s02x2aan - x = 0. Therefore, the displacements on the...

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NAME PHYS-1200 PHYSICS II QUIZ 2 MARCH 20, 2002 SOLUTIONS PART A. 1. D 7. D 2. D 8. A 3. B 4. B 5. C 6. D PART B . 1. a) OUT OF THE PAGE. The direction of the dispacement current is the direction of increasing electric field. b) A MAGNETIC FIELD. Maxwell’s addition to Ampere’s law says that a time varying electric field produces a magnetic field. 2. a) THERE IS NO DISPLACEMENT CURRENT WITHIN THE CIRCULAR REGION. A displacement current is the result of a time varying electric field. This problem has a time varying magnetic field. b) AN ELECTRIC FIELD. Faraday’s law says that a time varying magnetic field produces an electric field. 3. a) The wavelength can be read directly from the graph on the left. λ = 3.0 m b) v = /T. is known from part a), and from the graph on the right, T = 0.60 s. Then, v = (3.0 m)/(0.60 s) v = 5.0 m/s c) NEGATIVE x DIRECTION. The graph on the right shows that the displacement at x = 0 becomes negative as time increases from t = 0. The graph on the left shows that the negative values of displacement are on the positive side of
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Unformatted text preview: x = 0. Therefore, the displacements on the positive side of x = 0 must move toward x = 0. The wave moves in the negative x direction. NAME PART C . 1. a) N B = Li , so 5 A) H)(1.5 10 . 4 ( 6- = = N Li B B = 1.2 10-6 Wb b) 2 6 2 1 2 2 1 A) H)(1.5 10 . 4 (- = = Li U B U B = 4.5 10-6 J = 4.5 J c) The magnitude is given by, A/s) H)(6.0 10 . 4 ( 6- = = dt di L E . E = 2.4 10-5 V = 24 V d) FROM RIGHT TO LEFT. It opposes the increase of current. 2. a) k m T 2 = , so k m T 2 2 4 = , and 2 2 2 2 4 N/m) 80 ( s) 0010 . ( 4 = = k T m m = 2.0 10-6 kg = 2.0 10-3 g = 2.0 mg b) 2 2 1 m kx E = , so k E x m 2 2 = , and N/m 80 ) J 10 . 4 ( 2 2 9- = = k E x m x m = 1.0 10-5 m = 10 m c) LC T 2 = , so LC T 2 2 4 = , and F) 10 . 1 ( 4 s) 0010 . ( 4 6 2 2 2 2- = = C T L L = 2.5 10-2 H = 25 mH d) C Q U 2 2 = , so CU Q 2 2 = , and ) J 10 . 4 )( F 10 . 1 ( 2 2 9 6-- = = CU Q Q = 8.9 10-8 C...
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s02x2aan - x = 0. Therefore, the displacements on the...

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