑄 4→5 | > | 𝑄𝑄 2→3 | = | 𝑄𝑄 3→4 | > | 𝑄𝑄 1→2 | b. | 𝑄𝑄 1→2 | > | 𝑄𝑄 4→5 | = | 𝑄𝑄 3→4 | > | 𝑄𝑄 2→3 | c. | 𝑄𝑄 1→2 | > | 𝑄𝑄 4→5 | > | 𝑄𝑄 2→3 | > | 𝑄𝑄 3→4 | d. | 𝑄𝑄 2→3 | > | 𝑄𝑄 3→4 | = | 𝑄𝑄 4→5 | > | 𝑄𝑄 1→2 | Solution: To evaluate everything and get numbers we can compare we do need one piece of information we didn’t get to cover explicitly in this course, which is the work done in an isothermal process. This can be obtained by integrating (using calculus; which is not required of you of course). Without that, it is still possible to make some qualitative arguments to make the comparison, but since we want things quantitative and definitive and since this is a good chance to introduce work done in an isothermal process (for your own general knowledge), here it is: 𝑊𝑊 𝑔𝑔𝑔𝑔𝑔𝑔 , 𝑖𝑖𝑔𝑔𝑖𝑖𝑖𝑖ℎ𝑒𝑒𝑒𝑒𝑒𝑒𝑔𝑔𝑒𝑒 = 𝑛𝑛𝑛𝑛𝑇𝑇 ln � 𝑉𝑉 𝑓𝑓 𝑉𝑉 𝑖𝑖 � = 𝑝𝑝𝑉𝑉 ln � 𝑉𝑉 𝑓𝑓 𝑉𝑉 𝑖𝑖 � ∆𝐸𝐸 th = 𝑄𝑄 − 𝑊𝑊 gas ⇒ 𝑄𝑄 = ∆𝐸𝐸 th + 𝑊𝑊 gas Also:

Name: _______________________________________ Student ID _____________ Score _________ last first Physics 115 – Practice Exam 3A 5 ∆𝐸𝐸 th = 3 2 𝑛𝑛𝑛𝑛𝑇𝑇 and 𝑝𝑝𝑉𝑉 = 𝑛𝑛𝑛𝑛𝑇𝑇 ⇒ ∆𝐸𝐸 th = 3 2 𝑝𝑝∆𝑉𝑉 Process ∆𝐸𝐸 th 𝑊𝑊 gas | 𝑄𝑄 | 1 → 2 3 2 � (3 𝑃𝑃 )( 𝑉𝑉 ) − (3 𝑃𝑃 )(3 𝑉𝑉 ) � = − 9 𝑃𝑃𝑉𝑉 − 6 𝑃𝑃𝑉𝑉 15 𝑃𝑃𝑉𝑉 2 → 3 3 2 ( −𝑃𝑃𝑉𝑉 ) 0 1.5 𝑃𝑃𝑉𝑉 3 → 4 0 𝑃𝑃𝑉𝑉 −𝑃𝑃𝑉𝑉 ln 2 𝑉𝑉 𝑉𝑉 1.4 𝑃𝑃𝑉𝑉 4 → 5 3 2 𝑃𝑃𝑉𝑉 𝑃𝑃𝑉𝑉 2.5 𝑃𝑃𝑉𝑉 9. [7 pts] The electric dipole moment of the heart at a particular instant is shown below right. Rank the following voltages from most positive to most negative: 𝑉𝑉 𝑋𝑋 = 𝑉𝑉 3 − 𝑉𝑉 1 , 𝑉𝑉 𝑌𝑌 = 𝑉𝑉 4 − 𝑉𝑉 1 and 𝑉𝑉 𝑍𝑍 = 𝑉𝑉 4 − 𝑉𝑉 2 . a. 𝑉𝑉 𝑍𝑍 > 𝑉𝑉 𝑋𝑋 > 𝑉𝑉 𝑌𝑌 b. 𝑉𝑉 𝑋𝑋 > 𝑉𝑉 𝑍𝑍 > 𝑉𝑉 𝑌𝑌 c. 𝑉𝑉 𝑍𝑍 > 𝑉𝑉 𝑌𝑌 > 𝑉𝑉 𝑋𝑋 d. 𝑉𝑉 𝑋𝑋 = 𝑉𝑉 𝑍𝑍 > 𝑉𝑉 𝑌𝑌 e. 𝑉𝑉 𝑋𝑋 > 𝑉𝑉 𝑌𝑌 > 𝑉𝑉 𝑍𝑍 Solution: Considering the direction of the dipole moment, the potential is positive at points 2 and 3 with 𝑉𝑉 3 > 𝑉𝑉 2 (since point 2 is closer) and negative and points 1 and 4. The potential at points 1 and 4 is approximately the same (positioned relatively the same relative to the dipole). 𝑉𝑉 𝑋𝑋 > 𝑉𝑉 𝑌𝑌 since because 𝑉𝑉 3 > 𝑉𝑉 4 , and 𝑉𝑉 𝑍𝑍 < 𝑉𝑉 𝑌𝑌 since 𝑉𝑉 2 > 𝑉𝑉 1 ⇒ −𝑉𝑉 2 < −𝑉𝑉 1 ⇒ 𝑉𝑉 4 − 𝑉𝑉 2 < 𝑉𝑉 4 − 𝑉𝑉 1 . 10. [7 pts] As shown in the figure, three small charges are equally spaced on the arc of a circle that is centered at the charge Q , where Q = +23 nC and all the other quantities are accurate to two significant figures. What is the magnitude of the net electric force on the charge Q due to the other three charges? a. 1.7 × 10 −4 N b. 2.1 × 10 −4 N c. 3.1 × 10 −4 N d. 2.6 × 10 −4 N Solution: First, we find the magnitudes of forces due to all three charges, let’s call them 1,2,3 from left to right, so that 𝑄𝑄 1 = 𝑄𝑄 3 = 2 nC, 𝑄𝑄 2 = − 6nC . Let 𝑟𝑟 = 0.05 m : 𝐹𝐹 on 𝑄𝑄 from 1 = 𝐹𝐹 on 𝑄𝑄 from 3 = 𝐾𝐾 | 𝑄𝑄 1 𝑄𝑄 | 𝑟𝑟 2 = (9 × 10 9 )(2 × 10 −9 )(23 × 10 −9 ) 0.05 2 = 1.7 × 10 −4 N 𝐹𝐹 on 𝑄𝑄 from 2 = 𝐾𝐾 | 𝑄𝑄 2 𝑄𝑄 | 𝑟𝑟 2 = (9 × 10 9 )(6 × 10 −9 )(23 × 10 −9 ) 0.05 2 = 5.0 × 10 −4 N The horizontal components of the forces due to 𝑄𝑄 1 and 𝑄𝑄 3 cancel. Adding all contributions:

Name: _______________________________________ Student ID _____________ Score _________ last first Physics 115 – Practice Exam 3A 7 (1.6 × 10 −19 C) × (2.45 × 10 4 V) + 1 2 (1.67 × 10 −27 ) 𝑣𝑣 i 2 = 0 ⇒ 𝑣𝑣 i = 2.2 × 10 6 m/s 13. [6 pts] A number of resistors are connected across points A and B as shown in the figure. What is the equivalent resistance between points A and B? a. 4 Ω b. 6 Ω c. 8 Ω d. 10 Ω e. 12 Ω Solution: We start by combining the resistors on the branch to the right in series: 𝑅𝑅 eq1 = 2 + 2 + 4 + 2 + 2 = 12 𝛺𝛺 This branch is in parallel with the 6 𝛺𝛺 resistor, so we combine them in parallel: 𝑅𝑅 𝑒𝑒𝑒𝑒2 = � 1 12 𝛺𝛺 + 1 6 𝛺𝛺 � −1 = 4 𝛺𝛺 This reduces the network to three resistors in series the last remaining two 2 𝛺𝛺 resistors and 𝑅𝑅 𝑒𝑒𝑒𝑒2 . So, we do the final combination of the entire collection in series: 𝑅𝑅 𝑒𝑒𝑒𝑒 = 𝑅𝑅 𝑒𝑒𝑒𝑒2 + 2(2 𝛺𝛺 ) = 8 𝛺𝛺 14. [6 pts] Series/Parallel Circuits: A 9-V battery is hooked up to two resistors in series. One has a resistance of 5 Ω, and the other has a resistance of 10 Ω. Several locations along the circuit are marked with letters, as shown in the figure. Through which resistor is energy being dissipated at the higher rate? a. the 10- Ω resistor b. the 5- Ω resistor c. Energy is being dissipated by both resistors at the same rate. d. Resistors do not dissipate energy, none of the above. e. Solution: Resistance represents a measure of how the conductor opposes the flow of current, like friction in mechanics, a higher resistance means a higher rate of energy dissipation into the environment, therefore, since the current is the same in both resistors (they’re in series), in comparing the energy dissipated by the two resistors in the same amount of time (i.e. power) we conclude that the 10 Ω will be dissipating more into the surroundings. 15. [6 pts] Three resistors connected in parallel have individual values of 4.0, 6.0 and 10.0 Ω , respectively. If this combination is connected in series with a 12-V battery and a 2.0- Ω resistor, what is the current in the 10- Ω resistor? a. 0.59 A b. 1.0 A c. 11 A d. 16 A Solution:

Name: _______________________________________ Student ID _____________ Score _________ last first Physics 115 – Practice Exam 3A 8 First we need to find the current put out by the battery, to do so we need to find the equivalent resistance connected to the battery, so we combine the ones connected in parallel first: 1 𝑅𝑅 𝑒𝑒𝑒𝑒1 = 1 4 + 1 6 + 1 10 ⇒ 𝑅𝑅 𝑒𝑒𝑒𝑒1 = 1.9 Ω Then we combine that with the resistor in series to get 𝑅𝑅 𝑒𝑒𝑒𝑒 = 1.9 Ω + 2.0 = 3.9 Ω , which means by Ohm’s law: 𝐼𝐼 𝑡𝑡𝑡𝑡𝑡𝑡 = 12 3.9 = 3.1 A We now find the voltage drop across the series resistor so as to use the remainder to find the voltage across the parallel combination: ∆𝑉𝑉 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑒𝑒𝑝𝑝 = 12 − 𝐼𝐼 𝑡𝑡𝑡𝑡𝑡𝑡 × 2.0 = 5.8 V This is the voltage across the parallel combination, and it is also the voltage across any of the resistors combined in parallel, which means: 𝐼𝐼 = 5.8 10 = 0.58 A 16. [6 pts] For the circuit shown in the figure, calculate the emf's ℰ 1 and ℰ 3 , assuming that the batteries are ideal. Note that two currents are shown. a. ℰ 1 = 44 V, ℰ 3 = 28 V b. ℰ 1 = 28 V, ℰ 3 = 44 V c. ℰ 1 = 14 V, ℰ 3 = 22 V d. ℰ 1 = 22 V, ℰ 3 = 14 V Solution: Let’s use the loop rule to find ℰ 1 using the bigger loop traversed clockwise starting from the lower junction: � ∆𝑉𝑉 𝑖𝑖 𝑖𝑖 = 0 = − (4.0 A)(4.0 𝛺𝛺 ) + ℰ 1 − (4.0 A)(4.0 𝛺𝛺 ) − 20 𝑉𝑉 + (4.0 𝐴𝐴 )(6.0 𝛺𝛺 ) We can solve this readily for ℰ 1 = 28 V . To find ℰ 3 we’ll use the smaller loop traversed from the same jnction clockwise: � ∆𝑉𝑉 𝑖𝑖 𝑖𝑖 = 0 = −ℰ 3 + (4.0 A + 4.0 A)(5.0 𝛺𝛺 ) − 20 𝑉𝑉 + (4.0 𝐴𝐴 )(6.0 𝛺𝛺 ) Solving for ℰ 3 gives 44 V. So, the answer is (b). 17. [6 pts] A beam of electrons is sent in the positive x -direction in a region with a uniform magnetic field B in the positive y -direction and a uniform electric field E in the positive z -direction. At which of the following speeds would the electrons be deflected in the positive z -direction? a. v < E/B b. v = E/B c. v > E/B d. There is no speed for which this will happen. Solution: The magnetic field will cause the electrons to be deflected in the negative 𝑧𝑧 direction (using the right hand rule), and the electric field will also cause a deflection in the negative 𝑧𝑧 , because the negative charge always moves spontaneously against the direction of the electric field. Therefore, there is no way the electron at that moment could be deflected along positive 𝑧𝑧 with the given fields. ℰ 3 ℰ 1

Name: _______________________________________ Student ID _____________ Score _________ last first Physics 115 – Practice Exam 3A 10 𝐵𝐵 𝑡𝑡𝑡𝑡𝑡𝑡 = 𝐵𝐵 𝑙𝑙𝑡𝑡𝑙𝑙𝑙𝑙𝑙𝑙 + 𝐵𝐵 𝑢𝑢𝑢𝑢𝑢𝑢𝑙𝑙𝑙𝑙 = 𝜇𝜇 0 𝐼𝐼 𝑙𝑙𝑡𝑡𝑙𝑙𝑙𝑙𝑙𝑙 2 𝜋𝜋𝑦𝑦 0 − 𝜇𝜇 0 𝐼𝐼 𝑢𝑢𝑢𝑢𝑢𝑢𝑙𝑙𝑙𝑙 2 𝜋𝜋 (20 − 𝑦𝑦 0 ) Setting this equal to zero, we can solve for 𝑦𝑦 0 ( 𝐼𝐼 𝑙𝑙𝑡𝑡𝑙𝑙𝑙𝑙𝑙𝑙 = 5.0 A) : 𝑦𝑦 0 = 20 𝐼𝐼 𝑙𝑙𝑡𝑡𝑙𝑙𝑙𝑙𝑙𝑙 𝐼𝐼 𝑙𝑙𝑡𝑡𝑙𝑙𝑙𝑙𝑙𝑙 + 𝐼𝐼 𝑢𝑢𝑢𝑢𝑢𝑢𝑙𝑙𝑙𝑙 = 7.7 cm Use the circuit at right to answer Questions 21 and 22 21. [6 pts] For the circuit shown in the figure and using the directions of current specified, choose the Kirchhoff current equation for the node labeled A. a. 𝐼𝐼 1 − 𝐼𝐼 2 + 𝐼𝐼 3 = 0 b. 𝐼𝐼 1 + 𝐼𝐼 2 + 𝐼𝐼 3 = 0 c. 𝐼𝐼 1 = 𝐼𝐼 2 + 𝐼𝐼 3 d. 𝐼𝐼 1 = −𝐼𝐼 2 + 𝐼𝐼 3 Solution: At junction A, using the directions of current specified, currents 𝐼𝐼 1 and 𝐼𝐼 3 are going into the junction, while current 𝐼𝐼 2 is exiting the junction, therefore: 𝐼𝐼 1 + 𝐼𝐼 3 = 𝐼𝐼 2 ⇒ 𝐼𝐼 1 − 𝐼𝐼 2 + 𝐼𝐼 3 = 0 22. [6 pts] Now choose the Kirchhoff loop equation for the entire outside loop. Notice the directions of the currents! a. 𝑉𝑉 1 − 𝐼𝐼 3 𝑅𝑅 3 + 𝑉𝑉 2 − 𝐼𝐼 1 𝑅𝑅 1 = 0 b. 𝑉𝑉 1 − 𝐼𝐼 3 𝑅𝑅 3 − 𝑉𝑉 3 + 𝐼𝐼 1 𝑅𝑅 1 = 0 c. 𝑉𝑉 1 + 𝐼𝐼 3 𝑅𝑅 3 − 𝑉𝑉 3 − 𝐼𝐼 1 𝑅𝑅 1 = 0 d. 𝑉𝑉 2 + 𝐼𝐼 3 𝑅𝑅 3 + 𝑉𝑉 3 − 𝐼𝐼 1 𝑅𝑅 1 = 0 Solution: Let’s pick a travel direction that starts from, say, point A, and goes counterclockwise. This means, we’d be going through the first battery from its positive to negative terminals (hence −𝑉𝑉 1 ) and then, we’re traveling against the direction of current through 𝑅𝑅 1 , so that we pick up a positive potential 𝐼𝐼 1 𝑅𝑅 1 , then the potential through the 3 rd battery is positive, since we’re going negative to positive, and finally we go through 𝑅𝑅 3 with the direction of current, so we experience a potential drop (hence the minus in −𝐼𝐼 3 𝑅𝑅 3 ): ∆𝑉𝑉 AA = ∑𝑉𝑉 𝑖𝑖 = −𝑉𝑉 1 + 𝐼𝐼 1 𝑅𝑅 1 + 𝑉𝑉 3 − 𝐼𝐼 3 𝑅𝑅 3 = 0

Name: _______________________________________ Student ID _____________ Score _________ last first Physics 115 – Practice Exam 3A 11 23. [6 pts] RC Circuits: A charged capacitor is connected in series with a resistor and an open switch. At time t = 0 s, the switch is closed. Which of the graphs below best describes the potential difference V across the capacitor as a function of time t ? Solution: Since the capacitor is originally charged, it means, it will start with some initial potential ( ∆𝑉𝑉 C ) 0 , and since we have no battery to replenish the charge on its plates, charge will be depleted through recombination and the potential will decay exponentially. Therefore, the answer is (B) 24. [6 pts] A circular metal loop of radius 10 cm and three long straight wires carry currents of 𝐼𝐼 1 = 20 A , 𝐼𝐼 2 = 50 A , 𝐼𝐼 3 = 20 A and 𝐼𝐼 4 = 40 A, as shown in the figure. Each of the straight wires is 20 cm from the center of the loop. The axes are shown in the figure, with the + z- axis coming out of the paper. What is the y component of the resultant magnetic field at the center of the loop? a. − 54 𝜇𝜇𝜇𝜇 b. − 60 𝜇𝜇𝜇𝜇 c. − 50 𝜇𝜇𝜇𝜇 d. +60 𝜇𝜇𝜇𝜇 e. +50 𝜇𝜇𝜇𝜇 Solution: Let’s look at the directions of the magnetic field (using the right-hand-rule) due to each wire at the center of the loop along with the magnitude of the field contribution there: Field designation Direction Magnitude 1 (loop) −𝑧𝑧 𝜇𝜇 0 𝐼𝐼 1 2 × 0.1 m = 0.126 mT 2 + 𝑧𝑧 𝜇𝜇 0 𝐼𝐼 2 2 𝜋𝜋 × 0.2 m = 50 𝜇𝜇 T 3 − 45 ∘ below + 𝑥𝑥 -axis 𝜇𝜇 0 𝐼𝐼 3 2 𝜋𝜋 × 0.2 m = 20 𝜇𝜇 T 4 − 90 ∘ below + 𝑥𝑥 -axis 𝜇𝜇 0 𝐼𝐼 4 2 𝜋𝜋 × 0.2 m = 40 𝜇𝜇 T Adding all contributions, we get the total field: 𝐵𝐵 �⃗ 𝑡𝑡𝑡𝑡𝑡𝑡 = 𝐵𝐵 �⃗ 1 + 𝐵𝐵 �⃗ 2 + 𝐵𝐵 �⃗ 3 + 𝐵𝐵 �⃗ 4 = �𝐵𝐵 tot , 𝑥𝑥 , 𝐵𝐵 tot , 𝑦𝑦 , 𝐵𝐵 tot , 𝑧𝑧 � = (0 + 0 + 𝐵𝐵 3 cos 𝜃𝜃 3 + 𝐵𝐵 4 cos 𝜃𝜃 4 , 0 + 0 + 𝐵𝐵 3 sin 𝜃𝜃 3 + 𝐵𝐵 4 sin 𝜃𝜃 4 , −𝐵𝐵 1 + 𝐵𝐵 2 + 0 + 0) Therefore: 𝐵𝐵 �⃗ 𝑡𝑡𝑡𝑡𝑡𝑡 = �𝐵𝐵 tot , 𝑥𝑥 , 𝐵𝐵 tot , 𝑦𝑦 , 𝐵𝐵 tot , 𝑧𝑧 � = (14.1 𝜇𝜇 T, − 54.1 𝜇𝜇 T, − 76 𝜇𝜇 T) We are only after the 𝑦𝑦 -component, therefore the answer is 𝐵𝐵 tot , 𝑦𝑦 = − 54.1 𝜇𝜇 T .

Name: _______________________________________ Student ID _____________ Score _________ last first Physics 115 – Practice Exam 3A 13 28. [6 pts] A cell membrane has a resistance and a capacitance and thus a characteristic time constant. What is the time constant of a 9.0-nm-thick membrane surrounding a 0.040-mm-diameter spherical cell? 𝜌𝜌 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 = 3.6 × 10 7 Ω ⋅ 𝑚𝑚 , 𝜅𝜅 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 = 9.0 a. 2.9 ms b. 4.2 ms c. 8.1 ms d. 12 ms e. 17 ms Solution: 𝑅𝑅 = 𝜌𝜌𝜌𝜌 𝐴𝐴 = (3.6 × 10 7 )(9.0 × 10 −9 ) 4 𝜋𝜋 (0.02 × 10 −3 ) 2 = 6.45 × 10 7 Ω 𝐶𝐶 = 𝜅𝜅𝜖𝜖 0 𝐴𝐴 𝑑𝑑 = (9.0)(8.85 × 10 −12 )(4 𝜋𝜋 (0.02 × 10 −3 ) 2 ) 9.0 × 10 −9 = 4.44 × 10 −11 F ⇒ 𝜏𝜏 = 𝑅𝑅𝐶𝐶 = (6.45 × 10 7 Ω )(4.44 × 10 −11 F) = 2.9 ms 29. [6 pts] Draw a circuit with two batteries, a resistor between them, and a capacitor in parallel with the resistor. The batteries are connected negative pole to positive pole.

Name: _______________________________________ Student ID _____________ Score _________ last first Physics 115 – Practice Exam 3A 14 30. [6 pts] A ring with a clockwise current (as viewed from above the ring) is situated with its center directly above another ring, which has a counter-clockwise current, as shown in the figure. In what direction is the net magnetic force exerted on the top ring due to the bottom ring? a. upward b. Downward c. to the left d. to the right e. The net force is zero. Solution: We need to determine the direction of the field due to the bottom ring at the position of the top one. Applying the right-hand-rule, we see that the field from the bottom one points up for any point above the bottom ring and points down for any point below the bottom ring (we only need points above it for this problem). To see that, curl the fingers of your right hand with the current of the bottom ring and your thumb will point along the field generated by that ring. Therefore, the bottom ring acts as a magnet, above it is a north pole, and below it is a south pole. Doing the same for the top loop, we notice that this time, below the top loop is a north pole and above it is a south pole, which means that the two loops would repel each other, so that the force from the bottom onto the top one is upward. 31. [6 pts] Consider an ideal solenoid of length L , N windings, and radius b ( L is much longer than b ). A current I is flowing through the wire windings. If the radius of the solenoid is doubled to 2b , but all the other quantities remain the same, the magnetic field inside the solenoid will a. remain the same. b. become twice as strong as initially. c. become one-half as strong as initially. d. become four times as strong as initially. e. become one-fourth as strong as initially. Solution: The magnetic field inside a solenoid depends on the ratio between the number of windings and the length, it does not depend on the radius of solenoid, therefore it will remain the same.