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HW8-Solutions

HW8-Solutions - Version PREVIEW HW8 feng(58230 This...

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Version PREVIEW – HW8 – feng – (58230) 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – Fnd all choices before answering. Holt SF 21Rev 31 001 10.0 points A proton moves eastward in the plane of Earth’s magnetic equator so that its distance from the ground remains constant. The acceleration of gravity is 9 . 81 m / s 2 and the charge on a proton is 1.60 × 10 19 . What is the speed of the proton if Earth’s magnetic Feld points north and has a magni- tude of 5 . 6 × 10 5 T? Correct answer: 0 . 0018313 m / s. Explanation: Let : m = 1 . 673 × 10 27 kg , B = 5 . 6 × 10 5 T , q e = 1 . 60 × 10 19 C , and g = 9 . 81 m / s 2 . The magnetic and gravitational forces are equal: F m = F g q v B = mg v = mg q B = (1 . 67262 × 10 27 kg) (9 . 81 m / s 2 ) (1 . 6 × 10 19 C) (5 . 6 × 10 5 T) = 0 . 0018313 m / s Circular Motion of a Particle 002 (part 1 of 3) 10.0 points Consider the circular motion of a positively charged particle in the plane of this paper, due to a constant magnetic Feld V B which points out of the paper. Neglect the e±ect due to gravity. B B What is the direction of the orbital motion of the particle? 1. clockwise correct 2. Insu²cient information 3. counterclockwise Explanation: To maintain a circular orbit, the magnetic force on the particle must be directed toward the center of the circle. ³rom the right hand rule we see that orbit must be in the clock- wise direction for the force to point radially inwards. 003 (part 2 of 3) 10.0 points What is the radius of the orbit? 1. r = mv q B correct 2. r = mv 2 q B 3. r = q v 2 mB 4. r = q m v B 5. r = q v mB 6. r = mB q v 7. r = v B q m 8. r = q B mv 2 9. r = q B mv

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Version PREVIEW – HW8 – feng – (58230) 2 10. r = v 2 B q m Explanation: From Newton’s second law F = q v B = m v 2 r r = mv q B . 004 (part 3 of 3) 10.0 points Suppose the tangential speed of the charged particle is increased to twice the original tan- gential speed. ( v new = 2 v old ). V B is ±xed. What is the ratio of the new frequency to the old frequency? Note: A unit of frequency is a revolution per second. 1. f new f old = 1 3 2. f new f old = 1 4 3. f new f old = 4 4. f new f old = 1 2 5. f new f old = 1 correct 6. f new f old = 3 7. f new f old = 2 8. f new f old = 1 6 9. f new f old = 5 10. f new f old = 1 5 Explanation: For uniform circular motion, the period ( i.e. , the time to complete one revolution) is de±ned by T = 2 π r v , and from Eq. 1 above, v r = q B m . Frequency is the reciprocal of the period ( i.e. , the number of revolutions per unit time), so f = 1 T = v 2 π r = 1 2 π q B m . Hence, if B is kept ±xed, the frequency of the motion of a charged particle is independent of its tangential speed; i.e. , f new f old = 1 .
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HW8-Solutions - Version PREVIEW HW8 feng(58230 This...

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