F05FinalExamSolutions

# F05FinalExamSolutions - PHYSICS 222 FINAL EXAM SOLUTIONS...

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PHYSICS 222 FALL 2005 FINAL EXAM SOLUTIONS DECEMBER 13, 2005 Problems 61-80 are worth 4 points each 61. A current-carrying circular coil with a magnetic moment µ is in a uniform magnetic field B . The maximum magnitude of the torque on the coil in the magnetic field occurs when the angle between µ and B is ______ degrees. A. 0 B. 45 C. 90 D. 135 E. 180 τ = µ B sin φ , so τ is maximum when φ = 90˚. 62. A circular one-turn wire loop with a radius of 1.00 cm is in a uniform magnetic field pointing perpendicular to the surface of the loop. The magnitude of the magnetic field is B = (0.400 T/s)t, where t is the time in seconds. The magnitude of the induced EMF in the loop is ____ mV. A. 0.13 B. 0.28 C. 0.56 D. 0.73 E. 0.95 |EMF| = N d Φ B dt = NA dB dt = (1)[ π (0.0100 m) 2 ](0.400 T/s) = 0.126 mV . 63. A series RLC circuit consists of a 10.0 µ F capacitor, a 5.00 mH inductor, and a 30.0 resistor. The circuit is driven by a sinusoidal voltage source with an amplitude of 10.0 V and an angular frequency of 2000 rad/s. The average power dissipated by the circuit is ____ W. A. 0.5 B. 0.6 C. 0.7 D. 0.8 E. 1.0 X C = 1 ω C = 1 (2.00 × 10 3 rad/s)(1.00 × 10 5 F) = 50.0 . X L = ω L = (2.00 × 10 3 rad/s)(5.00 × 10 3 H) = 10.0 . Z =  (X L X C ) 2 + R 2 =  (10 50 ) 2 + (30 ) 2 = 50 . I rms = V 2 Z = 10.0 V 2(50 ) = 0.141 A. P ave = I rms 2 R = (0.141 A) 2 (30.0 ) = 0.600 W . 64. A transverse wave on a stretched string has the wave function y(x,t) = (0.0100 m) sin[(5.00 rad/m)x] sin[(200 rad/s)t]. At time t = 15.7 ms, the transverse speed of a particle in the string at position x = 0.942 m is ____ m/s. A. 0.0 B. 1.0 C. 1.5 D. 2.0 E. 2.5 v y (x,t) = y(x,t) t = (0.0100 m)(200/s) sin[(5.00 rad/m)x] cos[(200 rad/s)t] = (2.00 m/s) sin[(5.00 rad/m)(0.942 m)] cos[(200 rad/s)(0.0157 s)] = 2.00 m/s .

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65. A laser emits 1.00 mW of average light power in a cylindrical beam of radius 5.00 mm. The intensity of the light at a point in the beam is ____ W/m 2 . A. 5 B. 7 C. 9 D. 11 E. 13 I = Average Power Area = 1.00 × 10 3 W π (5.00 × 10 3 m) 2 = 12.7 W/m 2 . 66. A glass converging lens has one flat side and one side with a radius of curvature of 20.0 cm. The focal length of the lens is ____ cm. (The index of refraction of the glass in the lens is 1.50) A. 10 B. 20 C. 30 D. 40 E. 50 1 f = (n 1) 1 R 1 1 R 2 = (1.50 1) 1 20.0 cm 1 = 1 40.00 cm , so f = 40.0 cm . 67. Light of wavelength 650 nm is normally incident on two narrow slits in a two-slit interference experiment. If the two slits are 2.50 µ m apart, the second-order interference intensity maximum of the light is at an angle of ____ degrees with respect to the direction of the incident light. A. 31 B. 42 C. 53 D. 64 E. 75 sin θ = m i λ d = 2(6.50 × 10 7 m) 2.50 × 10 6 m = 0.520, so θ = 31.3˚ . 68. In order for a telescope on the Earth to resolve two points on the Moon that are 500 m apart using light of wavelength 530 nm, the diameter of the objective lens in the telescope has to be at least ____ m. (The distance from the Earth to the Moon is 3.84 × 10 8 m) A. 0.2 B. 0.5 C. 0.8 D. 1.1 E. 1.4 D = 1.22 λ R y 1 = 1.22 (5.30 × 10 7 m)(3.84 × 10 8 m) 500 m = 0.497 m .
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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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F05FinalExamSolutions - PHYSICS 222 FINAL EXAM SOLUTIONS...

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