Homework 8-solutions

# Homework 8-solutions - kumar(vk4433 Homework 8...

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kumar (vk4433) – Homework 8 – markert – (55325) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points A positively charged particle moving at 45 angles to both the x -axis and y -axis enters a magnetic field (pointing out of of the page), as shown. ˆ ı is in the x -direction, ˆ is in the y -direction, and ˆ k is in the z -direction. x y v × z vector B vector B + q What is the initial direction of deflection? 1. hatwide F = 1 2 parenleftBig ˆ k ı parenrightBig 2. hatwide F = 1 2 ( ˆ ı ) 3. hatwide F = 1 2 parenleftBig ˆ k parenrightBig 4. hatwide F = 1 2 parenleftBig ˆ + ˆ k parenrightBig 5. hatwide F = 1 2 parenleftBig + ˆ k ˆ ı parenrightBig 6. hatwide F = 1 2 parenleftBig ˆ k ˆ ı parenrightBig 7. hatwide F = 1 2 ( ˆ ˆ ı ) 8. hatwide F = 1 2 (+ˆ ˆ ı ) correct 9. hatwide F = 1 2 parenleftBig ˆ ˆ k parenrightBig 10. vector F = 0 ; no deflection Explanation: The force is vector F = qvectorv × vector B , vector B = B ( ˆ ı ) , vectorv = 1 2 v (+ˆ ı + ˆ ) , and q > 0 , so vector F = + | q | vectorv × vector B = + | q | 1 2 v B bracketleftBig (+ˆ ı + ˆ ) × parenleftBig ˆ k parenrightBigbracketrightBig = + | q | 1 2 v B (+ˆ ˆ ı ) hatwide F = 1 2 (+ˆ ˆ ı ) . 002 10.0points What does Gauss’ law for magnetism tell us? 1. The net charge in any given volume. 2. That charges must be moving to produce magnetic fields. 3. The magnetic field of a current element. 4. That the line integral of a magnetic field around any closed loop vanishes. 5. That magnetic monopoles do not exist. correct Explanation: Gauss’ law for magnetism tells us that inte- grating time-independent magnetic field over any closed surface gives zero. Therefore there can’t be any magnetic charges or monopoles; i.e. , no sources nor sinks of magnetic field lines. 003(part1of2)10.0points A proton travels with a speed of 4 . 02 × 10 6 m / s at an angle of 37 . 5 with a magnetic field of 0 . 316 T pointed in the y direction. The charge of proton is 1 . 60218 × 10 19 C. What is the magnitude of the magnetic force on the proton? Correct answer: 1 . 239 × 10 13 N. Explanation: Let : v = 4 . 02 × 10 6 m / s , θ = 37 . 5 , and B = 0 . 316 T .

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kumar (vk4433) – Homework 8 – markert – (55325) 2 F = q v B sin θ = (1 . 60218 × 10 19 C)(4 . 02 × 10 6 m / s) × (0 . 316 T) sin 37 . 5 = 1 . 239 × 10 13 N. 004(part2of2)10.0points The mass of proton is 1 . 67262 × 10 27 kg What is the proton’s acceleration? Correct answer: 7 . 40752 × 10 13 m / s 2 . Explanation: Let : m = 1 . 67262 × 10 27 kg , F = m a . a = F m = 1 . 239 × 10 13 N 1 . 67262 × 10 27 kg = 7 . 40752 × 10 13 m / s 2 . 005 10.0points Consider a straight wire carrying current I , as shown. R I What is the direction of the magnetic field at point R caused by the current I in the wire?
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