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Unformatted text preview: Exam 2 Practice Problems The exam problems will be similar to the book problems with the addition that you may be asked to explain your answer. Keeping with this spirit, the practice problems are taken from the Additional Problems section at the end of each chapter. When solving these practice problems and during the exam, you should always draw a diagram or sketch to show your ideas. Make sure drawings are consistent with the rest of your work. Chapter 21 Problems 7277, 7982, 83 Chapter 22 Problems 70, 7273 Chapter 24 Problem 49 Chapter 25 Problems 3642 Chapter 26 Problems 105, 107108, 110111, 113 Exam 2 Practice Problems Solutions Chapter 21 72. REASONING The magnitude B of the magnetic field is given by sin F B q v = (Equation 21.1), and we will apply this expression directly to obtain B . SOLUTION The charge 6 8.3 10 C q =  travels with a speed v = 7.4 10 6 m/s at an angle of = 52 with respect to a magnetic field of magnitude B and experiences a force of magnitude F = 5.4 10 3 N. According to Equation 21.1, the field magnitude is ( 29 3 4 6 6 5.4 10 N 1.1 10 T sin 8.3 10 C 7.4 10 m/s sin 52 F B q v  = = =  Note in particular that it is only the magnitude q of the charge that appears in this calculation. The algebraic sign of the charge does not affect the result. 74. REASONING The angle between the electrons velocity and the magnetic field can be found from Equation 21.1, sin F q vB = According to Newtons second law, the magnitude F of the force is equal to the product of the electrons mass m and the magnitude a of its acceleration, F = ma . SOLUTION The angle is ( 29 ( 29 ( 29( 29( 29 31 14 2 1 1 19 6 4 9.11 10 kg 3.50 10 m/s sin sin 19.7 1.60 10 C 6.80 10 m/s 8.70 10 T ma q vB  = = = 75. SSM REASONING The magnitude of the torque that acts on a currentcarrying coil placed in a magnetic field is specified by = NIAB sin (Equation 21.4), where N is the number of loops in the coil, I is the current, A is the area of one loop, B is the magnitude of the magnetic field, and is the angle between the normal to the coil and the magnetic field. All the variables in this relation are known except for the current, which can, therefore, be obtained. SOLUTION Solving the equation = NIAB sin for the current I and noting that = 90.0 since is specified to be the maximum torque, we have 2 MAGNETIC FORCES AND MAGNETIC FIELDS ( 29 ( 29 ( 29 2 2 5.8 N m 2.2 A sin 1200 1.1 10 m 0.20 T sin 90.0 I NAB  = = = 76. REASONING The magnitude of the magnetic force acting on the particle is F = q vB sin (Equation 21.1), where q and v are the charge magnitude and speed of the particle, respectively, B is the magnitude of the magnetic field, and is the angle between the particles velocity and the magnetic field. The magnetic field is produced by a very long, straight wire, so its value is given by Equation 21.5 as very long, straight wire, so its value is given by Equation 21....
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This note was uploaded on 10/04/2009 for the course PHYS 750:204 taught by Professor Rollino during the Spring '09 term at Rutgers.
 Spring '09
 rollino
 Physics

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