Chapter 30. Forces and Torques in a Magnetic Field                              Physics, 6th Edition


           Chapter ...
Chapter 30. Forces and Torques in a Magnetic Field                                Physics, 6th Edition


30-6. What curren...
Chapter 30. Forces and Torques in a Magnetic Field                                   Physics, 6th Edition


30-10. A galva...
Chapter 30. Forces and Torques in a Magnetic Field                                             Physics, 6th Edition


*30-...
Chapter 30. Forces and Torques in a Magnetic Field                             Physics, 6th Edition




30-17. A solenoid ...
Chapter 30. Forces and Torques in a Magnetic Field                                  Physics, 6th Edition




30-21. A sole...
Chapter 30. Forces and Torques in a Magnetic Field                                            Physics, 6th Edition


*30-2...
Chapter 30. Forces and Torques in a Magnetic Field                              Physics, 6th Edition


*30-28. A commercia...
Chapter 30. Forces and Torques in a Magnetic Field                                 Physics, 6th Edition


                ...
Chapter 30. Forces and Torques in a Magnetic Field                               Physics, 6th Edition


*30-32. Verify the...
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  1. 1. Chapter 30. Forces and Torques in a Magnetic Field Physics, 6th Edition Chapter 30. Forces and Torques in a Magnetic Field Magnetic Torque on a Loop (α is angle that plane of loop makes with B field.) 30-1. A rectangular loop of wire has an area of 30 cm2 and is placed with its plane parallel to a 0.56-T magnetic field. What is the magnitude of the resultant torque if the loop carries a current of 15 A ? (1 cm2 = 1 x 10-4 m2) τ = NBIA cos α = (1)(0.56 T)(15 A)(30 x 10-4 m 2 ) cos 00 ; τ = 0.0252 N⋅m 30-2. A coil of wire has 100 turns, each of area 20 cm2. The coil is free to turn in a 4.0 T field. What current is required to produce a maximum torque of 2.30 N⋅m? τ 2.30 N ⋅ m I= = ; I = 2.88 A NBA cos α (100)(4.0 T)(20 x 10-4 m 2 )(1) 30-3. A rectangular loop of wire 6 cm wide and 10 cm long is placed with its plane parallel to a magnetic field of 0.08 T. What is the magnitude of the resultant torque on the loop if it carries a current of 14.0 A. [ A = (0.10 m)(0.06 m) = 6 x 10-3 m2 ] τ = NBIA cos α = (1)(0.08 T)(14 A)(6 x 10-3 m 2 ) cos 00 ; τ = 6.72 x 10-3 N⋅m 30-4. A rectangular loop of wire has an area of 0.30 m2. The plane of the loop makes an angle of 300 with a 0.75-T magnetic field. What is torque on the loop is the current is 7.0 A. τ = NBIA cos α = (1)(0.75 T)(7.0 A)(0.30 m 2 ) cos 300 ; τ = 1.36 N⋅m 30-5. Calculate the B field required to give a 100-turn coil a torque of 0.5 N m when its plane is parallel to the field. The dimensions of each turn is 84 cm2 and the current is 9.0 A. τ 0.500 N ⋅ m B= = ; B = 66.1 mT NIA cos α (100)(9.0 A)(84 x 10-4 m 2 ) cos 00 163
  2. 2. Chapter 30. Forces and Torques in a Magnetic Field Physics, 6th Edition 30-6. What current is required to produce a maximum torque of 0.8 N⋅m on a solenoid having 800 turns of area 0.4 m2? The flux density is 3.0 mT. What is the position of the solenoid in the field? (For solenoid, θ is the angle that coil axis makes with B field). The torque is a maximum when: θ = 90 τ 0.80 N ⋅ m I= = ; I = 833 mA NBA sin θ (800)(0.003 T)(0.4 m 2 ) sin 900 30-7. The axis of a solenoid, having 750 turns of wire, makes an angle of 340 with a 5-mT field. What is the current if the torque is 4.0 N m at that angle. The area of each turn of wire is 0.25 m2. τ 4.00 N ⋅ m I= = ; I = 7.63 A NBA sin θ (750)(0.005 T)(0.25 m 2 )sin 340 Galvanometers, Voltmeters, and Ammeters 30-8. A galvanometer coil 50 mm x 120 mm is mounted in a constant radial 0.2-T B field. If the coil has 600 turns, what current is required to develop a torque of 3.6 x 10-5 N m? A = (0.05 m)(0.120 m) = 6 x 10-3 m2; θ = 900; B = 0.200 T; τ = 3 x 10-5 N m τ 3.60 x 10-5 N ⋅ m I= = ; I = 50.0 µA NBA sin θ (600)(0.2 T)(6 x 10-3 m 2 ) sin 900 30-9. A galvanometer has a sensitivity of 20 µA per scale division. What current is required to give full-scale deflection with 25 divisions on either side of the central equilibrium position? I = 25 div(20 µ A/div); I = 500 µA 164
  3. 3. Chapter 30. Forces and Torques in a Magnetic Field Physics, 6th Edition 30-10. A galvanometer has a sensitivity of 15 µA per scale division. How many scale divisions will be covered by the deflecting needle when the current is 60 µA? 60 µ A Number = ; Number = 15 div 15 µ A/div 30-11. A certain voltmeter draws 0.02 mA for full-scale deflection at 50 V. (a) What is the resistance of the voltmeter? (b) What is the resistance per volt? 50 V Rv = = 2.5 x 106 Ω; Rv = 2.5 MΩ 2 x 10-5 A R 2.5 x 106 Ω R = ; = 50.0 kΩ/V V 50 V V *30-12. For the voltmeter in Problem 30-11, what multiplier resistance must be used to convert this voltmeter to an instrument that reads 100 mV full scale? VB − Vg 150 V - 50 V Rm = = ; Rm = 5.00 MΩ If 2 x 10-5 A *30-13. The coil of a galvanometer will burn out if a current of more than 40 mA is sent through it. If the coil resistance is 0.5 Ω, what shunt resistance should be added to permit the measurement of 4.00 A? I g Rg (40 x 10-3A)(0.500 Ω) Rs = = ; Rs = 5.05 Ω I − Ig 4 A - 0.04 A 165
  4. 4. Chapter 30. Forces and Torques in a Magnetic Field Physics, 6th Edition *30-14. A current of only 90 µA will produce full-scale deflection of a voltmeter that is designed to read 50 mV full scale. (a) What is the resistance of the meter? (b) What multiplier resistance is required to permit the measurement of 100 mV full scale? 50 x 10-3 V 0.100 V - 0.050 V (a) Rv = ; Rv = 555 Ω (b) Rm = ; Rm = 555 Ω 90 x 10-6 A 90 x 10-6 A Adding the same resistance as voltmeter resistance doubles the range of the meter. *30-15. An ammeter that has a resistance of 0.10 Ω is connected in a circuit and indicates a current of 10 A at full scale. A shunt having a resistance of 0.01 Ω is then connected across the terminals of the meter. What new circuit current is needed to produce full-scale deflection of the ammeter? Rg Ig Is IsRs = IgRg; Is = I – Ig; (I – Ig) Rs = IgRg IRs – IgRs = Ig Rg IRs = IgRs + IgRg Solve for I Rs I VB RL I g ( Rg + Rs ) (10 A)(0.1 Ω + 0.01 Ω) I= = ; I = 110 A Rs 0.01 Ω Challenge Problems 30-16. A wire of length 12 cm is made in to a loop and placed into a 3.0-T magnetic field. What is the largest torque the wire can experience for a current of 6 A? C = 2πR C 0.12 m R= = ; R = 0.0191 m; A = π R 2 = π (0.0191 m) 2 ; A = 1.16 x 10-4 m2 2π 2π τ = NBIA cos α = (1)(3 T)(6 A)(1.16 x 10-4 m 2 ) cos 00 ; τ = 2.09 x 10-3 N⋅m 166
  5. 5. Chapter 30. Forces and Torques in a Magnetic Field Physics, 6th Edition 30-17. A solenoid has 600 turns of area 20 cm2 and carries a current of 3.8 A. The axis of the solenoid makes an angle of 300 with an unknown magnetic field. If the resultant torque on the solenoid is 1.25 N⋅m, what is the magnetic flux density? τ 1.25 N ⋅ m B= = ; B = 548 mT NAI sin θ (600)(20 x 10-4 m 2 )(3.8 A) sin 300 30-18. A flat coil of wire has 150 turns of radius 4.0 cm, and the current is 5.0 A. What angle does the plane of the coil make with a 1.2-T B field if the resultant torque is 0.60 N⋅m? A = πR2 = π(0.04 m)2; A = 5.03 x 10-3 m2; I = 5 A; τ = 0.6 N⋅m τ 0.600 N ⋅ m cos α = = ; α = 82.40 NBIA (150)(1.2 T)(5 A)(5.03 x 10-3 m 2 ) 30-19. A circular loop consisting of 500 turns carries a current of 10 A in a 0.25-T magnetic field. The area of each turn is 0.2 m2. Calculate the torque when the plane of the loop makes the following angles with the field: 00, 300, 450, 600, and 900. (a) τ = NBIA cos α = (500)(0.25 T)(10.0 A)(0.20 m 2 ) cos 00 ; τ = 250 N⋅m (b) τ = (250 N⋅m) cos 300; τ = 217 N⋅m; (c) τ = (250 N⋅m) cos 450; τ = 177 N⋅m (d) τ = (250 N⋅m) cos 300; τ = 125 N⋅m; (e) τ = (250 N⋅m) cos 00; τ = 0 N⋅m 30-20. A solenoid of 100 turns has a cross-sectional area of 0.25 m2 and carries a current of 10 A. What torque is required to hold the solenoid at an angle of 300 with a 40-mT magnetic field? ( The angle of the solenoid axis is θ = 300 ) τ = NBIA sin θ = (100)(0.04 T)(10.0 A)(0.25 m 2 ) sin 300 ; τ = 5.00 N⋅m 167
  6. 6. Chapter 30. Forces and Torques in a Magnetic Field Physics, 6th Edition 30-21. A solenoid consists of 400 turns of wire, each of radius 60 mm. What angle does the axis of the solenoid make with the magnetic flux if the current through the wire is 6 A, the flux density is 46 mT, and the resulting torque is 0.80 N m? A = πR2 = π(0.06 m)2 = 0.0113 m2; τ = 0.80 N⋅m; B = 46 mT; I = 6 A; θ = ? τ 0.800 N ⋅ m sin θ = = ; θ = 39.80 NBIA (400)(0.046 T)(6 A)(0.0113 m 2 ) *30-22. A galvanometer having an internal resistance of 35 Ω requires 1.0 mA for full-scale deflection for a current of 10 mA. What multiplier resistance is needed to convert this device to a voltmeter that reads a maximum of 30 V? VR 30 V Rm = − Rg = − 35 Ω; Rm = 29,965 Ω Ig 0.001 A *30-23. The internal resistance of a galvanometer is 20 Ω and it reads full scale for a current of 10 mA. Calculate the multiplier resistance required to convert this galvanometer into a voltmeter whose range is 50 V. What is the total resistance of the resulting meter? VR 50 V Rm = − Rg = − 20 Ω; Rm = 4980 Ω Ig 0.01 A Rv = Rm + Rg = 4980 Ω + 20 Ω ; Rv = 5000 Ω *30-24. What shunt resistance is needed to convert the galvanometer of Problem 30-22 to an ammeter reading 10 mA full scale? I g Rg (0.001A)(35 Ω) Rs = = ; Rs = 3.89 Ω I − Ig 0.01 A - 0.001 A 168
  7. 7. Chapter 30. Forces and Torques in a Magnetic Field Physics, 6th Edition *30-25. A certain voltmeter reads 150 V full scale. The galvanometer coil has a resistance of 50 Ω and produces a full-scale deflection on 20 mV. Find the multiplier resistance of the voltmeter. VR Vg 0.020 V Rm = − Rg ; Ig = = = 4 x 10-4 A Ig Rg 50.0 Ω VR 150 V Rm = − Rg = − 50 Ω; Rm = 374,950 Ω Ig 4 x 10-4 A *30-26. A galvanometer has a coil resistance of 50 Ω and a current sensitivity of 1 mA (full scale). What shunt resistance is needed to convert this galvanometer to an ammeter reading 2.0 A full scale? I g Rg (0.001A)(50 Ω) Rs = = ; Rs = 0.025 Ω I − Ig 2 A - 0.001 A *30-27. A laboratory ammeter has a resistance of 0.01 Ω and reads 5.0 A full scale. What shunt resistance is needed to increase the range of the ammeter tenfold? I I R = 10; Rs = g g ; IRs − I g Rs = I g Rg ; IRs = I g ( Rs + Rg ); Ig I − Ig I R + Rg = s = 10; Rs + Rg = 10 Rs ; Rs (10 − 1) = Rg Ig Rs Rg 0.01 Ω Rs = = ; Rs = 1.11 mΩ 9 9 Thus, by adding 1.11 µΩ in parallel, the meter range will extend to 50 A full scale. 169
  8. 8. Chapter 30. Forces and Torques in a Magnetic Field Physics, 6th Edition *30-28. A commercial 3-V voltmeter requires a current of 0.02 mA to produce full-scale deflection. How can it be converted to an instrument with a range of 150 V? VB − Vg 150 V - 3 V Rm = = ; Rm = 7.35 Ω Ig 0.02 x 10-3A Adding Rm = 7.35 Ω in series will increase the range to 150 V. Critical Thinking Problems 30-29. Consider the rectangular, 4 cm by 6 cm loop of 60 turns of wire in Fig. 30-11. The plane of the loop makes an angle of 400 with a 1.6-T magnetic field B directed along the x-axis. The loop is free to rotate about the y-axis and the clockwise current in the coil is 6.0 A. What is the torque and will it turn toward the x axis or toward the z axis? y A = (0.06 m)(0.04 m) = 2.4 x 10-3 m2; α = 400 4 cm τ = NBIA cos α = (60)(1.6 T)(6 A)(0.0024 m2)cos 400 60 turns τ = 1.06 N⋅m, The downward direction of the current 6 cm B I in the near segment means that the force producing the 400 x torque will cause the loop to rotate AWAY from x axis. τ = 1.06 N⋅m, toward z-axis z *30-30. A voltmeter of range 150 V and total resistance 15,000 Ω is connected in series with another voltmeter of range 100 V and total resistance 20,000 Ω. What will each meter read when they are connected across a 120-V battery of negligible internal resistance? To determine what the readings will be we need to know the galvanometer currents Ig required for full-scale deflection of the galvanometer in each voltmeter. These are found by dividing the full-scale voltages by their galvanometer resistances (Ohm’s law). 170
  9. 9. Chapter 30. Forces and Torques in a Magnetic Field Physics, 6th Edition 150 V 100 V *30-30. (Cont.) For A: I gf = = 0.010 A; For B: I gf = = 0.005 A 15,000 Ω 20,000 Ω Since voltmeters are in series: Iv = Ig 150-V meter A 100-V meter RgA RgB GA GB 120 V Iv = = 3.43 mA ; Ig = 3.43 mA 15 kΩ + 20 kΩ RmA RmB Iv Consider reading of Voltmeter A (150-V full): VA = Iv(RmA + RgA) = (3.43 mA)(15 kΩ) 120-V source C VA = Iv(RmA + RgA) = (3.43 mA)(20 kΩ) VA = 51.5 V; VB = 68.6 Ω *30-31. The magnetic moment µ is a vector quantity whose magnitude for a coil of N turns of area A is given by NIA, where I is the current in the coil. The direction of the magnetic moment is perpendicular to the plane of the coil in the direction given by the right hand thumb rule (see Fig. 29-19). If such a coil is placed into a uniform B field, show that the torque has a magnitude given by: τ = µB sin θ This is sometimes written as a vector (cross) product τ = µ x B. For a coil or loop: τ = NBIA cos α [ cos α = sin θ ] Recall that the angle θ is the compliment of the angle α µ that the plane of the loop makes with the B field, thus the θ B α magnetic moment is perpendicular to the plane of the loop. Simply substitute µ = NIA into τ = (NIA)B sin θ, then write: τ = µB sin θ . 171
  10. 10. Chapter 30. Forces and Torques in a Magnetic Field Physics, 6th Edition *30-32. Verify the answer obtained for Problem 30-19 by applying the formula derived in the previous problem. Do not confuse the angle θ with the angle α that the plane of the loop makes with the B field. [ Recall that θ + α = 900 and cos α = sin θ ] Thus angles change from: α = 00, 300, 450, 600, and 900 to θ = 900, 600, 450, 300, and 00 The maximum torque is still 250 N⋅m, from τ = NBIA cos α or from τ = NBIA sin θ: µ = NIA = (500)(10.0 A)(0.20 m2) = 1000 A⋅m2 directed at angle θ Applying τ = µB sinθ = (1000 A⋅m2)(0.25 T) sin θ for each angle gives: (a) 250 N⋅m; (b) 217 N⋅m; (c) 177 N⋅m; (d) 125 N⋅m; and (e) 0 N⋅m *30-33. The internal wiring of a three-scale voltmeter is shown in Fig. 30-12. The galvanometer has an internal resistance of 40 Ω and a current of 1.00 mA will produce full-scale deflection. Find the resistances R1, R2, and R3 for the use of the voltmeter for 10, 50, and 100 V. (Multiplier resistances Rm) R2 R3 Rm = VR − Rg = 10 V − 40 Ω; R1 Ig 0.001A Rm = 9960 Ω Rg 50 V V 50 V G 10 V 100 V Rm = R − Rg = − 40 Ω; Rm = 49,960 Ω Ig 0.001A VR 100 V Rm = − Rg = − 40 Ω; Rm = 99,960 Ω Ig 0.001A 172

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