Practice HW 11 solution

Practice HW 11 solution - song (shs546) Practice HW 11...

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Unformatted text preview: song (shs546) Practice HW 11 Solutions weathers (22202) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 0.0 points A positively charged particle passes through a laboratory traveling in an easterly direction. There are both electric and magnetic field in the room and their effects on the charged particle cancel. If the electric field points upward, what must be the direction of the magnetic field? 1. east 2. west 3. south correct 4. north 5. upward 6. downward Explanation: The force on the object due to the magnetic field must point down in order to cancel the force due to the electric field. Using the right- hand rule, the magnetic field must point south in order for the magnetic force to be directed downward. 002 0.0 points An electron in a vacuum is first accelerated by a voltage of 81700 V and then enters a region in which there is a uniform magnetic field of 0 . 444 T at right angles to the direction of the electrons 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: 1 . 20591 10 11 N. Explanation: Let : V = 81700 V , B = 0 . 444 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 = radicalbigg 2 q e V m = radicalBigg 2 (1 . 60218 10 19 C)(81700 V) 9 . 11 10 31 kg = 1 . 6952 10 8 m / s . Then the force on it is f = q v B = (1 . 60218 10 19 C) (1 . 6952 10 8 m / s) (0 . 444 T) = 1 . 20591 10 11 N . 003 (part 1 of 2) 0.0 points An electron is projected into a uniform mag- netic field given by vector B = B z k + B x , where B z = 4 . 9 T and B x = 1 . 7 T. The magnitude of the charge on an electron is 1 . 60218 10 19 C . z x y v = 3 . 9 10 5 m / s electron 4 . 9 T 1 . 7 T B Find the direction of the magnetic force when the velocity of the electron is v , where v = 3 . 9 10 5 m / s. 1. hatwide F = correct 2. hatwide F = song (shs546) Practice HW 11 Solutions weathers (22202) 2 3. hatwide F = 4. hatwide F = 5. hatwide F = k 6. hatwide F = k 7. hatwide F = 1 2 parenleftBig k + parenrightBig 8. hatwide F = 1 2 parenleftBig k parenrightBig 9. hatwide F = 1 2 parenleftBig k parenrightBig Explanation: Let : q = 1 . 60218 10 19 C , B z = 4 . 9 T , and B x = 1 . 7 T . Basic Concepts: Magnetic force on a mov- ing charge is given by vector F = qvectorv vector B ....
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This note was uploaded on 04/22/2011 for the course PHYS 2220 taught by Professor Littler during the Spring '00 term at North Texas.

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Practice HW 11 solution - song (shs546) Practice HW 11...

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