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09 homework KIM, JI Due: Mar 23 2008, 4:00 am Question 1, chap 31, sect 2. part 1 of 1 10 points A square loop of wire of resistance R and side a is oriented with its plane perpendicular to a magnetic field B, as shown in the figure. B I a B What must be the rate of change of the magnetic field in order to produce a current I in the loop? I a2 dB = dt R dB Ra 2. = dt I dB 3. = I Ra dt dB IR 4. = 2 correct dt a Ia dB = 5. dt R Explanation: The emf produced is given by 1. E= d dB = a2 . dt dt 1 I0 = 31.6 A and a frequency f = 42.9 Hz. A loop that consists of N = 29 turns of wire links the toroid, as in the figure. N a b R Nl Determine the maximum E induced in the loop by the changing current I. Correct answer: 0.275601 V (tolerance 1 %). Explanation: Basic Concept: Faraday's Law E =- d B . dt Magnetic field in a toroid B= 0 N I . 2r Solution: In a toroid, all the flux is confined to the inside of the toroid B= 0 N I . 2r So, the flux through the loop of wire is B1 = = B dA Using Ohm's law we can solve for B dB IR =a dt dB IR = 2 . dt a 2 b+R 0 N I0 a dr sin( t) 2 r R b+R 0 N I0 a sin( t) ln = 2 R . Question 2, chap 31, sect 2. part 1 of 1 10 points A toroid having a rectangular cross section (a = 2.62 cm by b = 4.11 cm) and inner radius 2.34 cm consists of N = 210 turns of wire that carries a current I = I0 sin t, with Applying Faraday's law, the induced emf can be calculated as follows E = -N d B1 dt 0 N I0 b+R = -N a ln 2 R = -E0 cos( t) cos( t) homework 09 KIM, JI Due: Mar 23 2008, 4:00 am where = 2f was used. The maximum magnitude of the induced emf, E0 , is the coefficient in front of cos( t). E0 = -N d B1 dt Ohm's Law is I= V . R 2 The motional emf induced in the circuit is E = Bv = (2 T) (4 m) (4 m/s) = 32 V . From Ohm's law, the current flowing through the resistor is I= E 32 V = = 16 A . R 2 b+R a ln 2 R = -(29 turns) 0 (210 turns) (31.6 A) (42.9 Hz) (2.62 cm) (4.11 cm) + (2.34 cm) ln (2.34 cm) = -0.275601 V |E| = 0.275601 V . = -N 0 N I0 Question 3, chap 31, sect 1. part 1 of 2 10 points In the arrangement shown in the figure, the resistor is 2 and a 2 T magnetic field is directed out of the paper. The separation between the rails is 4 m . Neglect the mass of the bar. An applied force moves the bar to the left at a constant speed of 4 m/s . Assume the bar and rails have negligible resistance and friction. 2T m1 g 4m I 2 4 m/s Thus, the magnitude of the force exerted on the bar due to the magnetic field is FB = I B = (16 A)(4 m)(2 T) = 128 N . To maintain the motion of the bar, a force must be applied on the bar to balance the magnetic force F = FB = 128 N . Question 4, chap 31, sect 1. part 2 of 2 10 points At what rate is energy dissipated in the resistor? Correct answer: 512 W (tolerance 1 %). Explanation: The power dissipated in the resistor is P = I2 R = (16 A)2 (2 ) = 512 W . Note: Third of four versions. Question 5, chap 31, sect 1. part 1 of 1 10 points 2T Calculate the applied force required to move the bar to the left at a constant speed of 4 m/s. Correct answer: 128 N (tolerance 1 %). Explanation: Motional emf is E = Bv. Magnetic force on current is F =I B. homework 09 KIM, JI Due: Mar 23 2008, 4:00 am A copper wire of constant length is moving in a constant magnetic field, as shown. rect 5 4 3 2 1 0 0 1 2 3 4 5 6 7 8 910 Velocity (m/s) Figure: The wire and the velocity vector v are perpendicular (in the horizontal plane) to each other and are both perpendicular to the magnetic field B (vertical). Which of the following graphs best represents the magnitude of the emf E between the ends of the wire as a function of the speed v of the wire? 5 4 3 2 1 0 0 1 2 3 4 5 6 7 8 910 Velocity (m/s) 5 4 3 2 1 0 0 1 2 3 4 5 6 7 8 910 Velocity (m/s) 5 4 3 2 1 0 0 1 2 3 4 5 6 7 8 910 Velocity (m/s) 5 4 3 2 1 0 0 1 2 3 4 5 6 7 8 910 Velocity (m/s) E (V) E (V) E (V) 5 4 3 2 1 0 0 1 2 3 4 5 6 7 8 910 Velocity (m/s) 5 4 3 2 1 0 0 1 2 3 4 5 6 7 8 910 Velocity (m/s) E (V) E (V) 3 r v 5. B B 6. 7. 1. E (V) 2. Explanation: As the conductor moves with velocity v, the charge carriers experience a magnetic force Fmag . This force leads to a separation of charge carriers of opposite sign. This in turn creates an electric field EH that leads to an electric force FE opposing the magnetic force. In equilibrium these two forces balance out. Hence FE = Fmag q EH = q v B ; with EH = E one gets E = EH = v B , E v, or 3. 4. cor- since B and are constants. Alternative Solution: From Faraday's law, E = B v = E v , since B and are constants. Thus the correct graph is E (V) homework 09 KIM, JI Due: Mar 23 2008, 4:00 am 5 4 3 2 1 0 0 1 2 3 4 5 6 7 8 910 Velocity (m/s) Question 6, chap 31, sect 4. part 1 of 1 10 points Prior to 1960, magnetic field strength was measured by means of a rotating coil gaussmeter. This device used a small loop of many turns rotating on an axis perpendicular to the magnetic field at fairly high speed, which was connected to an AC voltmeter by means of slip rings, like those shown in figure. The sensing coil for a rotating coil gaussmeter has 300 turns, an area of 1.4 cm2 and rotates at 360 rpm. B B so the induced emf is (t) = - d m (t) dt d = - [N B A cos t] dt = N B A sin t . 4 E (V) Thus the maximum induced emf is max = N B A = 300 (0.55 T) (0.00014 m2 ) (37.6991 rad/s) = 0.870849 V . Question 7, chap 31, sect 1. part 1 of 2 10 points The figure below shows one of the blades of a helicopter which rotates around a central hub. The vertical component of the Earth's magnetic field is into the plane of the paper. 2 n turns B 0.55 T 1 10-5 T 2m v/ re s If the mangetic field strength is 0.55 T, find the maximum induced emf in the coil. Correct answer: 0.870849 V (tolerance 1 %). Explanation: Let : N = 300 , = 0.2 cm , A = 1.4 cm2 = 0.00014 m2 , f = 360 rpm , and B = 0.55 T . The angular velocity of this coil is = 2 f = 2 (360 rpm) = 37.6991 rad/s . The magnetic flux through this coil is m (t) = N B A cos t , 2 1 min rev 60 s 1 10-5 T What is the magnitude of the emf E induced between the blade tip and the central hub? Correct answer: 0.000251327 V (tolerance 1 %). Explanation: B f B homework 09 KIM, JI Due: Mar 23 2008, 4:00 am dA dt 2 = -B 2 1 |E| = B 2 , 2 = -B 5 Let : = 2 m, f = 2 rev/s , and B = 1 10-5 T . For a point on the blade, the velocity with which the point moves changes linearly with the distance from the point to the center of the hub. Then the effective velocity for the whole blade is the mean velocity, veff = 2 2f = 2 2 (2 rev/s) (2 m) = 2 = 12.5664 m/s , the same result as obtained above. Question 8, chap 31, sect 1. part 2 of 2 10 points The tip of the blade is 1. charged, but sign cannot be determined. 2. charged negative. 3. charged positive. correct 4. uncharged. Explanation: The magnetic force on a charge carrier of charge q is q v B where v is the local velocity of the blade. If q is positive, a force (use the cross product rule) is induced in the direction from the hub to the tip of the blade. Hence, positive charges move to the tip (leaving negative charges at the hub). Eventually an electric field induced by this charge separation points from the positively charged tip to the hub and the total force balance between magnetic and electric forces is established on the charge carriers in the blade. The net result is the tip is charged positively. If the charge q is negative, the magnetic force is towards the hub; negative charge accumulates there, leaving positive charge at the tip, with the same conclusion that the tip is positively charged. Question 9, chap 31, sect 2. part 1 of 1 10 points A metal spring (wrapped loosely around a cardboard doughnut as a toroid) has its ends and the induced emf in the blade is E = B veff 1 = B 2 2 1 = (1 10-5 T) 2 (12.5664 rad/s) (2 m)2 = 0.000251327 V . Alternative Solution: An alternative d B dA method is to calculate =B , where dt dt A is the area enclosed in the diagram below and B the enclosed flux in the figure. The area inside the triangle is A = 2 d 2 dA = = . and dt 2 dt 2 Therefore, E =- d B dt 2 , 2 homework 09 KIM, JI Due: Mar 23 2008, 4:00 am attached so that it forms a circle. It is placed in a uniform magnetic field, as shown. B B Question 10, chap 31, sect 3. part 1 of 1 10 points 6 In the figure shown, the magnet is first withdrawn upward from the loop of wire, then moved downward toward the loop of wire. N S up then down Counterclockwise I induced current B B Which of the following will NOT cause a current to be induced in the spring? 1. Moving the spring parallel to the magnetic field correct 2. Increasing the diameter of the circle by stretching the spring 3. Changing the magnitude of the magnetic field 4. Moving the spring in and out of the magnetic field 5. Rotating the spring about a diameter Explanation: There will be an induced current if there is an induced emf around the spring. By Faraday's law, we know that d B E =- , dt where B is the magnetic flux through the loop formed by the spring. To obtain a non-zero emf, we must change the magnetic flux. This can be accomplished by either changing the magnitude of the magnetic field or changing the area of the loop perpendicular to the magnetic field. Of the five choices, only moving the spring parallel to the magnetic field will NOT change the magnetic flux, thus will NOT cause a current to be induced in the spring. Clockwise induced I current As viewed from above, the induced current in the loop is 1. for both cases clockwise with increasing magnitude. 2. first counter-clockwise, then clockwise. correct 3. for both cases counterclockwise with increasing magnitude. 4. for both cases counterclockwise with decreasing magnitude. 5. for both cases clockwise with decreasing magnitude. 6. first clockwise, then counter-clockwise. Explanation: From Ohm's law and Faraday's law, the V 1 d current in magnitude is I = = - , R R dt where 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 withdrawn upward, as is the current direction according to the above eqaution. Using the right-hand-rule and from Lenz's law, we know that when the magnet is first withdrawn upward from the loop of wire, then moved downward toward the loop of wire, the current in the loop homework 09 KIM, JI Due: Mar 23 2008, 4:00 am is first counter-clockwise, then clockwise , as viewed from above. Question 11, chap 31, sect 3. part 1 of 1 10 points Assume: These are ideal coils. Given: Two coils are suspended around a common central axis as shown in the figure below. One of the coils is connected to a resistor R with ends labeled "a" and "b". The other coil is connected to a battery E through a switch S. Use Lenz's law to answer the following question concerning the direction of induced currents. a R b 3. 4. 5. 6. 7. 8. A1, B1, C1 A2, B2, C2 A2, B1, C2 A1, B1, C2 correct A1, B2, C2 A2, B1, C1 7 E + - S The direction of the magnetic field in the coils after the switch is closed for a long period of time is A1: right to left (= B). A2: left to right (B =). After switch S has been open for a long period of time, the switch is closed. The direction of the internal induced magnetic field by the larger diameter (secondary) coil just after the switch is closed is B1: left to right (Binduced =). B2: right to left (= Binduced ). The direction of the current induced in the resistor R just after the switch is closed is C1: from "a" through R to "b" (I ). C2: from "b" through R to "a" ( I). Choose the appropriate answer. 1. 2. A2, B2, C1 A1, B2, C1 Explanation: The smaller helical coil produces a counterclockwise current (looking from left-hand end of the coil). Using the right-hand rule, when the switch is the closed, magnetic field is directed from right to left (= B). The induced magnetic field depends on whether the flux is increasing or decreasing, since by Lenz's law the induced current will produce flux to oppose a change in flux. When the switch is open, there is no magnetic field. At the moment the switch is closed, the magnetic flux through the coils increases. The induced field through the coil in the circuit with a resistor must produce a magnetic field from left to right (Binduced =) to resist any change of flux in the coil (Lenz's law). The larger helical coil is wound clockwise (looking from left-hand end of the coil) from terminal a to terminal b. Using the right-hand rule, when the induced magnetic field is left to right (Binduced =), the induced current through the resistor R flows from "a" through R to "b" (I ). Note: There are eight different presentations of this problem and this is the fourth. Question 12, chap 31, sect 3. part 1 of 1 10 points A magnetic dipole is falling in a conducting metallic tube. Consider the induced current in an imaginary current loop when the magnet is moving away from the upper loop and when homework 09 KIM, JI Due: Mar 23 2008, 4:00 am the magnet is moving toward the lower loop. 8 Iabove ? Iabove y x v S N N Npole Ibelow ? Ibelow Spole dipole magnet Npole y x v S Spole dipole magnet z z Determine the directions of the induced currents Iabove and Ibelow in an imaginary loop shown in the figure, as viewed from above, when the loop is above the falling magnet and when the loop is below the falling magnet. 1. Iabove = clockwise and Ibelow = clockwise 2. Iabove = counter-clockwise and Ibelow = counter-clockwise 3. no current flow 4. Iabove = clockwise and Ibelow = counter-clockwise correct 5. Iabove = counter-clockwise Ibelow = clockwise and Question 13, chap 31, sect 3. part 1 of 1 10 points A pendulum consists of a supporting rod and a metal plate (see figure). The rod is pivoted at O. The metal plate swings through a region of magnetic field (directed out of the paper). Consider the case where the pendulum's metallic plate enters the magnetic field region from left to right. O eld entering fi B B Explanation: When the falling magnet is below the upper loop, - ind must be down to attract the falling magnet and slow it down; i.e., clockwise as viewed from above. Before reaching the lower loop, - ind must be up to oppose the falling magnet; i.e., counter-clockwise as viewed from above. B B The direction of the induced magnetic field at the center of the circulating eddy current and the direction of the eddy currect are homework 09 KIM, JI Due: Mar 23 2008, 4:00 am O 1. along the direction of swing and counterclockwise. 2. into the plane and counter-clockwise. 3. along the direction of swing and clockwise. 4. out of the plane and counter-clockwise. 5. opposite to the direction of swing and counter-clockwise. 6. opposite to the direction of swing and clockwise. 7. along the rod away from the pivot point and counter-clockwise. 8. out of the plane and clockwise. 9. into the plane and clockwise. correct 10. along the rod toward the pivot point and clockwise. 9 Fw B eld entering fi i B B B Alternative Solution: Use the result of Part 1, and the right hand rule on the current flow to determine that the induced magnetic field must be directed into the paper. Question 14, chap 30, sect 4. part 1 of 1 10 points When a sample of liquid is inserted into a solenoid carrying a constant current, the magnetic field inside the solenoid decreases by 0.014 %. What is the magnetic susceptibility of the liquid? Correct answer: -0.00014 (tolerance 1 %). Explanation: Let : P = 0.014 = 0.00014 . % The field inside the solenoid with the liquid sample present is Explanation: A conduction electron in the pendulum will experience a magnetic force opposite the direction of v B, so using the right hand rule, we can determine that the motion of the electrons will be in the counter-clockwise direction, which produces a clockwise current. Because the magnetic field is pointing out of the paper, the magnetic flux through the pendulum is increasing, so by Lenz's law we know that the induced magnetic field is in the opposite direction, or into the paper. B = Bapp (1 + m ) , where Bapp is the magnetic field in the absence of the liquid sample. Thus the magnetic susceptibility m is m = B = 0.00014 . Bapp Question 15, chap 30, sect 4. part 1 of 1 10 points homework 09 KIM, JI Due: Mar 23 2008, 4:00 am A toroidal solenoid has an average radius of 13.21 cm and a cross sectional area of 1.626 cm2 . There are 755 turns of wire on an iron core which has magnetic permeability of 4250 0 . The permeability of free space is 1.25664 10-6 N/A2 . ab perme ility of 42 re, 50 co n Thus magnetic flux density B is B = m H = m NI . 2r 10 r cross sectional area is 1.626 cm2 75 (For Ohanian and Markert 3rd.ed. user, the magnetic field in paramagnetic material is given by m Bexternal B= 0 where Bexternal is the field due to the current in the coil in vacuum. Please refer to section30.4 EXAMPLE 5 and 6 for detail disccusion.) Assuming no B field outside the solenoid and uniform field inside, the magnetic flux is m = B A = m NIA 2r Iro 0 I tu rn st u rn s o f wire i n t or oi Therefore, the required current is given by I= m 2 r m 2 r = m N A k 0 N A (0.000612 Wb) 2 = (4250)(1.25664 10-6 N/A2 ) (0.1321 m) (755 turns)(0.0001626 m2 ) = 0.774761 A . Question 16, chap 31, sect 2. part 1 of 1 10 points A solenoid with circular cross section produces a steadily increasing magnetic flux through its cross section. There is an octagonally shaped circuit surrounding the solenoid as shown below. i Figure 1: Calculate the current necessary to produce a magnetic flux of 0.000612 Wb through a cross section of the core. Correct answer: 0.774761 A (tolerance 1 %). Explanation: Let : r = 13.21 cm = 0.1321 m , A = 1.626 cm2 = 0.0001626 m2 , N = 755 turns , 0 = 1.25664 10-6 N/A2 , m = 4250 0 , and m = 0.000612 Wb. Basic concepts: NI Magnetic field strength H = . 2r Magnetic flux density B = m H. Magnetic permeability m = k 0 , where k = 1 + , and is the susceptibility. Solution: To determine the magnetic field strength H in free space H= NI , 2r d where r is radius of toroid. 5 B B X B B Y i homework 09 KIM, JI Due: Mar 23 2008, 4:00 am The increasing magnetic flux gives rise to a counterclockwise induced emf E. The circuit consists of two identical light bulbs of equal resistance, R, connected in series, leading to a loop equation E - 2 i R = 0. Now connect the points C and D with a wire CAD , (see the figure below). We label the currents i1 , i2 , and i3 as indicated. i3 Figure 2: D equation E - i3 R = 0 . The loop XDY CX leads to E - i3 R - i1 R = 0 . 11 Subtracting the former from the latter equation we find i1 = 0 . This makes sense physically since the i3 has a "choice" of having a fraction going through bulb Y (resistance R) and the remainder going through the wire at A (resistance zero); The competition is such that the entire current takes the route of zero resistance. This argument should only be viewed as a complement to the above analysis based on loop equations. We also have one node equation; e.g., at the junction C: i3 = i1 + i2 or i2 = i3 . Thus the complete answer is E - i3 R = 0; i2 = i3 , i1 = 0. Question 17, chap 31, sect 2. part 1 of 1 10 points Assume: The induced emf for the closed loop octagonal CXDY 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. Two identical resistors with resistance R (labeled X and Y ) are in the circuit. A wire connects points C and D. The ratio of the solenoid's area AL left of the wire CD and the solenoid's area AR right of the wire CD is AL = 2. AR i1 A Y i2 B B X B B i3 C The corresponding loop and node equations lead to 1. E - 2 i3 R = 0 : 2. E - 2 i3 R = 0 : 3. E - 2 i3 R = 0 : 4. E - 2 i3 R = 0 : 5. E - i3 R = 0 : 6. E - i3 R = 0; 7. E - i3 R = 0 : 8. E - i3 R = 0 : i2 = i3 i3 , i1 = 2 2 i2 = 0, i1 = i3 3 i3 i3 i2 = , i1 = 4 4 i2 = i3 , i1 = 0 i2 = 3 i3 i3 , i1 = 4 4 i2 = i3 , i1 = 0 correct i2 = 0, i1 = i3 i2 = i3 i3 , i1 = 2 2 2 i3 i3 9. E - i3 R = 0 : i2 = , i1 = 3 3 i3 2 i3 10. E - 2 i3 R = 0 : i2 = , i1 = 3 3 Explanation: Consider the loop XCADX (where "Y " indicates the position of bulb Y and "X" indicates the position of bulb X), with loop homework 09 KIM, JI Due: Mar 23 2008, 4:00 am i2 terms of A. D AL = 2A 3 AR = A . 3 d dt B 12 i3 E =- i1 B Y R AL B AR B X R Then we can compute the magnitude of the induced emf around the right and left loops. E R = AR A dB 1 dB = = E dt 3 dt 3 dB 2A dB 2 E L = AL = = E. dt 3 dt 3 C Figure: Let i1 , i2 , and i3 be defined as positive if the currents flow in the same direction as shown by the arrows in the figure (otherwise the currents i1 , i2 , and i3 have negative values). Also, let the induced emf be defined as positive, E > 0. The equations for the (right) loop CXDC and the (left) loop CDY C are respectively given by E 1. + i1 R = 0 and 3 rect 2. 2E + i2 R = 0 . cor3 The induced emf and the changing magnetic flux are related by E=- dB d = -A . dt dt 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 3 2 E + i2 R = 0 . 3 (1) (2) 2E E + i1 R = 0 and + i2 R = 0 . 3 3 2E E 3. - i1 R = 0 and + i2 R = 0 . 3 3 E 2E 4. - i1 R = 0 and - i2 R = 0 . 3 3 E 2E + i2 R = 0 . 5. - i1 R = 0 and 3 3 E 2E 6. - i1 R = 0 and - i2 R = 0 . 3 3 2E E 7. + i1 R = 0 and - i2 R = 0 . 3 3 2E E 8. + i1 R = 0 and - i2 R = 0 . 3 3 Explanation: By definition, the areas of the left and right loops are related by A = AL + AR . AL Since = 2, we can solve for AL and AR in AR Question 18, chap 31, sect 3. part 1 of 1 10 points Two infinitely long solenoids (seen in cross section in the figure below) thread a circuit. Given: a = 0.6 m , r1 = 0.2 m , r2 = 0.1 m , B = 59 T/s , t Rl = 7.5 , Rm = 4.9 , and Rr = 1.2 , as in the figure below. The magnitude of B inside each solenoid is the same and is 300 T at time t = 0. homework 09 KIM, JI Due: Mar 23 2008, 4:00 am a r1 a Rl Il B in Im Rm B out Ir a r2 Rr 13 and the direction of this emf is clockwise. (Numerical values are inserted into the equations to verify answers; our calculation has only 7 place accuracy.) Moving clockwise around the loops, circuit equations for these two loops are Loop 1: -Rl Il + Rm Im - E1 = 0 -(7.5 )(-0.700059 A) +(4.9 )(0.441575 A) -7.41416 V = 0 . Loop 2: -Rm Im + Rr Ir + E2 = 0 (2) -(4.9 )(0.441575 A) +(1.2 )(-0.258483 A) +1.85354 V = -0.62036 V . From the junction rule, we have (1) What is the magnitude of the current, Im , in middle resistor, Rm ? Correct answer: 0.441575 A (tolerance 1 %). Explanation: Basic Concept: Faraday's Law: -d B E= dt Ohm's Law: I= Junction Rule: n V R Ii = 0 i=1 Il + Im + Ir = 0 Solving these three simultaneous equations (Eqs. 1, 2, and 3) yields Im = 0.441575 A , Ir = -0.258483 A , Il = -0.700059 A . Solution: Note: The side-length, a, of the circuit loop is not necessary for this problem. Neither is the magnitude of B at time t = 0. From Faraday's law, the induced emf in the left loop (Loop 1) is |E1 | = d B dt dB = A1 dt 2 dB = r1 dt = 7.41416 V and and the direction of this emf is counterclockwise. Similarly, the induced emf in the right loop (Loop 2) is |E2 | = d B dt dB = A2 dt 2 dB = r2 dt = 1.85354 V ... View Full Document

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