Enm hw - Assign Ten – Homework 9 – Due 2:00 pm – Inst Richard Saenz This print-out should have 34 questions Multiple-choice questions may

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Unformatted text preview: Assign, Ten – Homework 9 – Due: Dec 11 2003, 2:00 pm – Inst: Richard Saenz This print-out should have 34 questions. Multiple-choice questions may continue on the next column or page – find all choices before making your selection. The due time is Central time. these problem are extra credit homework, but the material will be in the final 001 (part 1 of 2) 10 points A rectangular loop located a distance from a long wire carrying a current is shown in the figure. The wire is parallel to the longest side of the loop. 1 From Amp`re’s law, we know that the e strength of the magnetic field created by the current-carrying wire at a distance r from the wire is (see figure.) B= µ0 I , 2πr so the field varies over the loop and is directed ￿ perpendicular to the page. Since B is paral￿ lel to dA, we can express the magnetic flux through an area element dA as ￿ Φ ≡ B dA ￿ µ0 I = dA . 2πr ￿ Note: B is not uniform but rather depends on r, so it cannot be removed from the integral. In order to integrate, we express the area element shaded in the figure as dA = b dr. Since r is the only variable that now appears in the integral, we obtain for the magnetic flux ￿ a+c µ0 I dr ΦB = b 2π r c ￿a+c µ0 I b ￿ = ln r￿ 2π ￿c ￿ µ0 I b a+c = ln 2π c ￿ ￿ µ0 (0.0915 A)(0.18 m) a+c = ln 2π c µ0 (0.0915 A)(0.18 m) = (0.50148) 2π = 1.65187 × 10−9 Wb . 002 (part 2 of 2) 10 points What is the direction of the magnetic field through the rectangular loop? 0.0915 A 6.45 cm 4.2 cm Find the total magnetic flux through the loop. Correct answer: 1.65187 × 10−9 Wb. Explanation: Let : c = 6.45 cm , a = 4.2 cm , b = 18 cm , and I = 0.0915 A . dr I r b 1. out of the plane of the paper 2. into the plane of the paper 3. cannot be determined with information given correct c a 18 cm Assign, Ten – Homework 9 – Due: Dec 11 2003, 2:00 pm – Inst: Richard Saenz Explanation: The direction of the current is not given, hence the absolute direction of the magnetic field cannot be determined, although the magnetic field is perpendicular to the plane of the paper. For example if the current flows upward (downward) the magnetic field would be into (out of ) the plane of the paper. 003 (part 1 of 1) 10 points A coil is wrapped with 198 turns of wire on the perimeter of a square frame of sides 34.4 cm. Each turn has the same area, equal to that of the frame, and the total resistance of the coil is 1.74 Ω. A uniform magnetic field is turned on perpendicular to the plane of the coil. If the field changes linearly from 0 to 0.908 Wb/m2 in a time of 1.13 s, find the magnitude of the induced emf in the coil while the field is changing. Correct answer: 18.8274 V. Explanation: Basic Concept: Faraday’s Law is E =− d ΦB . dt Explanation: Basic Concept: Faraday’s Law: E = −N d ΦB dt ￿ ￿ ￿ ￿ 2 From Faraday’s Law, we get ￿ ￿ d ΦB |E | = ￿ − N ￿ dt dB =NA dt ∆B =NA ∆t B =NA t So, the time needed equals t= N AB E (420 turns) (0.00230722 m2 ) (0.297 T) = (12.3 kV) −5 = 2.33986 × 10 s . Solution: The magnetic flux through the loop at t = 0 is zero since B = 0. At t = 1.13 s , the magnetic flux through the loop is ΦB = B A = 0.107449 Wb . Therefore the magnitude of the induced emf is N · ∆ΦB ∆t (198 turns) [(0.107449 Wb) − 0] = (1.13 s) = 18.8274 V |E | = 18.8274 V . E= 004 (part 1 of 1) 10 points A magnetic field of 0.297 T exists in the region enclosed by a solenoid that has 420 turns and a diameter of 5.42 cm. Within what period of time must the field be reduced to zero if the average magnitude of the induced emf within the coil during this time interval is to be 12.3 kV? Correct answer: 2.33986 × 10−5 s. 005 (part 1 of 1) 10 points A circular conducting loop is held fixed in a uniform magnetic field that varies in time according to B (t) = B0 exp(−a t) where t is in s, a is in s−1 and B is the field strength in T at t = 0. At t = 0, the emf induced in the loop is 0.0659 V . At t = 3.5 s, the emf is 0.0221 V, . Find a. Correct answer: 0.31216 s−1 . Explanation: Basic concept Faraday’s Law E =A· dB dt Solution: Since the emf is E =A dB , dt since only the magnetic field is changing and dB = −a B exp (−at) , dt Assign, Ten – Homework 9 – Due: Dec 11 2003, 2:00 pm – Inst: Richard Saenz we have 2 equations from the 2 different times. They are at t = 0 , −a A B = 0.0659 V , at t = 1.5 s , −a A B exp[−a (3.5 s)] = 0.0221 V . and The power dissipated is then P= V2 (0.00107819 V)2 (106 ) = R 0.00365 Ω = 318.49 µW 3 Dividing the second equation by the first and then taking the natural logarithm, we have ￿ ￿ (0.0221 V) − ln (0.0659 V) a= = 0.31216 s−1 . (3.5 s) 006 (part 1 of 1) 10 points A(n) 43.2 cm length of wire when used as a resistor has a resistance of 0.00365 Ω. The ends of the wire are connected to form a circular loop, and the plane of the loop is positioned at right angles to a uniform magnetic field that is increasing at the rate of 0.0726 T/s. At what rate is thermal energy generated in the wire? Correct answer: 318.49 µW. Explanation: The changing magnetic field generates a current in the wire. The induced potential is V =A dB . dt i3 007 (part 1 of 3) 10 points Assume: The induced emf for the closed loop octagonal C X DY C is E . A solenoid (with magnetic field B ) produces a steadily increasing uniform magnetic flux through its circular cross section. A octagonal circuit surrounds the solenoid as shown in the figure. The wires connecting in the circuit are ideal, having no resistance. The circuit consists of two identical light bulbs (labeled X and Y ) in series. A wire connects points C and D. The ratio of the solenoid’s area AL left of the wire C D and the solenoid’s AL = 4. area AR right of the wire C D is AR i2 D i1 B B Y AL B AR B X C The radius is found from the circumference, (C = ￿), to be: r= C 43.2 cm = . 2π 2π The equations for the (right) loop C X DC and the (left) loop C DY C are respectively given by 1. 2. E + i1 R = 0 5 4E − i1 R = 0 5 4E − i1 R = 0 5 E − i1 R = 0 5 4E + i1 R = 0 5 E − i1 R = 0 5 4E 5 E and 5 E and 5 4E and 5 E and 5 4E and 5 and − i2 R = 0 . + i2 R = 0 . − i2 R = 0 . + i2 R = 0 . + i2 R = 0 . − i2 R = 0 . Hence the area is given by A = πr = π 2 ￿ C 2π ￿2 1 = π ￿ ￿2 C 2 3. 4. 5. 6. Then the induced potential is ￿ ￿2 1C dB V= π2 dt = 0.00107819 V Assign, Ten – Homework 9 – Due: Dec 11 2003, 2:00 pm – Inst: Richard Saenz E 7. + i1 R = 0 and 5 rect 4E + i2 R = 0 . cor5 2. i3 = − 3. i3 4. i3 5. i3 6. i3 7. i3 8. i3 3E correct 5R 4E =− 5R 3E =+ 4R E =+ 4R E =− 4R 3E =+ 5R 4E =+ 5R 4 4E E 8. + i1 R = 0 and − i2 R = 0 . 5 5 Explanation: By definition, the areas of the left and right loops are related by A = A L + AR . Since AL = 4, we can solve for AL and AR in AR terms of A. 4A AL = 5 A AR = . 5 9. i3 = 0 Explanation: From the loop Eqs. 1 and 2 in Part 1, we can solve for the currents i1 and i2 , 1 5 4 i2 = − 5 i1 = − i2 = i1 + i3 we have i3 = − 4 5 3 =− 5 E 1E + R 5R E . R E R E . R i3 = i2 − i1 , Then we can compute the magnitude of the induced emf around the right and left loops. ER = AR dB A dB 1 = =E dt 5 dt 5 dB 4A dB 4 EL = AL = = E. dt 5 dt 5 The induced emf and the changing magnetic flux are related by E =− dΦ dB = −A . dt dt Since charge is conserved at a junction ⇒ Since the magnetic flux is increasing, the induced emf is in the clockwise direction and the direction of the current is counter-clockwise, as shown in the figure. From Kirchoff ’s laws, the loop equations for the right and left loops respectively are right loop : left loop : 1 E + i1 R = 0 5 4 E + i2 R = 0 5 (1) (2) 008 (part 2 of 3) 10 points Note: i3 is defined as positive if it flows in the same direction as shown in the figure. What is the current i3 ? 1. i3 = − 3E 4R 009 (part 3 of 3) 10 points The ratio of the brightness of bulb Y to that brightnessY of bulb X , , is brightnessX brightnessY 1. =3 brightnessX brightnessY 2. =9 brightnessX brightnessY =4 3. brightnessX brightnessY 4. = 16 correct brightnessX Assign, Ten – Homework 9 – Due: Dec 11 2003, 2:00 pm – Inst: Richard Saenz brightnessY 5. =2 brightnessX Explanation: The brightness of a bulb is proportional to the power dissipated by it. If the resistance of the bulb is R, then PY PX 2 EC DY C R = 2 EC X DC R dB AL dt = dB AR dt ￿ ￿2 AL = AR = (4)2 = 16 . 5 Basic Concept: Motional emf E = B ￿ v. Magnetic force on current Ohm’s Law ￿ ￿ F = I ￿ × B. ￿ I= V . R 2 Solution: The motional emf induced in the circuit is E = B ￿v = (2 T) (7 m) (9 m/s) = 126 V . From Ohm’s law, the current flowing through the resistor is I= E R 126 V = 9Ω = 14 A . 010 (part 1 of 2) 10 points Given: Assume the bar and rails have negligible resistance and friction. In the arrangement shown in the figure, the resistor is 9 Ω and a 2 T magnetic field is directed into the paper. The separation between the rails is 7 m . Neglect the mass of the bar. An applied force moves the bar to the right at a constant speed of 9 m/s . 2T m￿1 g 7m I 9Ω 9 m/s Thus, the magnitude of the force exerted on the bar due to the magnetic field is FB = I ￿ B = (14 A)(7 m)(2 T) = 196 N . To maintain the motion of the bar, a force must be applied on the bar to balance the magnetic force F = FB = 196 N 011 (part 2 of 2) 10 points At what rate is energy dissipated in the resistor? Correct answer: 1764 W. Explanation: The power dissipated in the resistor is P = I2 R = (14 A)2 (9 Ω) = 1764 W . 2T Calculate the applied force required to move the bar to the right at a constant speed of 9 m/s. Correct answer: 196 N. Explanation: Assign, Ten – Homework 9 – Due: Dec 11 2003, 2:00 pm – Inst: Richard Saenz Note: Second of four versions. 012 (part 1 of 4) 10 points A bar of negligible resistance and mass m = 38 kg in the figure below is pulled horizontally across frictionless parallel rails, also of negligible resistance, by a massless string that passes over an ideal pulley and is attached to a suspended mass M = 210 g. The uniform magnetic field has a magnitude B = 640 mT, and the distance between the rails is ￿ = 91 cm. The rails are connected at one end by a load resistor R = 71 mΩ. Use g = 9.8 m/s2 . m 6 Mg (2) ￿B To find the induced current, we use Ohm’s law dΦ and substitute in the induced emf, E = − dt I= I= |E | 1 dΦ = R R dt (3) Note: We have ignored the minus sign from the induced emf E because we will eventually evaluate the magnitude of the terminal velocity. The flux is Φ = B A. So dΦ dA =B = B ￿v dt dt I= (4) R M B What is the magnitude of the terminal velocity (i.e., the eventual steady-state speed v∞ ) reached by the bar? Correct answer: 0.430785 m/s. Explanation: Basic Concepts: ￿ Fg = M ￿ g ￿ Fm = I ￿ × B ￿￿ ￿ = (M + m) ￿ = Fg − Fm a￿ E =IR=− d ΦB dt ￿￿ ΦB = B · A E = B ￿v B ￿v . (5) R Using (2) and (5) and noting that v is the terminal velocity v∞ Mg B ￿ v∞ = . ￿B R (6) Solving for the magnitude of the terminal velocity v∞ v∞ = M gR (7) ￿2 B 2 (0.21 kg)(9.8 m/s2 )(0.071 Ω) = (0.91 m)2 (640 mT)2 = 0.430785 m/s . ￿ Fnet Solution: It follows from Lenz’s law that the magnetic force opposes the motion of the bar. When the wire acquires steady-state speed, the gravitational force Fg is counter-balanced by the magnetic force Fm (see figure below) a T m Fm 013 (part 2 of 4) 10 points What is the acceleration when the velocity v = 0.75 s? Correct answer: −0.0399108 m/s2 . Explanation: To get the velocity as a function of time we need the acceleration a(v ) ≡ dv . dt (8) R T a M Fg B Apply Newton’s second law to the bar and the suspended mass separately (1) m a = T − Fm M a = Fg − T , and Fg = M g = Fm = ￿ I B Assign, Ten – Homework 9 – Due: Dec 11 2003, 2:00 pm – Inst: Richard Saenz where T is the tension in the string. Combine and solve for a a= Fg − Fm , m+M (9) 7 where Fg = M g and Fm = I ￿ B . Further, E the induced current I = and E = B ￿ v , so R I= and Fm = Since, from (1) Fg = M g . (11) B ￿v R B 2 ￿2 v . R (10) 015 (part 4 of 4) 10 points What is the horizontal speed of the bar at time t = 4.15907 s, assuming that the bar was at rest at t = 0 s? Correct answer: 0.174675 m/s. Explanation: Integrating (14) we have ￿ ln ￿ v 0 dv = v∞ − v ￿ − ln ￿ ￿ t 0 dt τ = t τ t −0 τ (17) 1 v∞ − v ln ￿ Thus, using Eqs. (9), (10), and (11), Eq. (8) reduces to the differential equation a= or Mg B 2 ￿2 − v, (m + M ) R (m + M ) M gR −v 22 dv a= = B￿ . R (m + M ) dt B 2 ￿2 (12) ￿￿ t − v∞ − v τ =e v∞ ￿ ￿ t − τ 1 − e v∞ − v v∞ ￿ 1 v∞ ￿ =− (13) v = v∞ (18) ￿ ￿ 4.15907 s 7.99821 s 014 (part 3 of 4) 10 points What is the time constant τ ? Correct answer: 7.99821 s. Explanation: Rewriting the differential equation, Eq. (12), in dimensionless form and isolating t on one side and v on the other, we get dv dt = M gR R (m + M ) −v B 2 ￿2 B 2 ￿2 or (14) = (0.430785 m/s) 1 − e = 0.174675 m/s . − dv dt = (15) v∞ − v τ where v∞ is defined in Part 1, (7), and the time constant τ is τ= R (m + M ) B 2 ￿2 (0.071 Ω)(38 kg + 0.21 kg) = (640 mT)2 (0.91 m)2 = 7.99821 s . (16) 016 (part 1 of 1) 10 points A light bulb is connected to a battery, turned on, and is visibly lit. An iron core is first rapidly thrust into the coil, then rapidly withdrawn. iron core light bulb S These two actions will temporarily Assign, Ten – Homework 9 – Due: Dec 11 2003, 2:00 pm – Inst: Richard Saenz 1. brighten the bulb both ways. 2. have no effect on the bulb’s brightness. 3. dim the bulb both ways. 4. brighten one way, dim the other. correct Explanation: While the core moves, current will be induced. Moving in one direction will increase the DC current, while moving in the other direction will decrease the DC current. 017 (part 1 of 8) 10 points A rectangular loop with resistance R has N turns, each of length L and width W as shown in the figure. The loop moves into a uniform magnetic field B (into the page) with speed v . B Ba L v B 0 B a B b x0 B v B 0 What is d Φtotal b B x0 d c R c W 3. 8 d Φtotal =NBW dt d Φtotal = N B Lv 4. dt d Φtotal 5. = B Lv dt d Φtotal 6. = N B W Lv dt d Φtotal 7. = B W Lv dt d Φtotal =NBL 8. dt d Φtotal 9. = N B W v correct dt d Φtotal 10. =NBW L dt Explanation: Basic Concepts: Magnetic flux is defined as: ￿￿ Φ=B·A ￿￿ Φtotal = N Φ = N [B · A] Given: R=the resistance of the loop, N =the number of turns, L=the length of each loop, W =the width of the loop. Find: d Φtotal (1) , dt (2) The current in the loop, (3) The force on the loop as it enters the field, (4) The force on the loop as it moves within the field. Solution: The magnetic flux is given by Φtotal = N (B · A) , where N and B are constant, but the area is changing. A = W (v t) initially, so Φtotal = N B W v t d Φtotal dt = N BW v. d (the time derivative of the dt flux for all turns of the loop) just after the front edge (side ab) of the loop enters the field? 1. d Φtotal = zero dt d Φtotal 2. =BW v dt Assign, Ten – Homework 9 – Due: Dec 11 2003, 2:00 pm – Inst: Richard Saenz 018 (part 2 of 8) 10 points d Φtotal dΦ Remember: E = − = −N . dt dt What is the magnitude of the current I in the loop just after the front edge (side ab) of the loop enters the magnetic field? 1. I = E R 2. I = R E E2 3. I = R E2 4. I = 3 R E 5. I = 2 R 6. I = zero 7. I = E R2 R E2 E2 9. I = 2 R E 10. I = correct R Explanation: 8. I = E =− d Φtotal = −N B W v dt 9 What is the direction of the current I in the loop just after it enters the magnetic field? 1. counter-clockwise correct 2. clockwise 3. No current Explanation: The direction is counter-clockwise. 020 (part 4 of 8) 10 points Given: I is the current as found in Part 2. What is the magnitude of the force F on the loop just after the front edge (side ab) of the loop enters the field? N 2 B2 W 2 L v ￿ 1. ￿F ￿ = R2 N 2 B2 W 2 R ￿ 2. ￿F ￿ = v 2 B2 W v N ￿ 3. ￿F ￿ = R ￿ 4. ￿F ￿ = zero N 2 B2 W 2 v ￿ 5. ￿F ￿ = correct R N 2 B2 R ￿ 6. ￿F ￿ = W2 v N 2 B2 W L v ￿ 7. ￿F ￿ = R N2 W2 R v ￿ 8. ￿F ￿ = B2 N 2 B2 W ￿ 9. ￿F ￿ = Rv 2 B2 W 2 L v N ￿ 10. ￿F ￿ = R Explanation: The force on the loop is given by F = N 2 B2 W 2 v NIBW = as the forces act on R the current within the field, and the horizontal currents have equal and opposite forces. Thus only the right hand vertical loops have a force acting on them. This force acts to oppose the movement of the coils, and must point right. E NBW v =− R R N Bwv so the current I = is in the counterR clockwise direction (from b up to a) in order that flux is created in the loop which opposes the increase of flux in the loop of the uniform magnetic field (into the page) as the loop moves into this magnetic field. Note: The minus sign merely indicates that the direction of the current will be set up in such a way so as to resist the increasing magnetic flux. I= 019 (part 3 of 8) 10 points Assign, Ten – Homework 9 – Due: Dec 11 2003, 2:00 pm – Inst: Richard Saenz 021 (part 5 of 8) 10 points What is the direction of the force F on the loop just after the front edge (side ab) of the loop enters the field? 1. indeterminate, since the force is zero 2. towards the bottom of the page 3. left 4. right correct 5. towards the top of the page Explanation: Since ￿ F = W [￿ × B ] , I￿ Use right hand rule of cross product, we see that the direction of the force is pointing towards right. 022 (part 6 of 8) 10 points aB v d B x0 0 What is the magnitude of the force F on the loop as it moves within the field? b B ￿ 1. ￿F ￿ = ￿ 2. ￿F ￿ = ￿ 3. ￿F ￿ = ￿ 4. ￿F ￿ = ￿ 5. ￿F ￿ = N 2 B2 W 2 R v N 2 B2 W v R 2 B2 W N Rv 2 B2 W L v N R 2 B2 W 2 L v N R Bc ˆˆˆ F =I ×B. ￿ 6. ￿F ￿ = 0 correct 10 N 2 B2 W 2 L v ￿ 7. ￿F ￿ = R2 N2 W2 R v ￿ 8. ￿F ￿ = B2 2 B2 R N ￿ 9. ￿F ￿ = W2 v N 2 B2 W 2 v ￿ 10. ￿F ￿ = R Explanation: Within the field, the magnetic flux is cond Φtotal stant, so − = 0. Thus, E = 0, I = 0, dt and no force opposes the motion. 023 (part 7 of 8) 10 points What is the direction of the force on the loop as it moves within the field? 1. towards the top of the page 2. right 3. towards the bottom of the page 4. left 5. indeterminate, since the force is zero correct Explanation: 024 (part 8 of 8) 10 points a v b B d B B c B x0 0 What is the direction of the force on the loop as it moves out of the field (edge cd is in the field while edge ab is out of the field)? 1. left Assign, Ten – Homework 9 – Due: Dec 11 2003, 2:00 pm – Inst: Richard Saenz 2. right correct 3. indeterminate, since the force is zero 4. towards the top of the page 5. towards the bottom of the page Explanation: The force on the loop is given by F = N 2 B2 W 2 v NIBW = as the forces act on R the current within the field, and the horizontal currents have equal and opposite forces. Thus only the right hand vertical loops have a force acting on them. This force acts to oppose the movement of the coils, and must point right. 025 (part 1 of 5) 10 points A rectangular loop with resistance 13.1 Ω has 62 turns. The loop’s length is 9.3 m and width is 2 m (as shown in the figure). The loop moves with a speed of 4.1 m/s into a region with a uniform magnetic field of 5.2 T (into the page). The field exists in the region 0 < x < 10.6 m . B Ba 9.3 m 4.1 m/s 5.2 T b 13.1 Ω c 2m B Ba ￿ v B 0 B b R = 13.1 Ω Explanation: Given: R = 13.1 Ω , the resistance of the loop, N = 62 , the number of turns, ￿ = 9.3 m , the length of the loop, and w = 2 m , the width of the loop. 11 c w d 10.6 m Basic Concepts: Magnetic flux is defined as: ￿￿ Φn = N (B · A) Solution: We will find: (1) d ΦN , dt (2) The current in the loop, (3) The force on the loop as it enters the field, (4) The force on the loop as it moves into the field. The magnetic flux is given by ΦN = N (B · A) where N and B are constant, but the area is changing. A = w (v t) initially. ΦN = N B w v t d ΦN = N Bwv dt d ΦN E =− = −N B w v dt E I= R N Bwv =− R (62) (5.2 T) (2 m) (4.1 m/s) =− (13.1 Ω) = −201.808 A , N Bwv is in the counterR clockwise direction (from b up to a) in order that flux is created in the loop which opposes the increase of flux in the loop of the uniform magnetic field of 5.2 T (into the page) as the loop moves into this magnetic field. so the current I = B 0 B d 10.6 m B a B v B b B d c 0 10.6 m When the edge ab of the loop just enters the field and is between 0 and 10.6 m, what is the magnitude of the induced current in the loop? Correct answer: 201.808 A. Assign, Ten – Homework 9 – Due: Dec 11 2003, 2:00 pm – Inst: Richard Saenz Note: The minus sign merely indicates that the direction of the current will be set up in such a way so as to resist the increasing magnetic flux. 026 (part 2 of 5) 10 points When the edge ab of the loop just enters the field and is between 0 and 10.6 m, what force is required to keep the loop moving with constant speed? Correct answer: 130126 N. Explanation: The force on the loop is given by F = N 2 B 2 w2 v NIBw = as the forces act on R the current within the field, and the horizontal currents have equal and opposite forces. Thus only the right hand vertical loops have a force acting on them. This force acts to oppose the movement of the coils, and must point left. 12 aB v b B Bc d B 0 10.6 m Just after cd passes 10.6 m, while the coil is within the region of the field, in what direction does the current flow between a and b? 1. There is no current correct 2. b up to a 3. a down to b Explanation: There is no change in flux, consequently the current is zero. 029 (part 5 of 5) 10 points a v b B d B B c B N 2 B 2 w2 v F= R 2 (5.2 T)2 (2 m)2 (4.1 m/s) (62) = (13.1 Ω) = 130126 N . 027 (part 3 of 5) 10 points When the edge ab of the loop just enters the field and is between 0 and 10.6 m, what is the direction of the induced current flow between a and b? 1. b up to a 2. a down to b correct 3. There is no current Explanation: See Part 1, the current is counter-clockwise. 028 (part 4 of 5) 10 points 0 10.6 m Just after edge ab exits the field, in what direction does the current flow between a and b? 1. a down to b 2. There is no current 3. b up to a correct Explanation: See Part 1, the current is now clockwise; i.e., the change in flux is opposite to that in Part 1. 030 (part 1 of 2) 10 points Given: g = 9.8 m/s2 . Assign, Ten – Homework 9 – Due: Dec 11 2003, 2:00 pm – Inst: Richard Saenz Assume: The rod remains in contact with the rails as it slides down the rails. The rod experiences no friction or air drag. The rails at each side and on the bottom have negligible resistance. A straight, horizontal rod slides along parallel conducting rails at an angle with the horizontal, as shown below. The rails are connected at the bottom by a horizontal rail so that the rod and rails forms a closed rectangular loop. A uniform vertical field exists throughout the region. 1m 58 g 6.1 Ω B ￿ m R v 13 sliding rod B 0Ω Viewed from above sliding rod 4.5 m/s v 0Ω Viewed from above Basic Concepts: E =− d ΦB dt θ Viewed from the side 0.42 T 0.42 T s m/ 4.5 20◦ Viewed from the side The movement of the rod decreases the area of the loop, so the flux through the loop is changing in time, and there is an induced emf E . If we denote the area by A, this induced emf is E =− dΦ d (B A cos θ) =− dt dt If the velocity of the rod is 4.5 m/s, what is the current through the resistor? Correct answer: 291.151 mA. Explanation: Let : ￿ = 1 m , m = 58 g , R = 6.1 Ω , v = 4.5 m/s , B = 0.42 T . dA . dt since the flux is B · A = B A cos θ, where θ is the angle between the magnetic field B and the normal vector to the area. The magnetic field and the angle are both constant and were pulled out of the differentiation. Now, if we call the distance from the rod to the resistor x, the emf becomes = −B cos θ E = −B cos θ d (￿ x) dt dx = −B ￿ cos θ dt = −B ￿ v cos θ . and Assign, Ten – Homework 9 – Due: Dec 11 2003, 2:00 pm – Inst: Richard Saenz Thus the current in the resistor is I= |E | B ￿ v cos θ = = R R (0.42 T) (1 m) (4.5 m/s) cos(20◦ ) = (6.1 Ω) = 0.291151 A = 291.151 mA . 14 terminal velocity, the power being lost in the resistor must equal the power being gained due to gravity. Thus d Ugrav dt E2 dz = mg R dt (B ￿ v0 cos θ)2 = m g v0 sin θ . R Solving this for v0 gives the same result as above. PR = 032 (part 1 of 1) 10 points In the figure shown, the north pole of the magnet is first moved down toward the loop of wire, then is withdrawn upward. N Counterclockwise 031 (part 2 of 2) 10 points What is the terminal velocity of the rod? Correct answer: 7.61317 m/s. Explanation: The terminal velocity is reached when the forces on the rod cancel, so it feels no more acceleration. The force from the induced current is, since the rod is perpendicular to the magnetic field, FB ,total = I ￿ B . However, this force is directed parallel to the ground. We need the component of this force parallel to the tracks, which is FB = I ￿ B cos θ . The component of the force of gravity parallel to the tracks is Fg = m g sin θ . At the terminal velocity, these forces are in equilibrium, FB = Fg , which yields m g sin θ = B ￿ v0 cos θ ￿ B cos θ R Clockwise As viewed from above, the induced current in the loop is 1. for both cases clockwise with increasing magnitude. 2. for both cases counterclockwise with decreasing magnitude. 3. for both cases counterclockwise with increasing magnitude. 4. for both cases clockwise with decreasing magnitude. 5. first clockwise, then counterclockwise. 6. first counterclockwise, then clockwise. correct Explanation: From Ohm’s law and Faraday’s law, the V 1 dΦ current in magnitude is I = =− , R R dt where Φ is the magnetic flux through the loop. We know the sign of the rate of change of the magnetic flux is changed when the magnet is where the expression for I from part 1 was used. We proceed to solve for v0 v0 = R m g sin θ [￿ B cos θ]2 (6.1 Ω) (0.058 kg) (9.8 m/s2 ) sin(20◦ ) = [(1 m) (0.42 T) cos(20◦ )]2 = 7.61317 m/s . Rather than worrying about force components, it might be easier to use a scalar quantity, such as power. When the rod is at its Assign, Ten – Homework 9 – Due: Dec 11 2003, 2:00 pm – Inst: Richard Saenz withdrawn upward, which, according to the equation the direction of the current is also changed. From Lenz’s law, we know when the magnet is moved down toward the loop, the current in the loop is counterclockwise as viewed from above. 033 (part 1 of 2) 10 points The counter-clockwise circulating current in a solenoid is increasing at a rate of 8.39 A/s. The cross-sectional area of the solenoid is 3.14159 cm2 , and there are 163 turns on its 18.4 cm length. What is the magnitude of the induced E produced by the increasing current? Correct answer: 0.478277 mV. Explanation: Basic Concepts: Faraday’s Law for solenoid E = −N dΦ dt dB = −N A . dt (1.25664 × 10−6 N/A2 ) (163)2 18.4 cm × (3.14159 cm2 )(8.39 A/s) ￿ ￿￿ ￿ 1m 3 mV × 10 V 10−2 cm = 0.478277 mV . = 034 (part 2 of 2) 10 points Choose the correct statement 15 1. The E tries to keep the current in the solenoid flowing in the counter-clockwise direction 2. The E does not effect the current in the solenoid 3. Not enough information is given to determine the effect of the E 4. By the right hand rule, the E produces magnetic fields in a direction perpendicular to the prevailing magnetic field 5. The E attempts to move the current in the solenoid in the clockwise direction correct Explanation: As the current is increasing in the counterclockwise direction, by Lenz’s law, the E will attempt to retard the current, which establishes an E that tries to counter the flow of the current, which in this case would be in the clockwise direction. Magnetic field produced by the changing current is B= µ0 N I L dB µ0 N d I = . dt L dt Faraday’s Law for solenoid E = −N dΦ dt d (B A) = −N dt 2A −N dI = µ0 . L dt Magnetic field induced by current B= µ0 N I . L Solution: Thus, the induced E is µ0 N 2 d I A |E | = L dt ...
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This note was uploaded on 01/20/2010 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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