Final_Exam_F_06_Solutions

# Final_Exam_F_06_Solu - PHYSICS 222 FINAL EXAM SOLUTIONS Problems 61-80 are worth 4 points each 61 FALL 2006 An electron of mass 9.11 10-31 kg and

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PHYSICS 222 FALL 2006 FINAL EXAM SOLUTIONS DECEMBER 11, 2006 Problems 61-80 are worth 4 points each 61. An electron of mass 9.11 × 10 31 kg and charge 1.60 × 10 19 C is moving with speed 3.20 × 10 6 m/s in a circle of radius 0.182 mm in vacuum due to a uniform magnetic field directed perpendicular to the plane of the orbit. The magnitude of the magnetic field is ____ T. A. 0.1 B. 0.2 C. 0.5 D. 3 E. 9 B = mv |q|R = (9.11 × 10 31 kg)(3.20 × 10 6 m/s) (1.60 × 10 19 C)(1.82 × 10 4 m) = 0.100 T . 62. A long thin wire carries a current of 9.55 A and is bent into two long straight sections and a circular arc of radius 1.00 cm as shown. All three sections lie in the x-y plane. Point P is at the center of the arc and the angle subtended by the arc at point P is ( π /3) rad. The magnetic field at point P is ____ . -1.0 -0.5 0.0 0.5 1.0 1.5 2.0 2.5 3.0 -2.0 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0 y (cm) x (cm) current P A. (100 µ T) k ˆ B. (100 µ T) k ˆ C. (5 µ T) k ˆ D. (5 µ T) k ˆ E. (50 µ T) i ˆ From the Biot-Savart law, the straight wire sections do not produce a magnetic field at point P. Using the circular right-hand rule, the direction of the magnetic field at P due to the circular current arc is in the k ˆ direction, which is pointed into the page.

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The magnitude of the field at P due to the circular arc is B = µ o I φ 4 π R = (4 π × 10 7 T·m/A)(9.55 A)[( π /3)rad] 4 π (0.0100 m) = 1.00 × 10 4 T = 100 µ T . 63. The ac current through a 2.00 µ F capacitor has an angular frequency of 5000 rad/s and an amplitude of 3.00 A. The amplitude of the ac voltage across the capacitor is ____ V. A. 50 B. 100 C. 200 D. 300 E. 400 X C = 1 ω C = 1 (5000 rad/s)(2.00 × 10 6 F) = 100 . V C = I X C = (3.00 A)(100 ) = 300 V . 64. A train is traveling along a straight track in the +x direction at 50 m/s. A car is traveling at 30 m/s in the +x direction towards the train on the road next to the track (the train is in front of the car). The train emits a whistle at 540 Hz. The frequency heard by the driver of the car is ____ Hz. (The speed of sound in air is 340 m/s) A. 512 B. 526 C. 540 D. 554 E. 568 The positive direction for the speeds of the source S and listener L is the direction from the listener to the source, which is the +x direction. Thus v L = +30 m/s and v S = +50 m/s. Then f L = f S v + v L v+ v S = (540 Hz) 340 m/s + 30 m/s 340 m/s + 50 m/s = 512 Hz . 65. A monochromatic light beam is incident from glass into air. The wavelength of the light beam in the glass is 400 nm. The wavelength of the light beam after passing into the air is ____ nm. (The index of refraction of the glass is 1.50 and the index of refraction of air is 1.00) A. 270 B. 400 C. 440 D. 530 E. 600 λ 0 = n λ = (1.50)(400 nm) = 600 nm . 66. In a two-slit interference experiment, monochromatic light of wavelength 600 nm is normally incident on the two slits. A point P is situated at the second order intensity maximum on the screen. If the distance from point P to slit 1 is r 1 and the distance from point P to slit 2 is r 2 , then |r 1 r 2 | = ___ nm.
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## This note was uploaded on 02/11/2012 for the course PHYSICS 222 taught by Professor Ogilvie during the Fall '05 term at Iowa State.

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Final_Exam_F_06_Solu - PHYSICS 222 FINAL EXAM SOLUTIONS Problems 61-80 are worth 4 points each 61 FALL 2006 An electron of mass 9.11 10-31 kg and

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