Physics Solutions Manual

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Chapter 24 continued 87. A force of 5.7810 16 N acts on an unknown particle traveling at a 90° angle through a magnetic field. If the velocity of the particle is 5.6510 4 m/s and the field is 3.2010 2 T, how many elementary charges does the particle carry? F qvB F q Bv 3.2010 19 C N (3.201019 1charge C) 1.601019 C 2 charges Mixed Review pages 667–668 Level 2 88. A copper wire of insignificant resistance is placed in the center of an air gap between two magnetic poles, as shown in Figure 24-34. The field is confined to the gap and has a strength of 1.9 T. 7.5 cm ■ Figure 24-34 a. Determine the force on the wire (direction and magnitude) when the switch is open. 0 N. With no current, there is no magnetic field produced by the wire and copper is not a magnetic material. b. Determine the force on the wire (direction and magnitude) when the switch is closed. Up, 0.62 N. The direction of the force is given by the third right-hand rule. I V R F ILB 5.7810 16 N (3.2010 2 T)(5.6510 4 m/s) N S VLB R 5.5 Ω 24 V 0.62 N c. Determine the force on the wire (direction and magnitude) when the switch is closed and the battery is reversed. Down, 0.62 N. The direction of the force is given by the third right-hand rule and the magnitude of the force is the same as in part b. d. Determine the force on the wire (direction and magnitude) when the switch is closed and the wire is replaced with a different piece having a resistance of 5.5 . Up, 0.31 N. The direction of the force is given by the third right-hand rule. I F ILB (24 V)(0.075 m)(1.9 T) 5.5 V R (24 V)(0.075 m)(1.9 T) 5.5 5.5 0.31 N 89. Two galvanometers are available. One has 50.0-A full-scale sensitivity and the other has 500.0-A full-scale sensitivity. Both have the same coil resistance of 855 . Your challenge is to convert them to measure a current of 100.0 mA, full-scale. a. Determine the shunt resistor for the 50.0-A meter. Find the voltage across the meter coil at full scale. V IR (50.0 A)(855 ) 0.0428 V Calculate the shunt resistor. V R I 0.428 VLB R 0.0428 V 100.0 mA 50.0 A b. Determine the shunt resistor for the 500.0-A meter. Find the voltage across the meter coil at full scale. V IR (500.0 A)(855 ) 0.428 V 496 Solutions Manual Physics: Principles and Problems Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc. Chapter 24 continued Calculate the shunt resistor. V R I 4.30 c. Determine which of the two is better for actual use. Explain. The 50.0-A meter is better. Its much lower shunt resistance will do less to alter the total resistance of the circuit being measured. An ideal ammeter has a resistance of 0 . 90. Subatomic Particle A beta particle (highspeed electron) is traveling at right angles to a 0.60-T magnetic field. It has a speed of 2.510 7 m/s. What size force acts on the particle? F Bqv (0.60 T)(1.610 19 C)(2.510 7 m/s) 2.410 12 N 91. The mass of an electron is 9.1110 31 kg. What is the magnitude of the acceleration of the beta particle described in problem 90? F ma F a m 2.610 18 m/s 2 92. A magnetic field of 16 T acts in a direction due west. An electron is traveling due south at 8.110 5 m/s. What are the magnitude and the direction of the force acting on the electron? F Bqv (16 T)(1.610 19 C)(8.110 5 m/s) 2.110 12 N, upward (right-hand rule—remembering that electron flow is opposite to current flow) 93. Loudspeaker The magnetic field in a loudspeaker is 0.15 T. The wire consists of 250 turns wound on a 2.5-cm-diameter cylindrical form. The resistance of the wire is 8.0 . Find the force exerted on the wire when 15 V is placed across the wire. I V R 0.428 V 100.0 mA 500.0 A 2.410 12 N 9.1110 31 kg L (# of turns)(circumference) nd F BIL F BVnd R Physics: Principles and Problems Solutions Manual 497 5.5 N 94. A wire carrying 15 A of current has a length of 25 cm in a magnetic field of 0.85 T. The force on a current-carrying wire in a uniform magnetic field can be found using the equation F ILB sin . Find the force on the wire when it makes the following angles with the magnetic field lines of a. 90° F ILB sin b. 45° c. 0° (15 A)(0.25 m)(0.85 T)(sin 90°) 3.2 N F ILB sin (15 A)(0.25 m)(0.85 T)(sin 45°) 2.3 N sin 0° 0 so F 0 N 95. An electron is accelerated from rest through a potential difference of 20,000 V, which exists between plates P 1 and P 2 , shown in Figure 24-35. The electron then passes through a small opening into a magnetic field of uniform field strength, B. As indicated, the magnetic field is directed into the page. P 1 Electron (0.15 T)(15 V)(250)()(0.025 m) 8.0 P2 X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X ■ Figure 24-35 a. State the direction of the electric field between the plates as either P 1 to P 2 or P 2 to P 1 . from P 2 to P 1
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