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HMWK 3 Ch 23: P 17, 23, 26, 34, 52, 58, 59, 62, 64, 73 Ch 24: Q 17, 34; P 5, 17, 34, 42, 51, 52, 53, 57 Chapter 23 P23.17. Prepare: The connecting wires are ideal with zero resistance. We have to reduce the circuit to a single equivalent resistor by continuing to identify resistors that are in series or parallel combinations. Solve: For the first step, the resistors 30 Ω and 45 Ω are in parallel. Their equivalent resistance is 1 1 1 = + Req 1 30 Ω 45 Ω Req 1 = 18 Ω For the second step, resistors 42 Ω and Req 1 = 18 Ω are in series. Therefore, Req 2 = Req 1 + 42 Ω = 18 Ω + 42 Ω = 60 Ω For the third step, the resistors 40 Ω and Req 2 = 60 Ω are in parallel. So, 1 1 1 = + Req 3 60 Ω 40 Ω The equivalent resistance of the circuit is 24 Ω. Req 3 = 24 Ω Assess: Have a good understanding of how series and parallel resistors combine to obtain equivalent resistors. P23.23. Prepare: The battery and the connecting wires are ideal. The figure shows how to simplify the circuit in Figure P23.23 using the laws of series and parallel resistances. We have labeled the resistors as R1 = 6.0 Ω, R2 = 15 Ω, R3 = 6.0 Ω, and R4 = 4.0 Ω. Having reduced the circuit to a single equivalent resistance Req, we will reverse the procedure and “build up” the circuit using the loop law and the junction law to find the current and potential difference of each resistor. Solve: R3 and R4 are combined to get R34 = 10 Ω, and then R34 and R2 are combined to obtain R234: 1 1 1 1 1 = + = + R234 R2 R34 15 Ω 10 Ω R234 = 6 Ω Next, R234 and R1 are combined to obtain Req = R234 + R1 = 6.0 Ω + 6.0 Ω = 12 Ω From the final circuit, I= ε Req = 24 V = 2 .0 A 12 Ω Thus, the current through the battery and R1 is IR1 = 2.0 A and the potential difference across R1 is I(R1) = (2.0 A) (6.0 Ω) = 12 V. As we rebuild the circuit, we note that series resistors must have the same current I and that parallel resistors must have the same potential difference ∆V. In Step 1 of the above figure, Req = 12 Ω is returned to R1 = 6.0 Ω and R234 = 6.0 Ω in series. Both resistors must have the same 2.0 A current as Req. We then use Ohm’s law to find ∆VR1 = (2 .0A)(6.0 Ω) = 12 V∆VR234 = (2.0 A)(6.0 Ω) = 12 V As a check, 12 V + 12 V = 24 V, which was ∆V of the Req resistor. In Step 2, the resistance R234 is returned to R2 and R34 in parallel. Both resistors must have the same ∆V = 12 V as the resistor R234. Then from Ohm’s law, IR2 = 12 V = 0.8 A 15 Ω I R34 = 12 V = 1.2 A 10 Ω As a check, IR2 + IR34 = 2.0 A, which was the current I of the R234 resistor. In Step 3, R34 is returned to R3 and R4 in series. Both resistors must have the same 1.2 A as the R34 resistor. We then use Ohm’s law to find (∆V)R3 = (1.2 A)(6.0 Ω) = 7.2 V(∆V)R4 = (1.2 A)(4.0 Ω) = 4.8 V #23 cont. As a check, 7.2 V + 4.8 V = 12 V, which was ∆V of the resistor R34. Resistor Potential difference (V) R1 R2 R3 R4 12 12 7.2 4.8 The three steps as we rebuild our circuit are shown below. Current (A) 2.0 0.8 1.2 1.2 Assess: This problem requires a good understanding of how to first reduce a circuit to a single equivalent resistor and then to build up a circuit. P23.26. Prepare: The voltage drop across a resistor may be determined by ∆VR = IR. The power dissipated 2 by a resistor may be determined by P = I R = IV. The power supplied to a device is related to the rate at which it uses energy by P = ∆W/∆t. Solve: (a) The voltage drop across the wire is ∆VW = IRW = (10 A)(0.10 Ω) = 1.0 V. 2 2 (b) The power dissipated in the wire is PW = I RW = (10 A) (0.10 Ω) = 10 W. (c) The voltage drop across the device is determined by ∆Voutlet = ∆Vwire + ∆Vdevice or ∆Vdevice = ∆Voutlet − ∆Vwire = 120V − 1.0V = 119 V. (d) The rate at which the device uses energy is determined by ∆W /∆t = P = IVdevice = (10 A)(119 V) = 1190 W = 1.2k W. Assess: This problem uses several of the relationships between the electric potential difference, current, resistance, and power. It is important to be familiar with all of these basic relationships used in solving electricity problems. P23.34. Prepare: Please refer to figure P23.34. The pictorial representation shows how to find the equivalent capacitance of the three capacitors shown in the figure. Solve: Because C1 and C2 are in series, their equivalent capacitance Ceq 12 is 1 Ceq 12 = 1 1 1 1 1 + = + = C1 C2 20 µF 30 µF 12 µF Ceq 12 = 12 µF Then, Ceq 12 and C3 are in parallel. So, Ceq = Ceq 12 + C3 = 12 µF + 25 µF = 37 µF Assess: We must understand well how to combine series and parallel capacitance. P23.52. Prepare: The connecting wires are ideal with zero resistance. A visual overview of how to reduce the circuit to an equivalence resistance is shown below. Solve: In the first step, the resistors 100 Ω, 100 Ω, and 100 Ω in the top branch are in series. Their combined resistance is 300 Ω. In the middle branch, the two resistors, each 100 Ω, are in series. So, their equivalent resistance is 200 Ω. In the second step, the three resistors are in parallel. Their equivalent resistance is 1 1 1 1 = + + Req 300 Ω 200 Ω 100 Ω Req = 54.5 Ω The equivalent resistance of the circuit is 50 Ω. P23.58. Prepare: Please refer to Figure P23.58. We will assume ideal connecting wires. Because the ammeter we have shows a full-scale deflection with a current of 500 µA, we must not pass a current greater than this through the ammeter. Solve: The maximum potential difference is 5 V and the maximum current is 500 µA. Using Ohm’s law, ∆V = IAR 5.0 V = (500 × 10−6 A)R R = 10 kΩ. P23.59. Prepare: The figure shows how to simplify the circuit in Figure P23.59 using the laws of series and parallel resistances. Having reduced the circuit to a single equivalent resistance, we will reverse the procedure and “build up” the circuit to find the current and potential difference of each resistor. We will assume that the battery and the connecting wires are ideal. Solve: (a) From the last circuit in the figure and from Kirchhoff’s law, I = 100 V/10 Ω = 10 A. Thus, the current through the battery is 10 A. Now as we rebuild the circuit, we note that series resistors must have the same current I and that parallel resistors must have the same potential difference. In Step 1 of the above diagram, we return the 10 Ω resistor to the 4.0 Ω, 4.0 Ω, and 2.0 Ω resistors in series. These resistors must have the same 10 A current as the 10 Ω resistance. That is, the current through the 2.0 Ω and the 4.0 Ω resistors is 10 A. The potential differences are ∆V2 = (10 A)(2.0 Ω) = 20 V ∆V4 (left) = (10 A)(4.0 Ω) = 40 V ∆V4 (left) = (10 A)(4.0 Ω) = 40 V In Step 2, we return the left 4.0 Ω resistor to the 20 Ω and 5.0 Ω resistors in parallel. The two resistors must have the same potential difference ∆V = 40 V. From Ohm’s law, I5 = 40 V = 8.0 A 5.0 Ω I20 = 40 V = 2.0 A 20 Ω The currents through the various resistors are I2 = I4 = 10 A, I5 = 8.0 A, and I20 = 2.0 A. 2 2 (b) The power dissipated by the 20 Ω resistor is I20 (2 0 Ω) = (2 A) (20 Ω) = 80 W. (c) Starting with zero potential at the grounded point, we travel along the outside path to point a and add/subtract the potential differences on the way: 0 V + (20 Ω)I20 + (2 Ω)I2 = (20 Ω)(2 A) + (2 Ω)(10 A) = 60 V = Va P23.62. Prepare: Capacitors in parallel add to a greater capacitance compared to individual capacitances. On the other hand, capacitors in series add to a smaller capacitance compared to individual capacitances. Solve: (a) Three capacitors in series: 1 1 1 1 3 −1 = + + = ( µF) Ceq 12 µF 12 µF 12 µF 12 Ceq = 4.0 µF (b) Two capacitors in parallel and the third in series with this parallel combination: Ceq 12 = 12 µF + 12 µF = 24 µF 1 1 1 1 1 1 = + = + = Ceq Ceq 12 12 µ F 24 µF 12 µF 8.0 µF Ceq = 8.0 µF (c) Two capacitors in series and the third in parallel with this series combination: 1 Ceq 12 = 1 12 µF + 1 1 −1 = (µF) 12 µF 6.0 Ceq 12 = 6.0 µF Ceq = Ceq 12 + 12 µF = 6.0 µF + 12 µF = 18 µF (d) Three capacitors in parallel: Ceq = 12 µF + 12 µF +12 µF = 36 µF Assess: Learn how to combine series and parallel capacitances. P23.64. Prepare: Please refer to Figure P23.64. While the switch is in position A, the capacitors C2 and C3 are uncharged. When the switch is placed in position B, the charged capacitor C1 is connected to C2 and C3. C2 and C3 are connected in series to form an equivalent capacitor Ceq 23. We will also assume that the battery is ideal. Solve: While the switch is in position A, a potential difference of V1 = 100 V across C1 charges it to Q1 = C1V1 = (15 × 10−6 F)(100 V) = 1500 µC. When the switch is moved to position B, this initial charge Q1 is redistributed. The charge Q1′ goes on C1 and the charge Qeq 23 goes on Ceq 23. The voltage across C1 and Ceq 23 is the same and Q1′ + Qeq 23 = Q1 = 1500 µC. Combining these two conditions, we get Q1′ Qeq 23 = C1 Ceq 23 Since Ceq 23 = ( 1 30 µ F + 301µF 1500 µC − Qeq 23 15 µF 1500 µC − Qeq 23 C1 = Qeq 23 Ceq 23 ) = 1 5 µF, we can rewrite this equation as −1 = Qeq 23 15 µF Qeq 23 = 750 µC Q1′ = Q1 − Qeq 23 = 1500 µC − 750 µC = 750 µC Having found the charge Qeq 23, it is easy to see that Q2 = Q3 = 750 µC because Ceq 23 is a series combination of C2 and C3. Thus, ∆V2 = Q2 750 µC = = 25 V C2 30 µF ∆V3 = Q3 750 µC = = 25V C3 30 µF ∆V1 = Q1′ 750 µC = = 50 V C1 15 µF Assess: the potential differences across and charges on the three capacitors are consistent with the ideas of a closed circuit. P23.73. Prepare: The capacitance of a dielectric-filled capacitor Equation 21.21: C = κε 0 A/d and then Equation 21.17 gives the charge: Q = C∆VC . is −9 given in We are told the dielectric is water and Table 21.2 gives κ = 80. We are also given A = 6.0 ×1 0 m , d = 7.0 × 2 10−9 m, and ∆VC = 70 × 10−3 V. The sodium ions have a +1 charge. Solve: (a) We can combine both equations. Q = C∆VC = κε 0 A d ∆VC = (80)(8.85 × 10−12 C 2/N⋅ m2 )(6.0 ×10−9 m2) −3 −11 (70 ×10 V) = 4.2 ×10 C 7.0 ×1 0−9 m (b) 4.2 × 10 −11 C 1e 8 = 2.7 × 10 e 1.6 × 10−19 C 8 or 2.7 × 10 sodium ions. Assess: This tiny fraction of a Coulomb corresponds to 270 million sodium ions. There is a lot going on in each living cell! [Also accept κ = 9, answer 3x107 ions. This is correct answer based on example problem] Chapter 24 Q24.17. Reason: The direction of the magnetic force determines the direction that the particle will be deflected when it enters the magnetic field. Use the right-hand rule for forces to determine the direction of the force. See Figure 24.27 in the textbook. (a) The velocity points to the right and the magnetic field points toward the bottom of the page, so the force points into the page. (b) In this case the velocity and the magnetic field are pointing in the same direction, so there is no force. Assess: The reason there is no force for part (b) can be shown using the equation F = qvBsin α, where α is the angle between the velocity and the magnetic field. So if the velocity is parallel to the magnetic field, this angle is zero, which makes the sine and the force equal zero as well. Q24.34. Reason: When a charged particle enters a uniform magnetic field it moves in uniform circular motion. The radius of the curvature is given by r = mv/qB. The correct choice is D. Assess: Since path D has the largest circular radius, it has the largest mass. The problem stated that the particles have the same speed (v), same charge (q), and enter the same uniform magnetic field (B). P24.5. Prepare: Assume the wires are infinitely long. Find the contribution of the magnetic field due to each wire. Solve: The magnetic field strength at point 1 is r r B1 = B to p + Bb o t to m = B1 = µ 0I , out of page 2π d + top µ0 I , into page 2π d b o t to m µ0 I 1 1 1 1 −7 − = ( 2)(10 T ⋅ m/A)(10 A) − 2 × 1 0 −2 m 6 × 1 0 −2 m 2 π 2 cm (4 + 2) cm r B1 = (6 .7 × 10 −5 T, o ut of pa ge) At points 2 and 3, B2 = r B3 = µ0 I 2π (2 cm ) , into pa ge + µ 0I 2 π (2 cm) , into page = ( 2.0 × 10 −4 T, into pa ge) µ0 I µ0 I , into pa ge + , out of p age = (6.7 × 10 −5 T, out o f page ) 2π (6 cm ) 2 π (2 c m) Assess: Each point is affected by both wires, so the contributions must add according to the direction of the field points. The equation of the magnetic field does not give its direction, only its magnitude. To get the direction you must use the right-hand rule. If the fields are in the same direction, they add. If they are in different directions, they subtract. P24.17. Prepare: We are not given the length of this coil, so we assume the 100 turns are all close together and we treat it as an N-turn coil rather than a solenoid. The applicable equation is Equation 24.3. We are given N = 100, I = 1.5 A, and R = 3.5 m. Solve: (a) B= µ0 NI 2R = (4π ×10−7 T ⋅m/A)(100)(1.5 A) −5 = 2 .7 ×10 T 2(3.5 m) (b) Table 24.1 indicates that the field strength at the surface of the earth is about 5 × 10–5 T, and our answer is smaller than that by about half. Assess: The sharks are detecting a field strength that is about half as strong as the earth’s field. They can, therefore, presumably detect the earth’s field. Some other species may also be sensitive to magnetic fields. P24.34. Prepare: Assume that the magnetic field is uniform. Solve: The magnitude of the magnetic force is expressed as F = ILBsin α , where α is the angle the wire makes with the magnetic field. F = ILB sin α F = (15 A)(3 .0 m)(2.5 T) sin 30° F = 56 N Using the right-hand rule, we can determine that the direction of the force is into the page. Assess: Given that the current and the magnetic field strength are large, a force of 56 N is reasonable. P24.42. Prepare: The wire will create a magnetic field at the position of the loop, so there will be a torque on the loop. The magnetic field created by the wire at the position of the loop will be approximately uniform, since the distance between the wire and the loop is large compared to the size of the loop. It can be approximated by calculating the magnitude and direction of the field created by the wire at the center of the loop. Once this is known, the torque on the loop from this applied field can be calculated. The torque on a loop in a uniform magnetic field is given in Equation 24.18. Solve: (a) The expression for the field created by a wire at the location of the loop is given in Equation 24.1: B= µ0 I (4π ×10 −7 T ⋅ m/A)(2.0 A) −5 = = 2.0 ×10 T 2π r 2π × (0.02 m) Using the right-hand rule, the direction of the field will be pointing straight up at the center of the loop, relative to Figure P24.42. To determine the torque on the loop, we need to determine the angle between this field and the dipole moment of the loop. The dipole moment of the loop points directly left relative to Figure P24.42. So the angle between the applied field and the dipole moment of the loop is 90°. Using Equation 24.18: τ = (IA)Bsinθ = (0.20 A)(π (0.002 m)2 )(2.0 ×10−5 T) sin (90°) = 5 ×10−11 N ⋅m (b) The dipole will rotate until the angle between the dipole moment and the field is zero, therefore, it will rotate clockwise 90°. Once it has rotated to its new position, the angle between the dipole moment and the field will be zero, and the loop will feel no torque or net force. At this point, it will be in equilibrium. P24.51. Prepare: An electric and magnetic field exerts two independent forces on the moving electron. The magnitude of the electric force is FE = q E and the direction is the direction of the electric field. The magnitude of the magnetic force is FB = qvB and the direction is given by the right-hand rule for forces. Solve: The electric field is V 200 V = , down = (20,000 V/m, down) d 0.01 m r r −15 The force this field exerts on the electron is FE = qE = −e E = (3.2 ×10 N, up). The electron will pass through r r r without deflection if the magnetic field also exerts a force on the electron such that Fnet = FE + FB = 0 N or r −15 FE = − FB: i.e., the electric and magnetic forces cancel each other. So, the magnetic force is FB = (3.2 ×10 N, E= down). For a negative charge with v to the right to have FB down requires, from the right-hand rule for forces, that the direction of the magnetic field point into the page. The magnitude of the magnetic force on a moving charge is FB = qvB, so the needed field strength is B= FB 3.2 ×10 −15 N −3 = = 2.0 ×10 T = 2.0 mT eν (1.60 × 10−19 C)(1.0 ×107 m/s) Thus, the required magnetic field is B = (2.0 mT, into page). P24.52. Prepare: A magnetic field exerts a magnetic force on a length of current carrying wire. We ignore gravitational effects and focus on the magnetic effects. The figure shows a wire in a magnetic field that is directed out of the page. The magnetic force on the wire is therefore to the right and will stretch the springs. Solve: The direction of the magnetic force is to the right; whereas, the direction of the spring forces is to the left. In static equilibrium, the sum of the forces on the wire is zero: r r r FB + Fspring 1 + Fspring 2 = 0 N ILB+ (−k∆x) + (−k∆x) = 0 I= 2k∆x 2(10 N/m)(0.01m) = = 2.0 A LB (0.20 m)(0.5 T) P24.53. Prepare: Electric and magnetic fields exert forces on a moving charge. The fields are uniform throughout the region. Solve: (a) We will first find the net force on the antiproton, and then find the net acceleration using Newton’s second law. The magnitudes of the electric and magnetic forces are FE = eE = (1.60 × 10 −19 FB = ev B = (1.60 × 1 0 C )(1000 V/m) = 1.60 × 10 −1 9 −1 6 N C )(5 00 m/s)( 2.5 T) = 2.0 × 10 −1 6 N The directions of these two forces on the antiproton are opposite. FB points down, whereas FE points up, opposite the direction of the electric field. Hence, Fn e t = FB + FE = 2.0 × 10 −1 6 N (down) − 1.60 × 1 0 −1 6 N (up) = 0.40 × 10 −1 6 N (down) Fn e t = ma a= Fn et m = 0.40 × 10 −16 N = 2.4 × 101 0 m /s 2 (down) 1 .67 × 10 −27 k g (b) If v were reversed, both FE and FB will point up. Thus, −16 Fnet = (1.6 ×10 N + 2.0 ×10 −16 N, up) r a= 3.6 ×10−16 N 11 2 = 2.2 ×10 m/s , up 1.67 ×10−27 kg P24.57. Prepare: We must start with the normal equation for a coil, Equation 24.4. Once we’ve calculated the magnetic field, it is simply a matter of multiplying it by a factor of 100. Solve: Bc o il = µ 0 N L I= (4 π × 10 Bc o il = 0.01 T Bi n c o il = 0.01 T (100) = 1.0 T −7 T⋅m A )(240)(0.60 A) 0.018 m