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# HW7 - Version One Homework 7 Savrasov 39819 This print-out...

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Version One – Homework 7 – Savrasov – 39819 – May 15, 2006 1 This print-out should have 8 questions. Multiple-choice 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 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 0 . 1 T at right angles to the direction of the electron’s motion. The mass of the electron is 9 . 11 × 10 31 kg and its charge is 1 . 60218 × 10 19 C. What is the magnitude of the force on the electron due to the magnetic field? Correct answer: 9 . 96593 × 10 13 N. Explanation: Let : V = 11000 V , B = 0 . 1 T , m = 9 . 11 × 10 31 kg , q e = 1 . 60218 × 10 19 C . The kinetic energy K gained after acceler- ation is K = 1 2 m v 2 = q e V , so the velocity is v = 2 q e V m = 2 (1 . 60218 × 10 19 C)(11000 V) 9 . 11 × 10 31 kg = 6 . 22024 × 10 7 m / s . Then the force on it is f = q v B = (1 . 60218 × 10 19 C) × (6 . 22024 × 10 7 m / s) (0 . 1 T) = 9 . 96593 × 10 13 N . Force on an Electron 29:01, calculus, numeric, > 1 min, wording- variable. 002 (part 1 of 2) 1 points An electron is projected into a uniform mag- netic field given by B = B z ˆ k + B x ˆ ı , where B z = 3 . 7 T and B x = 1 . 4 T. The magnitude of the charge on an electron is 1 . 60218 × 10 19 C . z x y v = 370000 m / s electron 3 . 7 T 1 . 4 T B Find the direction of the magnetic force when the velocity of the electron is v ˆ ı , where v = 370000 m / s.

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