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Unformatted text preview: song (shs546) – Homework 11 – weathers – (22202) 1 This printout should have 12 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A negatively charged particle moving parallel to the zaxis enters a magnetic field (pointing out of of the page), as shown in the figure below. z x v y vector B vector B − q Figure: ˆ ı is in the xdirection, ˆ is in the ydirection, and ˆ k is in the zdirection. What is the initial direction of deflection? 1. hatwide F = − ˆ 2. hatwide F = − ˆ ı correct 3. hatwide F = +ˆ ı 4. hatwide F = + ˆ k 5. vector F = 0 ; no deflection 6. hatwide F = − ˆ k 7. hatwide F = +ˆ Explanation: Basic Concepts: Magnetic Force on a Charged Particle: vector F = qvectorv × vector B Righthand rule for crossproducts. 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 (+ˆ ) , vectorv = v parenleftBig − ˆ k parenrightBig , and q < , therefore , vector F = − q  vectorv × vector B = − q  v B bracketleftBigparenleftBig − ˆ k parenrightBig × (+ˆ ) bracketrightBig = − q  v B (+ˆ ı ) hatwide F = − ˆ ı . This is the eighth of eight versions of the problem. 002 10.0 points A proton moving at 1 . 7 × 10 6 m / s through a magnetic field of 7 . 1 T experiences a magnetic force of magnitude 1 . 5 × 10 − 12 N. The charge of proton is 1 . 60218 × 10 − 19 C and the mass of proton is 1 . 67262 × 10 − 27 kg. What is the angle between the proton’s velocity and the field? Correct answer: 50 . 8652 ◦ . Explanation: Let : E = 1 . 5 × 10 − 12 N , B = 7 . 1 T , v = 1 . 7 × 10 6 m / s , q p = 1 . 60218 × 10 − 19 C , and m p = 1 . 67262 × 10 − 27 kg . The Lorentz force acting on a moving charged particle in a magnetic field is F = q v B sin θ , where q is the charge, v is the speed, B is the magnetic field. So thus the angle between the velocity and the field is θ = arcsin bracketleftbigg F q v B bracketrightbigg = arcsin bracketleftbigg (1 . 5 × 10 − 12 N) (1 . 60218 × 10 − 19 C) × 1 (1 . 7 × 10 6 m / s) (7 . 1 T) bracketrightbigg = 50 . 8652 ◦ . song (shs546) – Homework 11 – weathers – (22202) 2 003 (part 1 of 3) 10.0 points The magnetic field over a certain range is given by vector B = B x ˆ ı...
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 Spring '00
 Littler
 Charge, Work

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