Exam2-draft-v5-final-solutions_v2

pdf

School

University of California, Berkeley *

*We aren’t endorsed by this school

Course

Subject

Physics

Date

Jan 9, 2024

Type

pdf

Pages

Uploaded by GrandPolarBear2829

Please WAIT to open the exam! When told to begin, you may tear off this first sheet. General instructions  No notes or electronics. Please silence and stow all phones (being sure to turn off notifications and alarms).  The exam consists of short (multiple choice) questions and five problems of varying length.  The entire exam is worth 100 points; the points for each problem are indicated. Once you are allowed to begin, you are encouraged to look through the whole exam before starting.  You will have 110 minutes to complete the exam. The exam is worth 100 points, so you have slightly more time than one minute per point.  Be sure to write your name on every separate sheet at the top!  Please refrain from taking out-of-seat breaks (e.g. to use the bathroom) if possible. During the exam you may ask for assistance by raising a hand. A proctor will listen to your question or concern, but will NOT answer ANY questions individually. If your question or concern requires a clarification or correction, we will provide it on the boards for all students.  Please look only at your own exam and protect your work from wandering eyes. All students will be asked to sign an integrity statement at the end. Instructions for the Problems  To receive full credit on the problems, all reasoning must be shown .  Explain your reasoning with words, brief phrases, arrows, and other common conventions so that your answer and reasoning are easily understood.  Start by writing down basic definitions or equations that you are using in your solution.  Only the equations given on the equation sheet can be used without derivation.  Box-in your final answers.  Try all parts —even if you feel you cannot answer a preceding part.  For directions of vectors, symbols or words are acceptable, e.g., 𝑥ො , 𝚤̂ , “in the +x direction,” “clockwise and circumferential,” “radially outward from the center”, 𝑟̂ , “radially inward toward the axis”, −𝑟̂ , etc  Be careful , be concise , but be complete .  No extra scratch paper is allowed – write all of your work in the space provided on the test. This may count towards partial credit!

Second Midterm Exam Equation Sheet Electromagnetism 𝐹 ⃗ = ଵ ସగఢ బ ொ భ ொ మ ௥ మ 𝑟̂ = 𝑘 ொ భ ொ మ ௥ మ 𝑟̂ 𝐹 ⃗ = 𝑞𝐸 ሬ ⃗ 𝐸 ሬ ⃗ = ொ ସగఢ బ ௥ మ 𝑟̂ (point charge) 𝐸 ሬ ⃗ = ఒ ଶగఢ బ ୰ 𝑟̂ (infinite line of charge) 𝐸 ሬ ⃗ = ఙ ଶఢ బ 𝚤̂ ( infinite sheet of charge) 𝜆 = ௗொ ௗℓ 𝜎 = ௗொ ௗ஺ 𝜌 = ௗொ ௗ௏ 𝛷 ா = ∫ 𝐸 ሬ ⃗ 𝒜 ⋅ 𝑑𝐴 ⃗ ∮ 𝐸 ሬ ⃗ 𝒜 ⋅ 𝑑𝐴 ⃗ = ொ ೐೙೎ ఢ బ 𝑈 = 𝑞𝑉 𝑑𝑉 = −𝐸 ሬ ⃗ ⋅ 𝑑𝑟⃗ Δ𝑉 = − ∫ 𝐸 ሬ ⃗ ⋅ 𝑑𝑟⃗ ௕ ௔ 𝑉(𝑟) = ୕ ସగఢ బ ୰ (point charge) 𝑉(𝑟) = − ఒ ଶగఢ బ ln ቀ ௥ ோ ቁ + 𝑉(𝑅) (infinite line of charge) 𝑉(𝑥) = 𝐸(𝑑 − 𝑥) + 𝑉(𝑑) (infinite sheet; parallel plate capacitor) 𝑄 = 𝐶𝑉 𝐶 ୮ୟ୰ୟ୪୪ୣ୪ = 𝜅𝜖 ଴ 𝐴/𝑑 𝐶 ௘௤ = 𝐶 ଵ + 𝐶 ଶ + ⋯ (parallel) ଵ ஼ ೐೜ = ଵ ஼ భ + ଵ ஼ మ + ⋯ (series) 𝑈 = ொ మ ଶ஼ 𝑉 = 𝐼𝑅 𝐼 = ௗொ ௗ௧ 𝑃 = 𝐼𝑉 = 𝐼 ଶ 𝑅 = 𝑉 ଶ /𝑅 𝑅 = ∫ ఘ ஺ 𝑑ℓ 𝑅 = 𝜌𝐿/𝐴 (solid of uniform cross section A and length L) 𝐼 = ∫ 𝚥⃗ 𝒜 ⋅ 𝑑𝐴 ⃗ 𝚥⃗ = 𝑛𝑞𝑣⃗ ௗ = ா ሬ ⃗ ఘ 𝐼 = ௗொ ௗ௧ = 𝑛𝑞𝐴𝑣 ௗ 𝑅 ௘௤ = 𝑅 ଵ + 𝑅 ଶ + ⋯ (series) ଵ ோ ೐೜ = ଵ ோ భ + ଵ ோ మ + ⋯ (parallel) ∑ 𝐼 ୧୬ ୨୳୬ୡ୲୧୭୬ = ∑ 𝐼 ୭୳୲ ୨୳୬ୡ୲୧୭୬ ∑ 𝑉 loop = 0 Charging/discharging a capacitor 𝑄(𝑡) = 𝐶V ୤ ൫1 − 𝑒 ି௧/(ோ஼) ൯ 𝑄(𝑡) = 𝐶V ଴ 𝑒 ି௧/(ோ஼) 𝑉(𝑡) = V ୤ ൫1 − 𝑒 ି௧/(ோ஼) ൯ 𝑉(𝑡) = V ଴ 𝑒 ି௧/(ோ஼) 𝐼(𝑡) = 𝐼 ଴ 𝑒 ି௧/(ோ஼) ∮ 𝐵 ሬ ⃗ ⋅ 𝑑𝑙 ⃗ = 𝜇 ଴ 𝐼 ௘௡௖ 𝐵 ሬ ⃗ = ఓ బ ସగ ∫ ூ ௗℓ ሬ ⃗ ×௥̂ ௥ మ 𝐵 = ఓ బ ூ ଶగ௥ (infinitely long wire) 𝐵 = ఓ బ ூ ଶோ (center of thin loop of radius R) 𝐵 = 𝜇 ଴ 𝑛𝐼 (solenoid with n loops/length) 𝜇⃗ ≡ 𝐼𝐴𝑛 ො 𝜏⃗ = 𝜇⃗ × 𝐵 ሬ ⃗ 𝑈 ஻ = −𝜇⃗ ∙ 𝐵 ሬ ⃗ 𝛷 ஻ = ∫ 𝐵 ሬ ⃗ ⋅ 𝑑𝐴 ⃗ = 𝐵𝐴 cos 𝜃 ℰ = 𝑣𝑙𝐵 ℰ = − ௗః ಳ ௗ௧ ℰ = 𝑁 ቚ ୼ః ୼௧ ቚ 𝐹 ⃗ = 𝑞𝑣 ⃗ × 𝐵 ሬ ⃗ 𝐹 ⃗ = ∫ 𝐼𝑑ℓ ሬ ⃗ × 𝐵 ሬ ⃗ EM Waves 𝑣 = 𝑓𝜆 = ఠ ௞ , 𝜔 = 2𝜋𝑓 , 𝑘 = ଶగ ఒ 𝐸 ଴ = 𝑐𝐵 ଴ 𝐸 ௬ (𝑥, 𝑡) = 𝐸 ଴ 𝑐𝑜𝑠(𝑘𝑥 ± 𝜔𝑡) 𝐵 ௭ (𝑥, 𝑡) = 𝐵 ଴ 𝑐𝑜𝑠(𝑘𝑥 ± 𝜔𝑡) 𝑆 ⃗ = 𝐸 ሬ ⃗ × 𝐵 ሬ ⃗ (intensity of EM wave) 𝐼 = 𝑃/𝑎 𝐼 = 𝑃/4𝜋𝑟 ଶ 𝐼 = ଵ ଶ ට ఢ బ ఓ బ 𝐸 ሬ ⃗ ଴ ଶ = ଵ ଶ 𝑐𝜖 ଴ 𝐸 ሬ ⃗ ଴ ଶ 𝐼 = 𝐼 ଴ cos ଶ 𝜃 < cos ଶ 𝜃 > = ଵ ଶ 𝑐 = ଵ ඥఢ బ ఓ బ = 3 × 10 ଼ m/s Optics 𝜃′ = 𝜃 𝑛 ଵ sin 𝜃 ଵ = 𝑛 ଶ sin 𝜃 ଶ 𝜃 ୡ = sin ିଵ ቀ ௡ మ ௡ భ ቁ 𝑛 = ௖ ௩ 𝑠 ᇱ = ௡ మ ௡ భ 𝑠 or 𝑑 ௜ = ௡ మ ௡ భ 𝑑 ௢ 𝑀 = ௛ ೔ ௛ ೚ = − ௗ ೔ ௗ ೚ = − ௦ ᇲ ௦ ଵ ௗ ೔ + ଵ ௗ ೚ = ଵ ௙ or ଵ ௦ + ଵ ௦ ᇲ = ଵ ௙ 𝑃 = ଵ ௙ 𝑓 = ோ ଶ (mirror) Some 8A Formulas 𝛥𝑥 = 𝑣 ଴ 𝛥𝑡 (constant 𝑣 ଴ ) Constant acceleration motion: o 𝛥𝑥 = 𝑣 ଴ 𝛥𝑡 + ଵ ଶ 𝑎(𝛥𝑡) ଶ o 𝑣 = 𝑣 ଴ + 𝑎 𝛥𝑡 o 𝑣 ଶ = 𝑣 ଴ ଶ + 2𝑎(𝛥𝑥) Uniform circular motion: o 𝜔 = ଶగ ் o 𝑣 = ଶగோ ் = 𝜔𝑅 o 𝑎 = 𝜔 ଶ 𝑅 = ௩ మ ோ 𝐹 ⃗ = 𝑚𝑎 ⃗ Torque: 𝜏 = 𝑟𝐹𝑠𝑖𝑛𝜃

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

𝛥𝐾 + 𝛥𝑈 = 𝑊 ௡௖ 𝑊 = ∫ 𝐹 ⃗ ⋅ 𝑑𝑟⃗ = −𝛥𝑈 𝐾 = ଵ ଶ 𝑚𝑣 ଶ , 𝑈 ௚ = 𝑚𝑔ℎ Derivatives ௗ ௗ௫ (𝑎𝑥 ௠ ) = 𝑎𝑚𝑥 ௠ିଵ ௗ ௗ௫ (ln(𝑎𝑥)) = ଵ ୶ ௗ ௗ௫ (𝑒 ௔௫ ) = 𝑎𝑒 ௔௫ ௗ ௗ௫ (sin(𝑎𝑥)) = a cos(𝑎𝑥) ௗ ௗ௫ (cos(𝑎𝑥)) = −a sin(𝑎𝑥) Integrals ∫ 𝑥 ௠ 𝑑𝑥 = ௫ ೘శభ ௠ାଵ for 𝑚 ≠ −1 ∫ ଵ ௫ 𝑑𝑥 = ln 𝑥 ∫ 𝑥 ௡ 𝑒 ି௫ 𝑑𝑥 ஶ ଴ = 𝑛! ∫ ௗ௫ ୶ିୟ = ln(𝑥 − 𝑎) ∫ ௗ௫ (୶ିୟ) మ = ିଵ ௫ି௔ ∫ 𝑥 ଶ 𝑒 ௔௫ = e ௔௫ ( ௫ మ ௔ − ଶ௫ ௔ మ + ଶ ௔ య ) Trigonometry cos (0 ∘ ) = − cos (180 ∘ ) = 1 sin (90 ∘ ) = − sin (270 ∘ ) = 1 cos (90 ∘ ) = cos(270 ∘ ) = 0 sin( 0) = sin (180 ∘ ) = 0 cos (30 ∘ ) = sin (60 ∘ ) = √ଷ ଶ = 0.866 cos (60 ∘ ) = sin (30 ∘ ) = ଵ ଶ cos (45 ∘ ) = sin (45 ∘ ) = ଵ √ଶ = 0.707 sin (90 ∘ − 𝜃) = cos (𝜃) cos (90 ∘ − 𝜃) = sin (𝜃) Geometry circumference circle: 𝑠 = 2𝜋𝑅 area of a disk: 𝐴 = 𝜋𝑅 ଶ surface area of a sphere: 4𝜋𝑅 ଶ volume of a sphere: ସగ ଷ 𝑅 ଷ volume of a cylinder: 𝜋𝑅 ଶ ℎ lateral area of a cylinder: 2𝜋𝑅ℎ Arclength on circle: 𝑠 = 𝑟𝜃 Sum of interior angles of a triangle equals 𝜋 radians Equation of line: 𝑦 = 𝑚𝑥 + 𝑏 Constants and Units 1 V=1 J/C 1 V/m = 1 N/C 𝜇 ଴ = 4𝜋 × 10 ି଻ T m/A 𝜀 ଴ = 9 × 10 ିଵଶ C 2 /N∙m 2 , k = 1/(4 πε 0 ) = 9 × 10 9 N∙m 2 /C 2 Elementary charge 𝑒 = 1.6 × 10 ିଵଽ C Mass of electron 𝑚 ௘ = 9 × 10 ିଷଵ kg 𝑒 ିଵ = 0.37 , 𝜋 = 3.142 , 4𝜋 = 12.57 𝑐 = 3 × 10 ଼ m/s Medium Index of refraction n Air 1.0003 Water 1.33 = 4/3 Glass 1.5 = 3/2  Fold page here to use as a ruler/straight edge You may tear off this sheet for your use during the exam!

Physics 8B-002 (Barsky) Midterm 2 Nov 10, 2022 Name________________________ 5 O O O O O Name: __________________ Student ID: ___________ Discussion/Lab Section: _______________ Multiple choice [20 points total: 10 questions x 2 points each] Please pick the single best choice for each question. Please clearly indicate your choice by writing or circling the letter of your choice. If your choice is ambiguous, or more than one choice is indicated, it will be marked wrong. No partial credit will be awarded. Please FILL IN THE BUBBLES AT LEFT! 1) Three particles travel through a region of space where the magnetic field is out of the page, as shown in the figure. The electric charge of each of the three particles is, respectively A) 1 is negative, 2 is neutral, and 3 is positive. B) 1 is positive, 2 is negative, and 3 is neutral. C) 1 is neutral, 2 is negative, and 3 is positive. D) 1 is negative, 2 is positive, and 3 is neutral. E) 1 is positive, 2 is neutral, and 3 is negative. 2) A negatively charged particle is moving to the right, directly above a wire having a current flowing to the right, as shown in the figure. In which direction is the magnetic force exerted on the particle? A) upward B) downward C) into the page D) out of the page E) The magnetic force is zero since the velocity is parallel to the current. 3) A long straight conductor has a constant current flowing to the right. A wire rectangle is situated above the wire, and also has a constant current flowing through it (as shown in the figure). Which of the following statements is true regarding the net force and net torque on the wire rectangle? A) The net force is down, and there is a non-zero net torque. B) The net force is up, and there is a non-zero net torque. C) The net force is down, and there is zero net torque. D) The net force is up, and there is zero net torque. E) The net force is zero, and there is zero net torque. O O O O O O O O O O A) Apply magnetic force (Lorentz) and right hand rule: F =q v x B. Sign of charge can reverse force. First find the magnetic field due to the wire, using the right hand rule (Thumb in direction of current). Then apply the magnetic force as above. Again, the negative charge reverses the force. 1 pt: Mag field from straight wire is out of page and grows weaker with distance from the wire. Force on lower wire of loop is downward and stronger than force on upper wire (which is upward). So net force is downward. The two side wires have equal and opposite forces. 1 pt: There is no net torque because all forces are in the plane of the loop.

4) In the figure, the inner loop carries a clockwise current I that is decreasing. The resistor R is in the outer loop and both loops are in the same plane. The induced current through the resistor R is A) From a to b . B) From b to a . C) There is no induced current. 5) In an electromagnetic wave, the electric and magnetic fields are oriented such that they are A) parallel to one another and parallel to the direction of wave propagation. B) perpendicular to one another and parallel to the direction of wave propagation. C) perpendicular to one another and perpendicular to the direction of wave propagation. D) parallel to one another and perpendicular to the direction of wave propagation. 6) A ray of light goes from one transparent material into another, as shown in the figure. What can you conclude about the indices of refraction of these two materials? A) n 2 > n 1 B) n 1 > n 2 C) n 1 = n 2 D) The relationship between cannot be determined from this diagram alone. 7) When light goes from one material into another material having a HIGHER index of refraction, A) its speed decreases but its frequency and wavelength stay the same. B) its speed decreases but its wavelength and frequency both increase. C) its speed increases, its wavelength decreases, and its frequency stays the same. D) its speed, wavelength, and frequency all decrease. E) its speed and wavelength decrease, but its frequency stays the same. 8) A negative magnification for a mirror means that A) the image is upright. B) the image is inverted. C) the image is on the opposite side of the mirror as the object. D) the image is reduced E) Two of the above are correct. O O O O O O O O O O O O O O O O O O O O O The magnetic field due to the small loop is into the page and decreasing. Lenz's law says that the outer loop will have an induced current that goes against the change; therefore, there will be a clockwise current induced in the outer loop, which means current will flow b to a. The light bends away from the normal so medium 2 must have a lower index of refraction according to Snell's law (or simply that changes in angle are inverse to changes in index of refraction. EM waves are transverse and the E and B fields are mutually perpendicular, so all three are mutually perpendicular. By definition the speed decreases. Given that and wavelength f = speed, we see that the wavelength must decrease. By definition a negative magnification means the image is inverted (relative to the object). None of the other choices are correct.

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Physics 8B-002 (Barsky) Midterm 2 Nov 10, 2022 Name________________________ 7 Please write your name at the top of every sheet of paper in the rest of the exam. 9) A single current-carrying circular loop of radius R is placed next to a long, straight wire, as shown in the figure. The current I in the wire flows to the right. In which direction must current flow in the loop to possibly produce a net magnetic field of zero at its center? A) clockwise B) counterclockwise C) It could be either clockwise or counterclockwise. D) The current in the loop should be zero. 10) A lighted candle is placed a short distance from a plane mirror, as shown in the figure. At which location will the image of the flame appear to be located? A) at A B) at B C) at C D) at M (at the mirror) O O O O O O O O The magnetic field from the straight wire is out of the page in the loop. So the current in the loop must be clockwise to counter the field from the straight wire. The image appears to be directly opposite the object as we often experience in the bathroom.

Please read all the instructions for the problems on the cover sheet! Most importantly: To receive full credit on the problems, all reasoning must be shown . 1) (24 pts) You move a rigid square loop of wire with constant velocity v 0 in the x-direction into a constant, uniform magnetic field. The magnetic field is zero for x < 0. For x > 0 the magnetic field has magnitude B 0 in the +z direction. The field extends well beyond the loop along the y-axis. Each edge of the loop has length L and resistance R, and the loop lies in the x-y plane. The following questions a) – c) concern the situation where the leading edge is moving into the field (beyond x=0), and the trailing edge is outside the field (before x=0), as shown in the figure. For the following, give your answer in terms of the symbols defined here (such as B 0 , L, v 0 ), as well as any known constants, such as ε 0 , π, etc. You must show your work and explain your reasoning. a) What is the electric current (magnitude and direction), if any, through segment AB? (For direction, state “A to B” or “B to A.”) dE -EB ↓ As the loop moves into the magnetic field, the flux through the loop changes with time. Therefore, there will be an induced EMF and induced current in the loop (according to Faraday's law). Y x00 (dEll B) * E3=BAcosE & area of the loop ↓ inside B E = BLx D a ***** Me * I = ERloop = BLUo 4 R & total resistance of the loop is R Lenz's law: There will be a Bind in the loop to oppose the change in flux.. It is increasing, so Bind must point in the -z direction => I flows Clockwise viewed from above, so the direction of I is B to A. b

b) What is the net force (magnitude and direction), if any, on the magnetic loop? After some time the whole loop will be in the magnetic field, its velocity still v 0 in the x-direction. c) In this situation, what is the electric current (magnitude and direction), if any, through segment AB? d) In this situation, how much work must you do, if any, to push the loop a distance L in the x direction? The net force on the loop is a because the loop is moving at constant velocity. Therefore the force are to the magnetic field and the force you apply have the same magnitude and point in opposite directions. Y C The magnetic force on segments (B and x AD cancel. The total magnetic force is just Is the free or segment st ↓ answer to part a FB =1([xB) D R IF31= ILBsinod 900 Direction of on each segment shown in red. I F3 = BZvo UR #points in the -X dir. (by right hand rule), so the applied force is in the + x dir. Now the magnetic flux through the loop is not changing with time, so there is no induced EMF and no current. E = BL2 z = fB = O I = = = 0 I there is no current in the loop, then the force due to the magnetic field is 0 (E3=I(Ex) = 0) . You do not need to apply a force on the loop to keep it moving at constant velocity, so you do no work, * * *The answer "0" would earn full credit! Most students computed the magnetic (or applied) force for which they would also get full credit!

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Physics 8B-002 (Barsky) Midterm 2 Nov 10, 2022 Name While the loop is fully in the magnetic field you begin to rotate the loop with rotational speed ! O about its central axis by lifting the segment CD and pushing down the segment AB. e) Derive an expression for the electric current (magnitude and direction) as a function of time through segment AB? (You may set " = 0 to be a moment in which the loop is in the original orientation.) z B d A As the loop rotates, Ep will change, so there will be an induced EMF and current. D ↓ o = not x * f is 13 = BAcos(E) = Bcos (not 3 = - d #B=-Bla (cost] on the - S Formwke - rusin(not) sheet. z = BLwsin (wt) I = = Bo (rsin(ut i MR Initially I will be decreasing, so current will flow A-B to create a Bind that increases the flux. After half a period of rotation (8>1809, E will increase, so the current will reverse and flow BA. The current will keep alternating as the loop rotates. When I is positive, current flows A-B. When I is negative, current flows BA.

f) Extra credit (+3 pts): Derive an expression for the power required to accomplish this rotation. 9 The power required to rotate the loop will be dissipated in the resistance of the loop. The power dissipated in the loop is ~or ITReoop Or be on P= I S - BoRwsin (cot Boliwsin(wt) MR P = Bo L"wisin (with 4 R

2) (18 pts) Suppose you have a concave mirror with a radius of curvature R = 6 cm. You place a small object 2 cm away from the mirror. Our goal will be to characterize the image produced by this arrangement. a) Draw the mirror in the space below, and ray trace this situation using the space provided below. Be sure that you label the mirror, object, image, foci, and R . And include all three principal (or “special”) rays. (Note you can use the ruler provided on the formula sheet!) b) Use the mirror equation to algebraically solve the problem of where the image appears and reconcile with your ray tracing. c) With justification from the figure and your algebra, explain how we know if the image is real or virtual. d) Calculate the size and the magnification of the image. e) With justifications from the figure and your algebra, explain how we know if the image is upright or inverted. The image distance is negative, which for a mirror, means the image is on the opposite side of the mirror from the object. This is consistent with the ray-tracing which shows that you can only see the image by looking into the mirror. This is a "cosmetic mirror" situation. The magnification is positive; therefore, the image is upright (un-inverted). This is consistent with the ray- tracing.

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

Physics 8B-002 (Barsky) Midterm 2 Nov 10, 2022 Name________________________ 11 3. [ 12 pts] a) What is the wavelength of KALX (91 MHz), calculated to one significant figure? (Assume that transmission in air is equivalent to that in the vacuum.) b) Write an expression for the E field transmission along a line from the antenna to you. Use a coordinate system where the antenna is at the origin and a straight line to you is the x-axis. Define any symbols you use. c) If the transmission is 5 km from your radio, assuming the transmitted energy goes equally in all directions, what is the power of the station if your radio receives 1 microwatt using an antenna with an effective surface area of 10 sq cm? (Again, do the calculations to one sig. fig.) The final answer was erroneously written as "30 cm" in the original posting. Corrected Dec 29).

4. [12 pts] You have been asked to find a glowing toy in a pond of water as you look into the pond from the bank as in the figure. (In this problem ignore the possibility of light scattering or reflecting off other objects, dust, etc.) a) If the toy sat on the bottom, 4 m below the surface, and near the edge of the pool closest to you, about how far below the surface would the toy appear to be? b) In the situation shown above, draw a ray of light from the toy to your eyes. Include any reflections at the surface and label the relevant angles (incident, reflected, etc.). Write the relationship between the angles using simply an equal sign (=) or inequalities (< or >) in the following space: c) Assuming perfect clarity of the water (meaning you could see through any amount of water), could you see the light from the toy regardless of how far it is from the edge where you are standing? Explain in words and using your diagram. (No credit here for “yes” or “no.” You must make a physical/ mathematical argument.) When objects in a medium of high index of refraction n are view from a medium of lower n , they look magnified. This was derived in lecture and in the textbook (Eq. 2.10), and it is on the formula sheet (circled at right). Therefore, the apparent depth of the toy is ( n air / n wat ) 4 m (1/(4/3)) 4 m = 3 m. No, there is a distance at which the incident angle is the critical angle. At this distance--and all greater distances--the refracted light disappears; there is "total internal reflection," and the toy cannot be seen, no matter how low you crouch, as long as you remain outside the water. Students were not asked to calculate the distance, but it is related to the depth by geometry. Use Snell's law and tan (critical angle) = max_dist/depth. This gives max_dist = depth/sqrt(1.33^2 - 1) ~ 5 m.

Physics 8B-002 (Barsky) Midterm 2 Nov 10, 2022 Name________________________ 13 5. [14 pts] Two concentric, very long conducting cylinders are shown in the figure. The inner cylinder is a solid conductor and carries a current I , uniformly distributed over its volume . The outer cylinder is a conducting shell that carries a total current -2 I . Treat I and the radii R 1 , R 2 , and R 3 as known. In the following show your work/logic carefully and report the 𝐵 ሬ⃗ field magnitude and direction. The points (A, B, C) should be described by the radii: R A , R B , R C . Your answers to the following can use any symbols defined above as well as known constants such as k , ε 0 , π, etc. a. What is the 𝐵 ሬ⃗ field at point A? b. What is the 𝐵 ሬ⃗ field at point B? c. What is the 𝐵 ሬ⃗ field at point C? Use Ampere's law for all three points. The integral becomes just the magnetic field times a circle's circumference since the B-field is constant and circumferential. The total current enclosed is - I since I - 2 I = - I . The direction is circumferential and clockwise in the cross section. The current through the inner conductor is to the right. (Out of the page in the cross-section.) The direction is circumferential and counter- clockwise in the cross section. We again apply Ampere's law but now we need to divide I_0 by the fraction of area enclosed. The direction is circumferential and counter-clockwise in the cross section.

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

On my honor I have neither given nor received any unfair assistance during this exam. Signed ______________________________ If you run out of space on a problem, you may use the space below. You must indicate very carefully what problem you address here and also in the place where the problem originated.

Exam2-draft-v5-final-solutions_v2

Related Documents