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Unformatted text preview: dzifanu (sd26397) hw 9 opyrchal (121102) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 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 . 697 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: 6 . 94626 10 12 N. Explanation: Let : V = 11000 V , B = 0 . 697 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)(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 . 697 T) = 6 . 94626 10 12 N . 002 10.0 points A negatively charged particle moving paral- lel to the z-axis enters a magnetic field (point- ing toward the right-hand side of the page), as shown in the figure below. z x v y vector B vector B q Figure: is in the x-direction, is in the y-direction, and k is in the z-direction. What is the initial direction of deflection? 1. hatwide F = 2. hatwide F = + 3. hatwide F = 4. vector F = 0 ; no deflection correct 5. hatwide F = + k 6. hatwide F = k 7. hatwide F = + Explanation: Basic Concepts: Magnetic Force on a Charged Particle: vector F = qvectorv vector B Right-hand rule for cross-products. hatwide F vector F bardbl vector F bardbl ; i.e. , a unit vector in the F direc- tion. Solution: The force is vector F = qvectorv vector B . vector B = B parenleftBig + k parenrightBig , vectorv = v parenleftBig k parenrightBig , and q < , therefore , vector F = | q | vectorv vector B = | q | v B bracketleftBigparenleftBig k parenrightBig parenleftBig + k parenrightBigbracketrightBig = | q | v B (0) hatwide F = 0 no deflection . dzifanu (sd26397) hw 9 opyrchal (121102) 2 This is the seventh of eight versions of the problem. 003 (part 1 of 2) 10.0 points A proton moves perpendicularly to a uniform magnetic field, B , with a speed of 5 . 4 10 7 m/s and experiences an acceleration of 5 . 1 10 13 m / s 2 in the positive x direction when its velocity is in the positive z direction. Find the magnitude of the field. Correct answer: 0 . 00987535 T....
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This note was uploaded on 09/16/2011 for the course PHYS PHYS 121 taught by Professor Prodan during the Spring '09 term at NJIT.
- Spring '09