Exam 2_20230203 Solutions

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PHYS 102 Exam 2 Spring 2023-02-16 Two parts worth a total of 25 points: multiple choice (4 questions; 4 points) and written answer (2 questions; 11 and 10 points respectively). Multiple-choice answers are to be written on the bubble sheet at the end of the test. All multiple- choice questions have only one correct answer. You may detach the bubble sheet when the test starts. Written answer questions must be answered on the page(s) following the question. Partial marks will be awarded for solutions that are partially correct. Solutions Exam advice: • Write your name and student number on the front page. • Read or skim the exam to the end before starting. • Do the easier questions first. • Draw diagrams even if it’s not demanded by the question. • Show your work and write explanations neatly. Use full sentences! • Include units and reasonable significant figures for all numerical answers and directions for all vector quantities. • Put a box around your final answer in all cases. Possibly-useful facts: • Permittivity of free space: • Coulomb’s constant:

1. A flashlight with a bulb of resistance 50 Ω is powered by a 4.5 V battery. Which of the following changes would increase the brightness of the bulb? A) Add a second identical bulb to the circuit in parallel with the existing light bulb. B) Add a second identical bulb to the circuit in series with the existing light bulb. C) Add a second identical battery to the circuit in parallel with the existing battery. D) Add a second identical battery to the circuit in series with the existing battery. E) More than one of these is correct. A) A second bulb in parallel would increase the current draw from the battery, but this current would be split between the batteries, leaving the brightness unchanged. B) A second bulb in series would decrease the current through both bulbs, decreasing the brightness. C) A second battery in parallel would not change the current or voltage through the bulb. Each battery would supply half as much current. D) A second battery in series would double the potential difference across the bulb. According to Ohm’s law that would double the current, which increases the brightness.

2. A capacitor with capacitance C 1 is fully charged through a resistor with resistance R 1 using a battery with potential V bat = 12 V. A total of 100 mJ of energy is stored on the capacitor in this manner. The battery is then removed and replaced with a device that behaves like a resistor of resistance R L = 1 kΩ. After 1.0 seconds, the capacitor has discharged half of its energy . What is the capacitance C 1 and the resistance R 1 ? A) C 1 = 1.38 mF R 1 = 2077 Ω B) C 1 = 720 F R 1 = 2077 Ω C) C 1 = 1.38 mF R 1 = 3784 Ω D) C 1 = 720 F R 1 = 1077 Ω E) C 1 = 1.38 mF R 1 = 1077 Ω To find the capacitance, use the relation U = ½CV 2 reaaranged to solve for C. C 1 = 2U/V 2 , plugging in the values given for U and V we get C 1 = 1.39 mF. (there is a rounding error in the options provided) The voltage on a discharging capacitor follows the function V(t) = V 0 exp(-t/τ). In our case V 0 = V bat and τ = (R 1 +R L )C. Using the same relation as above, we can write the function for energy: U(t) = ½C (V bat exp(-t/τ)) 2 = ½CV bat 2 exp(-2t/τ) We are told that U(t 1 ) = U(0)/2 where t 1 is 1.0s, so we can write: U(0)/2 = ½½CV bat 2 exp(0) = ½CV bat 2 exp(-2t 1 /τ) which simplifies to: ½ = exp(-2t 1 /τ). Solving for τ and substituting the resistors, we obtain R 1 = 2t 1 /(C ln 2) - R L. Putting in the values given we get R1 = 1077 Ω .

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3. In which direction is the force exerted on a negative charge in the following uniform magnetic field? A) X B) ⊙ C) At a different angle, in the plane of the page D) At a different angle, out of the plane of the page E) F = 0 This is a negative charge, so we either use the right-hand-rule and reverse the direction, or use our left-hand. The result is out of the page. The velocity and the magnetic field are perpendicular to each other, so the force is not zero. The force is always perpendicular to both the magnetic field and velocity, so it had to be directly into or out of the page, not at some other angle.

4. Rank the magnitude of torque exerted on the loops below. In all four cases, the magnetic field strength is of the same magnitude, B, in the direction indicated, and the current in each loop is the same. Sizes are depicted to scale. A. τ a < τ b < τ c < τ d B. τ a = τ b < τ c < τ d C. τ b < τ a = τ c < τ d D. τ c < τ b < τ a < τ d E. None of these options is correct The torque is given by τ = NIAB sin ɸ where N is the number of loops, I is the current, A is the area of the loop, B is the magnetic field magnitude, and ɸ is the angle between the loop’s normal vector and the magnetic field direction. We note immediately that ɸ b = 0 so τ b = 0, so the answer can only be C or E. In the remaining cases I, B and ɸ are the same, so we only need to consider N and A to decide between C and E. The drawings are to scale. We can see that the diameters of the circles in a and b are the same as the side length of the squares in c and d. A circle of diameter L has an area A = π(L/2) 2 , whereas a square with side length L has an area L 2 . π/4 < 1 so the square has larger area. Thus A c = A d > A a = A b . N a = N c = 1, so we get that τ c > τ a . This eliminates option C leaving E as the only option. Finally note that N d = 3 and all the others are 1, but A d = A c so τ d = 3 τ c , or τ d > τ c . The proportions relative to τ a would be τ b = 0, τ c = 4/π τ a , τ d = 12/π τ a , for a ranking of τ b < τ a < τ c < τ d confirming our choice of E.

Written-answer questions (2 questions) 5. A worker bumblebee transports pollen from a flower to the hive. Assume the mass of the worker bumblebee is 0.17 g, that it carries pollen grains with a total charge of +0.20 pC, and that when you observe it, it is flying west at a speed of 4.0 m/s. At this location, the earth’s magnetic field is 40 μT (pointing north) and the local electric field is 100 V/m (pointing down towards the earth). a) (9 points) Determine the gravitational, electrical and magnetic forces acting on the bee, clearly indicating the magnitude and direction of each. b) (2 points) Determine the net force on the bee, specifying the direction relative to its velocity vector. a) F g = mg = 1.67x10 -3 N, pointing down towards the earth. F E = qE = 2.0x10 -11 N, pointing down towards the earth (since the electric field points in a direction a positive charge would go). F M = qvB sin θ = 3.2x10 -17 N. The B field is pointing north and the bee is flying west, so sin θ = 1 and the force is down towards the earth according to the right-hand-rule. b) F G , F E , and F M are all pointing down, so the total force is F net = F G + F E + F M = 1.67x10 -2 N pointing down towards the earth, 90° down from its velocity vector. The charge on the bee is very small, the electric field is not very strong, and neither is the magnetic field, so it is expected that the gravitational force dominates. Common mistakes: using 0.17kg accidentally as the mass. SI prefix conversion errors. 3 3 3 2

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6. (10 points) You are designing a portable mosquito-zapper device to help stop the spread of dengue fever. At your disposal are: a single 9 V battery, 0.1 F capacitors, and 100 Ω resistors. Schematically the device looks like this: Note that the resistance R and capacitance C in the diagram each represent a collection of one or more individual resistors or capacitors arranged in series or parallel . a) (4 points) Suppose you go with the simplest design and choose a single resistor and capacitor for R = 100 Ω and C = 0.1 F. Without the mosquito present, while the capacitor is first charging from Q = 0 to its maximum, what would be the: ◦ maximum current draw from the battery? ◦ maximum power through the resistor? ◦ maximum voltage on the capacitor? ◦ time constant for charging? Note: this is not how real bug zappers work. Mosquito image from https://svgsilh.com/image/2023316.html The maximum current draw from the battery is when the capacitor first starts charging from Q = 0. When Q = 0, V C = 0, the entire voltage from the battery is drop across the resistor. Thus I max = V/R = 0.09 A. The maximum power through the resistor is at the same time as the maximum current, and P = V 2 /R = 0.81 W. The maximum voltage on the capacitor is when it has finished charging. Finished charging means the current is zero, so the voltage drop on the resistor is also zero. Thus the entire voltage from the battery is dropped across the capacitor and V C = 9V. The time constant is τ = RC = 10 s. 1 1 1 1

b) (3 points) Now a mosquito flies in and connects to the two points indicated on the diagram above. If the mosquito has very low resistance of 1 Ω, you can reasonably ignore the battery and resistor branch of the circuit and only consider the capacitance C and the mosquito. With such a simplification done, determine the following quantities: ◦ The time constant for discharging the capacitors through the mosquito. ◦ The maximum current through the mosquito. ◦ The energy delivered to the mosquito in 100 milliseconds. Define R skeeter = 1 Ω. The time constant for discharging is τ = R skeeter C = 0.1 s The maximum current through the mosquito is at the initial moment of contact, when V C = 9V. Using Ohm’s law we get that I max = V/R skeeter = 9 A. The voltage of a discharging capacitor is written as V(t) = V 0 exp(- t/τ). Thus the change in energy in the capacitor from t = 0 to t 1 = 0.1 is: ΔU = U 1 - U 0 = ½CV(t 1 ) 2 - ½CV 0 2 ΔU = ½CV 0 2 (exp(-2t 1 /τ) - 1) = -3.5 J Thus the energy delivered to the mosquito is 3.50 J. Note that the total energy on the capacitor was U = ½CV 0 2 = 4.05 J. 1 1 1

◦ c) (1 point) You read the specifications for your components more carefully and discover: ◦ The 9 V battery has a maximum safe current output of 50 mA. ◦ The 0.1 F capacitors are rated for up to 3.3V maximum. ◦ The 100 Ω resistors are rated for 0.5W of power dissipation. Does your design respect these limits? Which components exceed them? No the simple design does not respect these limits. The maximum current from the battery is 90 mA, greater than the limit. The maximum voltage on the capacitor is 9V, much greater than the limit. The maximum power dissipated through the resistor is 0.81W, greater than the limit. 0.5 0.5

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d) (2 points) Challenge question: This is worth relatively few points given the difficulty, you should only spend time on this part if you are confident you have completed the rest of the exam to the best of your ability. Choose a combination of 100 Ω resistors and 0.1 F capacitors arranged in series or parallel so that each individual component is below the maximum limits. The equivalent circuit should still look like the one in the figure above ( i.e. combine all the resistors together and all the capacitors together, but do not mix them). Draw a circuit diagram for your chosen design. Show that your chosen arrangement meets the limits for each component. A design that meets the limits uses two 100 Ω resistors in series and three 0.1 F capacitors in series. The equivalent resistance is R eq = 2R = 200 Ω and the equivalent capacitance is C eq = C/3 = 0.33 F. The maximum current drawn from the battery is still when the capacitors are initially uncharged, with all the voltage dropped across the equivalent resistance. I max = V/R eq = 0.045 A (less than 50 mA). The maximum voltage on each capacitor is still when they are fully charged, but now the battery’s voltage is dropped across three identical capacitors, so each gets only V max = V/3 = 3 V (less than 3.3V). The maximum power dissipated by each resistor is given by P max = I max 2 R where R is the individual resistor, not R eq , so P max = 0.20 W (less than 0.5 W). Note that the charging speed of the capacitor is reduced (by 2/3). The discharging speed and the energy stored are both reduced by 1/3. So it’s a less powerful device overall. 1 1