Collin CollegePHYS 2426 .]CheatSheetFinal exam, previously tested in the exam

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Quiz 7: Electric Force and Electric (1402112) 1 This print-out should have 36 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. ... 001 10.0 points A strong lightning bolt transfers about 25 C to Earth. How many electrons are transferred? The charge on an electron is −1.60218 × 10−19 C. Correct answer: 1.56038 × 1020 . Explanation: Let : q = 25 C and qe = −1.60218 × 10−19 C . The charge is proportional to the number of electrons, so q = n |qe| n = q |qe| = 25 C |−1.60218 × 10−19 C| = 1.56038 × 1020 . 002 10.0 points Two charged particles of equal magnitude (+Q and −Q) are fixed at opposite corners of a square that lies in a plane (see figure below). A test charge −q is placed at a third corner. +Q −q −Q What is the direction of the force on the test charge due to the two other charges? 1. 2. correct 3. 4. 5. 6. 7. 8. Explanation: The force between charges of the same sign is repulsive. The force between charges with opposite signs is attractive. +Q −q −Q The resultant force is the sum of the two vectors in the figure. 003 10.0 points Suppose that 4.6 g of hydrogen (H2) is separated into electrons and protons, and that the protons are placed at the Earth’s North Pole and the electrons are placed at the South Pole. What is the resulting compressional force on the Earth? The radius of the Earth is 6.37 × 106 m, Avogadro’s number is 6.02214× 1023 and the molar mass of the H-atom is 1.00782 g/mol. Correct answer: 1.07392 × 107 N. Explanation: sapaugh (js67869) – Quiz 7: Electric Force and Electric Field – balasubramanya – (1402112) 2 Let : m = 4.6 g = 0.0046 kg , RE = 6.37 × 106 m , NA = 6.02214 × 1023 mol−1 , qe = 1.60218 × 10−19 C , and ke = 8.98755 × 109 N · m2 /C 2 . The mass m is proportional to n, the number of hydrogen atoms. The number of moles of hydrogen can be expressed two ways: m MH = n NA n = NA MH m =  6.02214 × 1023 mol−1 1.00782 g/mol  (4.6 g) = 2.74867 × 1024 atoms. Since each of these atoms is split into a proton and an electron, there will be n protons at the North Pole and n electrons at the South Pole, and the force is F = ke  n qe 2 RE 2 = 8.98755 × 109 N · m2 /C 2
×  2.74867 × 1024 2 (6.37 × 106 m)2 × 1.60218 × 10−19 C
2 = 1.07392 × 107 N . 004 10.0 points Four point charges are placed at the four corners of a square. Each side of the square has a length L. b b b b b q1 = −q q3 = q q2 = q q4 = q P L L Find the magnitude of the electric force on q2 due to all three charges q1, q3 and q4 for L = 1 m and q = 1.04 µC. Correct answer: 0.0145814 N. Explanation: F1 F4y F4 F3 F4x q2 From the figure, F1x = − k q2 L2 = − 8.98755 × 109 N · m2 /C 2
× (1.04 × 10−6 C)2 (1 m)2 = −0.00972094 N , F3y = − k q2 L2 = − 8.98755 × 109 N · m2 /C 2
× (1.04 × 10−6 C)2 (1 m)2 = −0.00972094 N , F4x = k q2 2 L2 1 √ 2 = 8.98755 × 109 N · m2/C 2 2 · (1 m)2 × (1.04 × 10−6 C)2 √ 2 = 0.00343687 N , and F4y = − k q2 2 L2 1 √ 2 = − 8.98755 × 109 N · m2/C 2 2 · (1 m)2 × (1.04 × 10−6 C)2 √ 2 = −0.00343687 sapaugh (js67869) – Quiz 7: Electric Force and Electric Field – balasubramanya – (1402112) 3 Fx = F1x + F4x = −0.00972094 N + 0.00343687 N = −0.00628407 N and Fy = F3y + F4y = −0.00972094 N + (−0.00343687 N) = −0.0131578 N , so kF~ k 2 = F 2 x + F 2 y = (−0.00628407 N)2 + (−0.0131578 N)2 = 0.000212617 N2 kF~ k = √ 0.000212617 N2 = 0.0145814 N . 005 10.0 points Two equal charges exert equal forces on each other. What if one charge has twice the magnitude of the other? 1. The bigger charge will exert a force four times as strong. 2. The forces will be equal. correct 3. The smaller charge will exert a force twice as strong. 4. The bigger charge will exert a force twice as strong. 5. The smaller charge will exert a force four times as strong. Explanation: The forces they exert on each other are still the same; Newton’s third law applies to all forces. 006 10.0 points Three identical point charges hang from three strings, as shown. 45◦ 45◦ 12.0 cm 12.0 cm + + + +q +q +q 0.10 kg 0.10 kg 0.10 kg What is the value of q? The Coulomb constant is 8.98755 × 109 N · m2 /C 2 , and the acceleration due to gravity is 9.81 m/s 2 . Correct answer: 7.92912 × 10−7 C. Explanation: Let : m = 0.10 kg , L = 12.0 cm = 0.12 m , θ = 45◦ , and ke = 8.98755 × 109 N · m2 /C 2 . r = 2 L sin θ = 2 L sin 45◦ = 2 L √ 2 2 = L √ 2 FT,x = FT sin θ FT,y = FT cos θ Each sphere is in equilibrium horizontally Fe − FT,x = 0 Fe − FT sin θ = 0 and vertically FT,y − Fg = 0 FT cos θ − Fg = 0 FT = Fg cos θ . sapaugh (js67869) – Quiz 7: Electric Force and Electric Field – balasubramanya – (1402112) 4 From the horizontal equilibrium, Fe =  Fg cos θ  sin θ = Fg tan θ = Fg (tan 45◦ ) = Fg . For either of the end charges, Fe = ke q 2 r 2 + ke q 2 r 2
2 = ke q 2 r 2 + 4 ke q 2 r 2 = 5 ke q 2 r 2 5 ke q 2 r 2 = m g , so |q| = s r 2 m g 5 ke = s (L √ 2)2 m g 5 ke = L r 2 m g 5 ke = (0.12 m) × s 2(0.1 kg)(9.81 m/s 2 ) 5(8.98755 × 109 N · m2/C2 ) = 7.92912 × 10−7 C . 007 10.0 points An electron remains suspended between the surface of the Earth (assumed neutral) and a fixed positive point charge, at a distance of 3.48 m from the point charge. Determine the charge required for this to happen. The acceleration due to gravity is 9.8 m/s 2 and the Coulomb constant is 8.98755 × 109 N · m 2 /C 2 . Correct answer: 7.50796 × 10−20 C. Explanation: Let : ke = 8.98755 × 109 N · m2 /C 2 and r = 3.48 m . For the electron to remain suspended, we must have ke q qe r 2 = me g q = me g r2 ke qe =  9.10939 × 10−31 kg 8.98755 × 109 N · m2/C2  × (9.8 m/s 2 ) (3.48 m)2 1.60218 × 10−19 C = 7.50796 × 10−20 C . This charge is not an integer times e. Since charge is quantized to be a multiple of e , the above observation is not possible. 008 (part 1 of 2) 10.0 points A charge of 7 µC is at the origin, and a charge of 7 µC is on the x axis at x = 1 m. Find the force on charge q2. The Colulomb constant is 8.98755 × 109 N · m 2 /C 2 . Correct answer: 0.44039 N. Explanation: Let : k = 8.98755 × 109 N · m2 /C 2 , q1 = 7 µC = 7 × 10−6 C , q2 = 7 µC = 7 × 10−6 C , x1 = 0 m , and x2 = 1 m . The force on charge q2 is F~ 1,2 = k q1 q2 r 2 1,2 ˆr1,2 = k q1 q2 (x2 − x1) 2 ˆr1,2 = (8.98755 × 109 N · m 2 /C 2 ) × (7 × 10−6 C) (7 × 10−6 C) (1 m − 0 m)2 ˆı = 0.44039 N ˆı . 009 (part 2 of 2) 10.0 points Find the force on q1. Correct answer: −0.44039 N. Explanation sapaugh (js67869) – Quiz 7: Electric Force and Electric Field – balasubramanya – (1402112) 5 The force on charge q1 is F~ 2,1 = −F~ 1,2 = −0.44039 N ˆı . 010 10.0 points As the distance from a charged particle decreases, the strength of the electric field 1. increases. correct 2. remains the same. 3. Not enough information is given. 4. decreases. Explanation: 011 (part 1 of 3) 10.0 points Five equal negative point charges −q are placed symmetrically around a circle of radius R as shown. O y x R −q −q −q −q −q a b c d e Use a rough approximation to express the magnitude of the electric field E~ at a point on the positive x axis a distance 5 R from the center of the circle. 1. kE~ k ≈ 0 2. kE~ k ≈ ke q 2 R2 3. kE~ k ≈ ke q R2 4. kE~ k ≈ ke q R3 5. kE~ k ≈ q R 6. kE~ k ≈ ke 5 q R2 correct 7. kE~ k ≈ 5 q 2 R 8. kE~ k ≈ 5 ke q R2 9. kE~ k ≈ ke q 2 R 10. kE~ k ≈ ke q R Explanation: Assume that all five negative charges are located at their symmetric center. The electric field is E~ = ke 5 q (5 R) 2 (−ˆı) kE~ k ≈ ke 5 q R2 . 012 (part 2 of 3) 10.0 points What is the direction of the electric field E~ at a point on the positive x axis a distance 5 R from the center of the circle? 1. In the xy plane pointing in an arbitrary direction away from the center of the circle 2. ⊙, along the positive z axis, up out of the plane of the circle 3. ↑, along the positive y axis 4. ⊗, along the negative z axis, down out of the plane of the circle 5. ←, along the negative x axis correct 6. ↓, along the negative y axis 7. →, along the positive x axis 8. None of these 9. In the xy plane pointing in an arbitrary direction towards the center of the circle Explanation: The direction of the electric field points away from a positive charge towards a negative charge. Since all the charges are negative and to the left of a point on the x axis at sapaugh (js67869) – Quiz 7: Electric Force and Electric Field – balasubramanya – (1402112) 6 a distance 2 R from the center of the circle, the electric field must point towards the left (toward the negative charges). From symmetry, the electric field E~ points only along the negative x axis and has no y component (←). 013 (part 3 of 3) 10.0 points Since the charges are distributed about the center in a symmetrical fashion, the electric field at the center is likely to be very small. If charges b and e are removed, what is the direction of the electric field E~ at the center of the circle due to charges a, c, and d? 1. None of these 2. ⊙, along the positive z axis, up out of the plane of the circle 3. In the xy plane pointing in an arbitrary direction away from the center of the circle 4. In the xy plane pointing in an arbitrary direction towards the center of the circle 5. ↑, along the positive y axis 6. ↓, along the negative y axis 7. →, along the positive x axis 8. ⊗, along the negative z axis, down out of the plane of the circle 9. ←, along the negative x axis correct Explanation: The electric field due to a point charge q at a distance r from it is given by E~ = ke q r 2 ˆr , where ˆr is the unit vector directed from the charge to the point. The total electric field due to the group of 5 charges is the vector sum of the electric fields of all the charges at that point. E~ = ke X i qi r 2 i ˆri . Since the charges are symmetrically placed, the angle made at the center by any two charges is 72◦ . Calculate the electric field at the center due to each charge and resolve them into x and y components as we calculate. Starting from the one at zero degrees and going counterclockwise, E1 = ke q R2 ˆı E2 = ke q R2 (cos 72◦ ˆı + sin 72◦ ˆ) E3 = ke q R2 (− cos 36◦ ˆı + sin 36◦ ˆ) E4 = ke q R2 (− cos 36◦ ˆı − sin 36◦ ˆ) E5 = ke q R2 (cos 72◦ ˆı − sin 72◦ ˆ). Ex = E1x + E2x + E3x + E4x + E5x = ke q R2 (1 + cos 72◦ − cos 36◦ − cos 36◦ + cos 72◦ ) = ke q R2 (1 + 0.309017 − 0.809017 − 0.809017 + 0.309017) = 0 . Ey = E1y + E2y + E3y + E4y + E5y = ke q R2 (sin 72◦ + sin 36◦ − sin 36◦ − sin 72◦ ) = 0 If charges a, c, and d were removed instead of charges b and e, the electric field E~ b,e would point along the positive x axis (→), towards the symmetric center of negative charge (b and e). Since the magnitude of the electric field due to all the charges E~ a,b,c,d,e is zero, the electric field due to charges a, c, and d must be equal and opposite. Consequently, the electric field E~ a,c,d due to charges a, c, and d points along the negative x axis (←). 014 (part 1 of 2) 10.0 points On a dry winter day, if you scuff your feet across a carpet, you build up a charge and get a shock when you touch a metal doorknob. sapaugh (js67869) – Quiz 7: Electric Force and Electric Field – balasubramanya – (1402112) 7 In a dark room you can actually see a spark about 2 cm long. Air breaks down at a field strength of 3 × 106 N/C. How much charge have you built up? Assume that just before the spark occurs, all the charge is in your finger, drawn there by induction due to the proximity of the doorknob. Approximate your fingertip as a sphere of diameter 1.36 cm, and assume that there is an equal amount of charge on the doorknob 2 cm away. Correct answer: 1.54347 × 10−8 C. Explanation: Let : x = 2 cm , d = 1.36 cm , and Ebreakdown = 3 × 106 N/C . In order for air to break down, there must be a field strength of Ebreakdown just outside your fingertip. The field due to a spherically symmetric charge distribution is the same as the field due to a point charge at its center. Therefore the field just outside the fingertip is E = ke q r 2 where r = 1.36 cm 2 = 0.0068 m is the radius of the approximating sphere. At the point when you have built up enough charge to make a spark jump, this field E is equal to Ebreakdown, so q = Ebreakdown r 2 ke = (3 × 106 N/C) (0.0068 m)2 8.98755 × 109 N · m2/C2 = 1.54347 × 10−8 C . 015 (part 2 of 2) 10.0 points How many electrons does this correspond to? The elemental charge is 1.60218 × 10−19 C. Correct answer: 9.63357 × 1010 . Explanation: Let : qe = 1.60218 × 10−19 C . For N electrons, q = N qe N = q qe = 1.54347 × 10−8 C 1.60218 × 10−19 C = 9.63357 × 1010 . 016 10.0 points Consider three electric field patterns. (Some of these patterns are physically impossible.) Assume these electric field patterns are due to static electric charges outside the regions shown. (a) (b) (c) Which electrostatic field patterns are physically possible? 1. (a) and (b) 2. (a) and (c) 3. (b) and (c) 4. (c) only 5. (b) only correct sapaugh (js67869) – Quiz 7: Electric Force and Electric Field – balasubramanya – (1402112) 8 6. (a) only Explanation: (a) Electrostatic lines of force do not intersect one another. Otherwise at the crossing point there would be an unphysical situation. A charged particle placed at the crossing point would not experience a unique physical force. Therefore (a) is not possible. (b) The electric charges at which the lines of force begin and end are out of the region on the left and right, respectively. Therefore (b) is possible. (c) In electrostatics lines of force begin and end at electric charges (or at infinity). In a localized region that contains no charges, therefore, no closed loop pattern is possible. Therefore (c) is not possible. 017 (part 1 of 2) 10.0 points Consider the electric field lines for two point charges separated by a small distance. q1 q2 What are the signs of q1 and q2? 1. q1 > 0 and q2 < 0 correct 2. q1 < 0 and q2 < 0 3. q1 > 0 and q2 > 0 4. q1 < 0 and q2 > 0 Explanation: Field lines are directed toward negative charges, so q1 > 0 and q2 < 0. 018 (part 2 of 2) 10.0 points What relationship does the ratio R = q1 q2 have? 1. R < −1 2. 0 < R < 1 3. R > 1 4. −1 < R < 0 correct Explanation: A stronger field is depicted by field lines of higher density, so |q2| > |q1| 1 > |q1| |q2| > 0 −1 < − |q1| |q2| < 0 −1 < q1 q2 < 0 since q1 < 0 and q2 > 0 . 019 (part 1 of 2) 10.0 points An electron and a proton are each placed at rest in an electric field of 334 N/C. What is the velocity of the electron 54 ns after being released? Consider the direction parallel to the field to be positive. The elementary charge is 1.60218 × 10−19 C and the mass of an electron is 9.10939 × 10−31 kg . Correct answer: −3.17221 × 106 m/s. Explanation: Let : E = 334 N/C , qe = −1.60218 × 10−19 C , me = 9.10939 × 10−31 kg , t = 54 ns = 5.4 × 10−8 s, and v0 = 0 m/s. The acceleration of each particle is a = FE m = q E m where q and m are different for the two particles (yielding different accelerations). Since they start from rest, v = v0 + a t = q E t m . sapaugh (js67869) – Quiz 7: Electric Force and Electric Field – balasubramanya – (1402112) 9 For the electron ve = qe E t me = (−1.60218 × 10−19 C) (334 N/C) 9.10939 × 10−31 kg × (5.4 × 10−8 s) = −3.17221 × 106 m/s . 020 (part 2 of 2) 10.0 points What is the velocity of the proton 54 ns after being released? The mass of the proton is 1.67262 × 10−27 kg. Correct answer: 1727.64 m/s. Explanation: Let : qp = 1.60218 × 10−19 C and mp = 1.67262 × 10−27 kg . For the proton vp = ve E t mp = 1.60218 × 10−19 C
(334 N/C) 1.67262 × 10−27 kg × (5.4 × 10−8 s) = 1727.64 m/s . 021 10.0 points The electron gun in a television tube is used to accelerate electrons (mass of 9.11 × 10−31 kg and charge of −1.6 × 10−19 C) from rest to 3 × 107 m/s within a distance of 1.4 cm. What electric field is required? Correct answer: 1.83013 × 105 N/C. Explanation: Let : me = 9.11 × 10−31 kg , qe = 1.6 × 10−19 C , v = 3 × 107 m/s, and d = 1.4 cm . The magnitude of the force is F = qe E = me a a = qe E me The final velocity is v 2 f = v 2 i + 2 a d = 2 a d since vi = 0, so v 2 = 2 d qe E me E = v 2 me 2 d qe =  3 × 107 m/s 2  9.11 × 10−31 kg
2 (1.4 cm) (1.6 × 10−19 C) = 1.83013 × 105 N/C . 022 (part 1 of 3) 10.0 points A proton has an initial velocity of 2.94 × 107 m/s in the horizontal direction. It enters a uniform electric field of 7500 N/C directed vertically. Ignoring gravitational effects, find the time it takes the proton to travel 0.053 m horizontally. The mass of the proton is 1.67262 × 10−27 kg . The charge of the proton is 1.60218 × 10−19 C . Correct answer: 1.80272 ns. Explanation: Let : vx = 2.94 × 107 m/s, E = 7500 N/C , and x = 0.053 m . The electric field E~ is in the vertical (y) direction, so the electric force F~ elec = q E~ exerted by the field on the proton is also in the y-direction, with no component in the xdirection. Hence, the field can exert no force on the proton in the x-direction. This impli sapaugh (js67869) – Quiz 7: Electric Force and Electric Field – balasubramanya – (1402112) 10 a constant speed in the x-direction. Consequently, x = vx t t = x vx = 0.053 m 2.94 × 107 m/s · 109 ns s = 1.80272 ns . 023 (part 2 of 3) 10.0 points What is the vertical displacement of the proton after the electric field acts on it for that time? Correct answer: 0.00116735 mm. Explanation: In the vertical direction, the proton experiences an electric force with magnitude Felec = q E = m ay ay = q E m = 1.60218 × 10−19 C
(7500 N/C) 1.67262 × 10−27 kg = 7.18412 × 1011 m/s 2 . The vertical dispacement is ∆y = v0 t + 1 2 a t2 = 1 2 a t2 since vo = 0, so ∆y = 1 2 7.18412 × 1011 m/s 2
× (1.80272 × 10−9 s)2 × 1000 mm 1 m = 0.00116735 mm . 024 (part 3 of 3) 10.0 points What is the proton’s speed after being in the electric field for that time? Correct answer: 29400 km/s. Explanation: In the x-direction, the proton has constant velocity. In the y-direction, the proton is accelerating, so vy = v0 + ay t = ay t = (7.18412 × 1011 m/s 2 ) × (1.80272 × 10−9 s) = 1295.1 m/s. Thus the proton’s speed v = q v 2 x + v 2 y = q (2.94 × 107 m/s)2 + (1295.1 m/s)2 × 1 km 1000 m = 29400 km/s . 025 10.0 points An electron begins at rest, and then is accelerated by a uniform electric field of 800 N/C that extends over a distance of 9 cm. Find the speed of the electron after it leaves the region of uniform electric field. The elementary charge is 1.6 × 10−19 C and the mass of the electron is 9.11 × 10−31 kg. Correct answer: 5.029 × 106 m/s. Explanation: Let : e = 1.6 × 10−19 C , me = 9.11 × 10−31 kg , E = 800 N/C , and ∆x = 9 cm = 0.09 m . Because of the constant acceleration, v 2 = v 2 0 + 2 a ∆x . Since v0 = 0 and a = Fnet me = e E me , v = r 2 e E ∆x me = s 2 (1.6 × 10−19 C) (800 N/C) 9.11 × 10−31 kg × √ 0.09 m = 5.029 × 106 m/s sapaugh (js67869) – Quiz 7: Electric Force and Electric Field – balasubramanya – (1402112) 11 026 10.0 points A particle of mass 4.5 g, and charge 7.2 mC moves in a region of space where the electric field is uniform and is given by Ex = −5.2 N/C, Ey = 0, and Ez = 0. If the velocity of the particle at t = 0 is vx0 = 87 m/s, vy0 = 0, and vz0 = 0, what is the speed |v| of the particle at 2.75 s? Correct answer: 64.12 m/s. Explanation: Let : m = 4.5 g = 0.0045 kg , q = 7.2 mC = 0.0072 C , Ex = −5.2 N/C , Ey = Ez = 0 , vx0 = 87 m/s, and vy0 = vz0 = 0 . The force on the particle is F = q E = m a a = q E m and the velocity is given by v = v0 + a t = v0 + q E m t = 87 m/s + (0.0072 C)(−5.2 N/C)(2.75 s) 0.0045 kg = 64.12 m/s, corresponding to a speed of 64.12 m/s . 027 10.0 points A 62 cm diameter loop is rotated in a uniform electric field until the position of maximum electric flux is found. The flux in this position is measured to be 7.78 × 105 N · m2 /C. What is the electric field strength? Correct answer: 2.57695 × 106 N/C. Explanation: Let : r = 31 cm = 0.31 m and Φ = 7.78 × 105 N · m2 /C . By Gauss’ law, Φ = I E~ ·dA . ~ The position of maximum electric flux will be that position in which the plane of the loop is perpendicular to the electric field; i.e., when E~ · dA~ = E dA. Since the field is constant, Φ = E A = E π r2 E = Φ π r2 = 7.78 × 105 N · m2/C π (0.31 m)2 = 2.57695 × 106 N/C . 028 (part 1 of 3) 10.0 points A plane is parallel to the field. A vertical What is the magnitude of the gravitational field flux through the planar surface of area A? Let g be the magnitude of the uniform gravitational field. Answer in terms of g and A. 1. None of these 2. √ 2 g A 3. 0 correct 4. 1 √ 2 g A 5. g A Explanation: The gravitational field flux is Φ = g A cos θ , sapaugh (js67869) – Quiz 7: Electric Force and Electric Field – balasubramanya – (1402112) 12 where θ is the angle between the normal of the surface and the direction of the gravitational field, so Φ = g A cos 90◦ = g A (0) = 0 . 029 (part 2 of 3) 10.0 points The plane is tilted at 45◦ to the field. A tilted at 45◦ What is the magnitude of the gravitational field flux through the planar surface of area A? 1. √ 2 g A 2. 0 3. 1 √ 2 g A correct 4. g A 5. None of these Explanation: Φ = g A cos 45◦ = g A 1 √ 2 . 030 (part 3 of 3) 10.0 points The plane is horizontal. A horizontal What is the magnitude of the gravitational field flux through the planar surface of area A? 1. 1 √ 2 g A 2. √ 2 g A 3. 0 4. g A correct 5. None of these Explanation: Φ = g A cos 0◦ = g A (1) = g A . 031 10.0 points A cubic box of side a, oriented as shown, contains an unknown charge. The vertically directed electric field has a uniform magnitude E at the top surface and 2 E at the bottom surface. a E 2 E How much charge Q is inside the box? 1. Qencl = 1 2 ǫ0 E a2 2. Qencl = 3 ǫ0 E a2 3. insufficient information 4. Qencl = 2 ǫ0 E a2 5. Qencl = 0 6. Qencl = ǫ0 E a2 correct Explanation: Electric flux through a surface S is, by convention, positive for electric field lines going out of the surface S and negative for lines going in. No flux passes through the vertical sides. sapaugh (js67869) – Quiz 7: Electric Force and Electric Field – balasubramanya – (1402112) 13 The top receives Φtop = −E a2 (inward is negative) and the bottom Φbottom = 2 E a2 , so the total electric flux is ΦE = −E a2 + 2 E a2 = E a2 . Using Gauss’s Law, the charge inside the box is Qencl = ǫ0 ΦE = ǫ0 E a2 . 032 10.0 points Pictured below is a distribution of 6 point charges and their surrounding electric field. -Q +Q -Q +Q -Q +Q Gaussian surface What is the total electric flux through the closed Gaussian surface shown? 1. −2 Q ǫ0 correct 2. 2 Q ǫ0 3. 0 4. Q ǫ0 5. −6 Q ǫ0 6. 6 Q ǫ0 7. −Q ǫ0 Explanation: The total charge within the Gaussian surface is −2 Q, so the total electric flux is φ = −2 Q ǫ0 . 033 10.0 points The nucleus of 208 82Pb (“lead-208”) has 82 protons within a sphere of radius 6.34 × 10−15 m. Each electric charge has a value of 1.60218 × 10−19 C. Calculate the electric field at the surface of the nucleus. The Coulomb constant is 8.98755 × 109 N · m2 /C 2 . Correct answer: 2.93756 × 1021 N/C. Explanation: Let : n = 82 , r = 6.34 × 10−15 m , q = 2.93756 × 1021 N/C , and ke = 8.98755 × 109 N · m2 /C 2 . Φ = Q ǫ0 = n q ǫ0 = 4 π r2 E E = n q r 2 1 4 π ǫ0 = ke n r 2 q = 8.98755 × 109 N · m2/C 2
(82) (6.34 × 10−15 m)2 × (1.60218 × 10−19 C) = 2.93756 × 1021 N/C . 034 10.0 points A uniformly charged conducting plate with area A has a total charge Q which is positive. The figure below shows a cross-sectional view of the plane and the electric field lines due to the charge on the plane. The figure is not drawn to scale. E E +Q + + + + + + + + + + + P Find the magnitude of the field at point P, which is a distance a from the plate. Assume that a is very small when compared to th sapaugh (js67869) – Quiz 7: Electric Force and Electric Field – balasubramanya – (1402112) 14 dimensions of the plate, such that edge effects can be ignored. 1. kE~ k = Q 4 π ǫ0 a 2 2. kE~ k = 4 π ǫ0 a Q 3. kE~ k = ǫ0 Q A 4. kE~ k = 2 ǫ0 Q A 5. kE~ k = Q 2 ǫ0 A correct 6. kE~ k = Q ǫ0 A 7. kE~ k = Q 4 ǫ0 A 8. kE~ k = 4 π ǫ0 a 2 Q 9. kE~ k = Q 4 π ǫ0 a 10. kE~ k = ǫ0 Q a2 Explanation: Basic Concepts Gauss’ Law, electrostatic properties of conductors. Solution: Let us consider the Gaussian surface shown in the figure. E +Q + + + + + + + + + + + E S Due to the symmetry of the problem, there is an electric flux only through the right and left surfaces and these two are equal. If the cross section of the surface is S, then Gauss’ Law states that ΦTOTAL = 2 E S = 1 ǫ0 Q A S ,so E = Q 2 ǫ0 A . 035 (part 1 of 2) 10.0 points +Q −Q The dotted line or surface in the figure above 1. is an equipotential line or surface. 2. is not an equipotential line or surface. correct 3. cannot be determined from the information given. Explanation: Consider the electric field: + − The dotted line is an electric field line. 036 (part 2 of 2) 10.0 points +Q −Q The dotted line or surface in the figure above 1. is not an equipotential line or surface. 2. is an equipotential line or surface. correct 3. cannot be determined from the information given. Explanation: sapaugh (js67869) – Quiz 7: Electric Force and Electric Field – balasubramanya – (1402112) 15 An equipotential line or surface is perpendicular to the electric field lines. sapaugh (js67869) – Quiz 8: Electric Potential and Circuits – balasubramanya – (1402112) 1 This print-out should have 27 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Two negatively charged spheres with different radii are shown in the figure below. −Q −Q The two conductors are now conneted by a wire. Which of the following occurs when the two spheres are connected with a conducting wire? 1. Negative charge flows from the smaller sphere to the larger sphere until the electric potential of each sphere is the same. correct 2. No charge flows. 3. Negative charge flows from the larger sphere to the smaller sphere until the electric field at the surface of each sphere is the same. 4. Negative charge flows from the smaller sphere to the larger sphere until the electric field at the surface of each sphere is the same. 5. Negative charge flows from the larger sphere to the smaller sphere until the electric potential of each sphere is the same. Explanation: When the wire is connected, charge will flow until each surface is at the same potential. When disconnected the potential of each sphere is given by V = ke q r . The smaller sphere is at a more negative potential than the larger sphere, so negative charge will flow from the smaller sphere to the large one until they are at the same potential. 002 (part 1 of 2) 10.0 points A positron is accelerated from rest between two points due to a fixed electrostatic potential difference, and acquires a speed of 20 % percent of the speed of light, which is 3 × 108 m/s. (Ignore relativistic corrections for this problem.) Denote m1 as the mass of a positron, e the charge of a positron, and v1 the final velocity of the positron. Find the magnitude of the potential difference |∆V | between the two points. 1. |∆V | = 1 2 m1 v 2 1 2. |∆V | = e m2 v 2 1 3. |∆V | = e m1 v 2 1 4. |∆V | = m2 v 2 1 e 5. |∆V | = 1 2 m1 v 2 1 e correct 6. |∆V | = 1 2 m2 v 2 1 7. |∆V | = 1 2 m2 v 2 1 e 8. |∆V | = 1 2 e m1 v 2 1 9. |∆V | = m1 v 2 1 e 10. |∆V | = 1 2 e m2 v 2 1 Explanation: According to energy conservation, the potential energy lost ∆U by the positron between two points is equal to the kinetic energy sapaugh (js67869) – Quiz 8: Electric Potential and Circuits – balasubramanya – (1402112) 2 ∆K gained by the positron, ∆U = −∆K Since the positron starts from rest, ∆K = 1 2 m1 v 2 1 , ∆U = e ∆V Therefore e ∆V = − 1 2 m1 v 2 1 , where ∆V is the potential difference between two points. Thus the potential difference is |∆V | = 1 2 m1 v 2 1 e 003 (part 2 of 2) 10.0 points A proton is also accelerated from rest between the same two points. Use 9.11 × 10−31 kg for the mass of the positron, with the mass of the proton given as 1.67 × 10−27 kg. What final speed will be reached by this proton? Correct answer: 1.40137 × 106 m/s. Explanation: Let : p = 0.2 , c = 3 × 108 m/s, m1 = 9.11 × 10−31 kg , and m2 = 1.67 × 10−27 kg . The final speed of the positron is given by v1 = 0.2 c = 0.2 (3 × 108 m/s) = 6 × 107 m/s. Since the proton has the same charge as the positron and moves through the same electrostatic potential difference, the change in its kinetic energy will equal that of the positron: 1 2 m2 v 2 2 = 1 2 m1 v 2 1 v2 = rm1 m2 v1 = s 9.11 × 10−31 kg 1.67 × 10−27 kg (6 × 107 m/s) = 1.40137 × 106 m/s . where v2 is the speed of the proton. 004 (part 1 of 2) 10.0 points +Q −Q The dotted line or surface in the figure above 1. is an equipotential line or surface. 2. cannot be determined from the information given. 3. is not an equipotential line or surface. correct Explanation: Consider the electric field: + − The dotted line is an electric field line. 005 (part 2 of 2) 10.0 points +Q −Q The dotted line or surface in the figure above sapaugh (js67869) – Quiz 8: Electric Potential and Circuits – balasubramanya – (1402112) 3 1. is an equipotential line or surface. correct 2. cannot be determined from the information given. 3. is not an equipotential line or surface. Explanation: An equipotential line or surface is perpendicular to the electric field lines. 006 10.0 points Two charges are located along the x-axis. One has a charge of 6.2 µC, and the second has a charge of −3.3 µC. If the electrical potential energy associated with the pair of charges is −0.045 J, what is the distance between the charges? The value of the Coulomb constant is 8.98756 × 109 N · m2 /C 2 , and the acceleration due to gravity is 9.81 m/s 2 . Correct answer: 4.08634 m. Explanation: Let : q1 = 6.2 µC = 6.2 × 10−6 C , q2 = −3.3 µC = −3.3 × 10−6 C , Ue = −0.045 J , and ke = 8.98756 × 109 N · m 2 /C 2 . Ue = ke q1 q2 r r = ke q1 q2 Ue = (8.98756 × 109 N · m 2 /C 2 ) × (6.2 × 10−6 C)(−3.3 × 10−6 C) −0.045 J = 4.08634 m . 007 (part 1 of 2) 10.0 points At some distance from a point charge, the electric potential is 639.0 V and the magnitude of the electric field is 155.0 N/C. The value of the Coulomb constant is 8.98755 × 109 N · m2 /C 2 and the acceleration of gravity is 9.81 m/s 2 . a) Determine the distance from the charge. Correct answer: 4.12258 m. Explanation: Let : ∆V = 639.0 V, E = 155.0 N/C , and ke = 8.98755 × 109 N · m2 /C 2 . ∆V E = ke q r ke q r 2 = r r = ∆V E = 639 V 155 N/C = 4.12258 m . 008 (part 2 of 2) 10.0 points b) Determine the charge. Correct answer: 2.93029 × 10−7 C. Explanation: q = ∆V · r ke = (639 V) (4.12258 m) 8.99 × 109 N · m2/C2 = 2.93029 × 10−7 C . 009 (part 1 of 2) 10.0 points An object with a charge 7 C and a mass 0.2 kg accelerates from rest to a speed of 11 m/s. Calculate the kinetic energy gained. Correct answer: 12.1 J. Explanation: sapaugh (js67869) – Quiz 8: Electric Potential and Circuits – balasubramanya – (1402112) 4 Let : m = 0.2 kg and v = 11 m/s. The kinetic energy is K = 1 2 m v2 = 1 2 (0.2 kg) (11 m/s)2 = 12.1 J . 010 (part 2 of 2) 10.0 points Through how large a potential difference did the object fall? Correct answer: 1.72857 V. Explanation: Let : q = 7 C . The potential difference is ∆V = K q = 12.1 J 7 C = 1.72857 V . 011 10.0 points Consider two points A and B in a constant electric field E~ as shown. B ℓ A 300 E What is the magnitude of the potential difference between A and B? 1. √ 2 E ℓ 2. 2 E ℓ 3. 0 4. E ℓ 2 5. E ℓ √ 2 6. 2 E ℓ √ 3 7. None of these 8. E ℓ 9. E ℓ 10. √ 3 E ℓ 2 correct Explanation: The potential difference between two points A and B is ∆V = Z B A E~ · d~s. In this case, we can choose a path which goes vertically up from A, then horizontally to B. Along the vertical path, E is perpendicular to the path, but along the horizontal part of the path, they are parallel, so that ∆V = E s = E ℓ cos 30◦ = r 3 E ℓ 2 . In a non-calculus way, ∆V = E~ · AB~ = E ℓ cos θ , for a uniform field and a straight line segment AB~ , where θ is the angle between the electric field E~ and the segment AB~ . 012 10.0 points When you touch a friend after walking across a rug on a dry day, you typically draw a spark of about 2 mm. The magnitude of the electric field for which dielectric breakdown occurs in air is about 3 MV/m. sapaugh (js67869) – Quiz 8: Electric Potential and Circuits – balasubramanya – (1402112) 5 Estimate the potential difference between you and your friend before the spark. Correct answer: 6000 V. Explanation: Let : r = 2 mm = 0.002 m and Eb = 3 MV/m = 3 × 106 V/m . The potential difference is V = Eb r = (3 × 106 V/m) (0.002 m) = 6000 V . 013 10.0 points Two parallel conducting plates separated by a distance d are connected to a battery of voltage E. Which of the following is correct if the plate separation is doubled while the battery remains connected? 1. The capacitance is unchanged. 2. The electric charge on the plates is doubled. 3. The potential difference between the plates is halved. 4. The electric charge on the plates is halved. correct 5. The potential difference between the plates is doubled. Explanation: The capacitance of the two parallel conducting plates is given by C = ǫ0 A d , so when the separation d is doubled, the capacitance is halved. The battery remains connected during the whole process, so the potential difference remains the same throughout. Q = C V,so the electric charge on the plates is also halved. 014 10.0 points An air-filled capacitor consists of two parallel plates, each with an area A, separated by a distance d. A potential difference V is applied to the two plates. The magnitude of the surface charge density on the inner surface of each plate is 1. σ = ǫ0 (V d) 2 2. σ = ǫ0  d V 2 3. σ = ǫ0 V d 4. σ = ǫ0  V d 2 5. σ = ǫ0(V d) 2 6. σ = ǫ0 V d correct 7. σ = ǫ0V d 8. σ = ǫ0d V Explanation: Use Gauss’s Law. We find that a pillbox of cross section S which sticks through the surface on one of the plates encloses charge σ S. The flux through the pillbox is only through the top, so the total flux is E S. Gauss’ Law gives σ = ǫ0 E = ǫ0 V d Alternatively, we could just recall this result for an infinite conducting plate (meaning we sapaugh (js67869) – Quiz 8: Electric Potential and Circuits – balasubramanya – (1402112) 6 neglect edge effects) and apply it. 015 (part 1 of 2) 10.0 points The muon (with mass 209 me) acts as a heavy electron. The muon can bind to a proton to form a muonic atom. Calculate the ionization energy of this atom. The value of ¯h is 1.05457 × 10−34 J · s ; the Rydberg constant for hydrogen is 1.09735 × 107 m−1 ; the Bohr radius is 5.29177 × 10−11 m ; and the ground state energy for hydrogen is 13.6057 eV . Correct answer: 2.8424 keV. Explanation: Let : mµ = 209 me , Z = 1 , E0H = 13.6057 eV , a0H = 5.29177 × 10−11 m , and ¯h = 1.05457 × 10−34 J · s. α = 1 137 = 0.00729719 . E0H = me 2 (α c) 2 E = mµ 2 (Z α c) 2 = 209 me 2 (α c) 2 = 209 E0H = 209 (13.6057 eV) · 1 keV 1000 eV = 2.8424 keV . 016 (part 2 of 2) 10.0 points Calculate the radius of the muonic atom in its ground state. Ignore reduced-mass effects. Correct answer: 2.53195 × 10−13 m. Explanation: r0 = ¯h m c α = a0H mµ me = 5.29177 × 10−11 m 209 = 2.53195 × 10−13 m . keywords: 017 10.0 points An electron in chromium makes a transition from the n = 4 state to the n = 1 state without emitting a photon. Instead, the excess energy is transferred to an outer electron in the n = 5 state, which is ejected by the atom. (This is called an Auger process, and the ejected electron is referred to as an Auger electron). Use the Bohr theory to find the kinetic energy of the Auger electron. Correct answer: 7030.66 eV. Explanation: Call the energy available from the n = 4 to n = 1 transition ∆E: ∆E = (13.6 eV)Z 2 1 n 2 f − 1 n 2 i ! = (13.6 eV)(24)2  1 − 1 4 2  = 7344 eV. The kinetic energy K of the Auger electron is equal to 7344 eV minus the energy required to ionize an electron in the n = 5 state, Eionization, Thus, K = 7344 eV − (13.6 eV)242 5 2 = 7030.66 eV. keywords: 018 10.0 points In a gas of hydrogen under normal conditions, the interatomic spacing is 1.61 × 10−8 m . Assume the gas is made of atomic, not molecular, hydrogen. For what n-value of the hydrogen atoms is the size of the atom comparable to the interatomic spacing? The value of ¯h is 1.05457 × 10−34 J · s, the Bohr radius is sapaugh (js67869) – Quiz 8: Electric Potential and Circuits – balasubramanya – (1402112) 7 5.29177 × 10−11 m , and the Rydberg constant for hydrogen is 1.09735 × 107 m−1 . Correct answer: 17. Explanation: Let : ¯h = 1.05457 × 10−34 J · s, a0 = 5.29177 × 10−11 m , and RH = 1.09735 × 107 m−1 . n 2 a0 = 1.61 × 10−8 m . n = s 1.61 × 10−8 m a0 = s 1.61 × 10−8 m 5.29177 × 10−11 m = 17.4426 ≈ 17 . keywords: 019 (part 1 of 4) 10.0 points A battery has an emf of 12 V and an internal resistance of 0.12 Ω. Its terminals are connected to a load resistance of 3 Ω. Find the current in the circuit. Correct answer: 3.84615 A. Explanation: Let : E = 12 V , R = 3 Ω , and r = 0.12 Ω . The total resistance is R + r, so I = E R + r = 12 V 3 Ω + 0.12 Ω = 3.84615 A . 020 (part 2 of 4) 10.0 points Calculate the terminal voltage of the battery. Correct answer: 11.5385 V. Explanation: The terminal voltage V of the battery is equal to Vr = E − I r = 12 V − (3.84615 A) (0.12 Ω) = 11.5385 V . 021 (part 3 of 4) 10.0 points Find the power dissipated in the load resistor. Correct answer: 44.3787 W. Explanation: The power dissipated in the load resistor is PR = I 2 R = (3.84615 A)2 (3 Ω) = 44.3787 W . 022 (part 4 of 4) 10.0 points Find the power dissipated in the battery. Correct answer: 1.77515 W. Explanation: The power dissipated in the battery is Pr = I 2 r = (3.84615 A)2 (0.12 Ω) = 1.77515 W . 023 10.0 points A 22.1 V emf is placed across a series combination of three resistors of size 10 Ω, 46.367 Ω and 44 Ω. At what rate is heat generated in the 46.367 Ω resistor? Correct answer: 2.24808 W. Explanation: Let : R1 = 10 Ω , R2 = 46.367 Ω , and R3 = 44 Ω . sapaugh (js67869) – Quiz 8: Electric Potential and Circuits – balasubramanya – (1402112) 8 The equivalent resistance is RT = R1 + R2 + R3 , so the current across each resistance is I = E RT = E 10 Ω + 46.367 Ω + 44 Ω = 0.220192 A and P = I 2 R2 = (0.220192 A)2 (46.367 Ω) = 2.24808 W . 024 (part 1 of 2) 10.0 points Two identical light bulbs A and B are connected in series to a constant voltage source. Suppose a wire is connected across bulb B as shown. E A B Bulb A 1. will burn half as brightly as before. 2. will go out. 3. will burn twice as brightly as before. 4. will burn nearly four times as brightly as before. correct 5. will burn as brightly as before. Explanation: The electric power is given by P = I 2 R . Before the wire is connected, IA = IB = V 2 R , so that PA =  V 2 R 2 · R = V 2 4 R . After the wire is connected, I ′ A = V R and I ′ B = 0 , so P ′ A =  V R 2 · R = V 2 R = 4 PA . 025 (part 2 of 2) 10.0 points and bulb B 1. will go out. correct 2. will burn as brightly as before. 3. will burn twice as brightly as before. 4. will burn half as brightly as before. 5. will burn nearly four times as brightly as before. Explanation: Since there is no potential difference between the two ends of bulb B, it goes out. 026 10.0 points A length of wire is cut into 5 equal pieces. The 5 pieces are then connected parallel, with the resulting resistance being 2 Ω. What was the resistance r of the original length of wire? Correct answer: 50 Ω. Explanation: Let : n = 5 and Rp = 2 Ω . The resulting resistance Rp of n equal piece resistors of resistance r connected parallel is Rp = r n , but when they are in series, the total resistance Rs is Rs = n r = n 2 Rp = 52 (2 Ω) = 50 Ω . sapaugh (js67869) – Quiz 8: Electric Potential and Circuits – balasubramanya – (1402112) 9 027 10.0 points Two identical parallel-wired strings of 33 bulbs are connected to each other in series. If the equivalent resistance of the combination is 163.0 Ω when it is connected across a potential difference of 114.0 V, what is the resistance of each individual bulb? Correct answer: 2689.5 Ω. Explanation: Let : Req = 163.0 Ω and N = 33 For each string, 1 Req, string = N R Req, string = R N When connected in series, Req = R N + R N = 2R N R = NReq 2 = 33(163 Ω) 2 = 2689.5 Ω . sapaugh (js67869) – Quiz 9: Magnetism – balasubramanya – (1402112) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A charged particle is projected with its initial velocity parallel to a uniform magnetic field. What is the resulting path? 1. straight line perpendicular to the field. 2. straight line parallel to the field. correct 3. circular arc. 4. parabolic arc. 5. spiral. Explanation: The force on a moving charge due to a magnetic field is given by F~ = q~v × B . ~ If ~v and B~ are parallel, then ~v × B~ = 0 . Hence the force on the particle is zero, and the particle continues to move in a straight line parallel to the field. 002 10.0 points The magnetic field at the equator points north. If you throw a positively charged object (for example, a baseball with some electrons removed) to the east, what is the direction of the magnetic force on the object? 1. Toward the west 2. Downward 3. Toward the east 4. Upward correct Explanation: Use the right-hand rule: point your index finger east and your middle finger north. Your thumb points upward (representing the force on a positively charged object). 003 10.0 points An electron in a vacuum is first accelerated by a voltage of 91800 V and then enters a region in which there is a uniform magnetic field of 0.611 T at right angles to the direction of the electron’s motion. The mass of the electron is 9.11 × 10−31 kg and its charge is 1.60218 × 10−19 C. What is the magnitude of the force on the electron due to the magnetic field? Correct answer: 1.75907 × 10−11 N. Explanation: Let : V = 91800 V , B = 0.611 T , m = 9.11 × 10−31 kg , qe = 1.60218 × 10−19 C . The kinetic energy K gained after acceleration is K = 1 2 m v2 = qe V , so the velocity is v = r 2 qe V m = s 2 (1.60218 × 10−19 C)(91800 V) 9.11 × 10−31 kg = 1.79694 × 108 m/s. Then the force on it is f = q v B = (1.60218 × 10−19 C) × (1.79694 × 108 m/s) (0.611 T) = 1.75907 × 10−11 N . 004 10.0 points What surrounds a stationary electric charge? What surrounds a moving electric charge? (Ignore the gravitational field.) sapaugh (js67869) – Quiz 9: Magnetism – balasubramanya – (1402112) 2 1. electric field; electric and magnetic fields correct 2. electric field; magnetic field 3. magnetic fields for both 4. electric fields for both 5. It cannot be predicted. 6. magnetic field; electric field Explanation: An electric field surrounds a stationary electric charge. An electric field and a magnetic field surround a moving electic charge. In reality, a gravitational field also surrounds both. 005 10.0 points Two long, parallel wires are separated by a distance 2 d, as shown below. Wire #1 carries a steady current I out of the plane of the page while wire #2 carries a steady current I out of the page. d d P I I S S ′ wire #1 wire #2 At what points in the plane of the page (besides points at infinity), is the magnetic field due to the currents zero? 1. At all points on a circle of radius d centered at either wire. 2. At only point P. correct 3. At all points on the line SS′ , a perpendicular bisector of a line connecting the two wires. 4. At no points. 5. At all points on the line connecting the two wires. Explanation: The only way that the total magnetic field would be zero is if the magnetic fields due to the two wires have the same magnitude but opposite directions at the same point. Only at points on the line SS′ do the magnetic fields have the same magnitude. Only at point P are the magnetic fields parallel (aligned with the vertical axis). Using the right hand rule, they are in opposite direction. Thus, at only point P (besides points at infinity) is the magnetic field due to the currents zero. 006 10.0 points A current is flowing clockwise around a loop placed on your desk. What would be the direction of the resulting magnetic field inside the loop? 1. no magnetic field was generated. 2. upward 3. downward correct 4. clockwise 5. counterclockwise Explanation: By the right-hand rule, the field points down. 007 10.0 points A wire carrying a current 30 A has a length 0.1 m between the pole faces of a magnet at an angle 60 ◦ (see the figure). The magnetic field is approximately uniform at 0.5 T. We ignore the field beyond the pole pieces. sapaugh (js67869) – Quiz 9: Magnetism – balasubramanya – (1402112) 3 θ ℓ I B What is the force on the wire? Correct answer: 1.29904 N. Explanation: Let : I = 30 A , ℓ = 0.1 m , θ = 60 ◦ , and B = 0.5 T . we use F = I ℓ B sin θ, so F = I ℓ B sin θ = (30 A) (0.1 m) (0.5 T) sin 60 ◦ = 1.29904 N . 008 10.0 points Two identical parallel sections of wire are connected parallel to a battery as shown. The two sections of wire are free to move. b b When the switch is closed, the wires 1. will heat up, and remain motionless. 2. will accelerate away each other. 3. will accelerate towards each other. correct Explanation: The currents in both rods move downward, so they are parallel currents that attract, causing them to accelerate toward one another. 009 (part 1 of 2) 10.0 points A long, straight wire carries a current of 19.7 A. An electron travels at 1.88 × 105 m/s parallel to the wire, 57.3 cm from the wire. The permeability of free space is 1.25664 × 10−6 N/A 2 and the charge on an electron is 1.6 × 10−19 C. What force does the magnetic field of the current exert on the moving electron? Correct answer: 2.06833 × 10−19 N. Explanation: Let : µ0 = 1.25664 × 10−6 N/A 2 , v = 1.88 × 105 m/s, q = 1.6 × 10−19 C , I = 19.7 A , and r = 57.3 cm = 0.573 m . Magnetic field produced by a straight, current-carrying wire is B = µ0 I 2 π r = (4 π × 10−7 ) (19.7 A) 2 π (0.573 m) = 6.87609 × 10−6 T . The electron moves in the direction which is perpendicular to the direction of the magnetic field caused by the long wire. So the force on the electron in a magnetic field is F = B v q = (6.87609 × 10−6 T) (1.88 × 105 m/s) × (1.6 × 10−19 C) = 2.06833 × 10−19 N . 010 (part 2 of 2) 10.0 points Which of the following statements is correct ? sapaugh (js67869) – Quiz 9: Magnetism – balasubramanya – (1402112) 4 1. The force on the electron is perpendicular to the electron’s motion and perpendicular to the plane in which both the electron position and the wire lie, with the direction determined by the right hand rule. 2. The force on the electron is perpendicular to the electron’s motion and perpendicular to the plane in which both the electron position and the wire lie, with not enough information given to determine the direction of the force. 3. The force on the electron is directed along the electron’s motion. 4. The force on the electron is directed opposite to the electron’s motion. 5. The force on the electron is directed perpendicular to the electron’s motion and directed away from the wire. 6. The force on the electron is directed perpendicular to the electron’s motion but with the limited information given we cannot determine if this force is directed towards or away from the wire. correct 7. The force on the electron is directed perpendicular to the electron’s motion and must be directed towards the wire. Explanation: As the electron moves, its current is opposite to the direction of its motion, and if this electron is in the same direction as the wire’s current, the force on the electron is directed perpendicular to the electron motion and towards the wire, and if the electron’s current is in the opposite direction as the wire’s current, the force is directed oppositely from the first situation. But the problem does not give the direction of the current and we do not know if the electron is repelled or attracted to the wire. 011 10.0 points Calculate the magnitude of the magnetic field at a point 104 cm from a long, thin conductor carrying a current of 1.24 A. The permeability of free space is 1.25664 × 10−6 T · m/A. Correct answer: 2.38462 × 10−7 T. Explanation: Let : µ0 = 1.25664 × 10−6 T · m/A , r = 104 cm = 1.04 m , and I = 1.24 A . The magnetic field of the wire is B = µ0 I 2 π r = (1.25664 × 10−6 T · m/A) (1.24 A) 2 π (1.04 m) = 2.38462 × 10−7 T . [Show More]

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