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Spring 11 Phys 121 - HW 09

# Spring 11 Phys 121 - HW 09 - dzifanu(sd26397 hw 9...

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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 10.0points 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 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: 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 10.0points 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: ˆ ı is in the x -direction, ˆ is in the y -direction, and ˆ k is in the z -direction. What is the initial direction of deflection? 1. hatwide F = ˆ ı 2. hatwide F = +ˆ 3. hatwide F = ˆ 4. vector F = 0 ; no deflection correct 5. hatwide F = + ˆ k 6. hatwide F = ˆ k 7. hatwide F = +ˆ ı 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 + ˆ k parenrightBig , vectorv = v parenleftBig ˆ k parenrightBig , and q < 0 , therefore , vector F = −| q | vectorv × vector B = −| q | v B bracketleftBigparenleftBig ˆ k parenrightBig × parenleftBig + ˆ k parenrightBigbracketrightBig = −| q | v B (0) hatwide F = 0 no deflection .

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dzifanu (sd26397) – hw 9 – opyrchal – (121102) 2 This is the seventh of eight versions of the problem. 003(part1of2)10.0points 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. Explanation: Let: q = 1 . 60 × 10 19 C , a = 5 . 1 × 10 13 + x , v = 5 . 4 × 10 7 + z , and m = 1 . 673 × 10 27 kg . F net = F mag m a = q v B B = m a q v = (1 . 673 × 10 27 kg) (5 . 1 × 10 13 m / s 2 ) (1 . 6 × 10 19 C) (5 . 4 × 10 7 m / s) = 0 . 00987535 T . 004(part2of2)10.0points What is its direction? 1. None of these 2. Positive z direction 3. Positive y direction 4. Negative y direction correct 5. Positive x direction 6. Negative x direction 7. Negative z direction Explanation: Right-hand rule (positive charge): Force directed out of the palm of the hand, fingers in the direction of the field, thumb in the direction of the velocity.
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Spring 11 Phys 121 - HW 09 - dzifanu(sd26397 hw 9...

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