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Unformatted text preview: Version One – Homework 7 – Savrasov – 39821 – May 14, 2007 1 This printout should have 9 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. The due time is Central time. Electron in a Magnetic Field 02 29:01, trigonometry, numeric, > 1 min, nor mal. 001 (part 1 of 1) 1 points An electron in a vacuum is first accelerated by a voltage of 11000 V and then enters a region in which there is a uniform magnetic field of . 1 T at right angles to the direction of the electron’s motion. What is the force on the electron due to the magnetic field? Correct answer: 9 . 96627 × 10 13 N. Explanation: Let : V = 6 . 22045 × 10 7 m / s and B = 0 . 1 T . The kinetic energy gained after acceleration is KE = 1 2 m e v 2 = q e V , so the velocity is v = r 2 q e V m = s 2(1 . 60218 × 10 19 C)(11000 V) 9 . 10939 × 10 31 kg = 6 . 22045 × 10 7 m / s . Then the force on it is f = qvB = (1 . 60218 × 10 19 C) × (6 . 22045 × 10 7 m / s)(0 . 1 T) = 9 . 96627 × 10 13 N . keywords: Magnetic Force on a Cable 29:03, trigonometry, numeric, > 1 min, nor mal. 002 (part 1 of 2) 1 points A long electric cable is suspended above the earth and carries a 50 A current parallel to the surface of the earth heading south. The earth’s magnetic field has a value of 5 . 6 × 10 5 T and makes an angle of θ =40 ◦ with respect to the cable. Earth N S W E B I θ cable What is the magnitude of the force exerted ona50mlengthofthecableduetotheearth’s magnetic field? Correct answer: 0 . 0899903 N. Explanation: Let : I = 50 A , B = 5 . 6 × 10 5 T , θ = 40 ◦ , and L = 50 m ....
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 Spring '10
 Dr.AndyGavrin
 Magnetic Field, 0.1 m, 2 m, 1 min, 0.23 m

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