11 - roofner(bar784 – Homework 11 – Weathers –(22202...

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Unformatted text preview: roofner (bar784) – Homework 11 – Weathers – (22202) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A negatively charged particle moving paral- lel to the z-axis enters a magnetic field (point- ing into 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 k 2. hatwide F = − ˆ ı 3. hatwide F = − ˆ k 4. hatwide F = +ˆ ı correct 5. vector F = 0 ; no deflection 6. hatwide F = +ˆ 7. hatwide F = − ˆ 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 ( − 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 seventh of eight versions of the problem. 002 10.0 points A proton moving at 4 . 9 × 10 6 m / s through a magnetic field of 6 . 9 T experiences a magnetic force of magnitude 1 . 3 × 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: 13 . 8858 ◦ . Explanation: Let : E = 1 . 3 × 10 − 12 N , B = 6 . 9 T , v = 4 . 9 × 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 . 3 × 10 − 12 N) (1 . 60218 × 10 − 19 C) × 1 (4 . 9 × 10 6 m / s) (6 . 9 T) bracketrightbigg = 13 . 8858 ◦ . roofner (bar784) – Homework 11 – Weathers – (22202) 2 003 (part 1 of 3) 10.0 points The magnetic field over a certain range is...
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This document was uploaded on 10/26/2011 for the course PHYS 2220 at North Texas.

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11 - roofner(bar784 – Homework 11 – Weathers –(22202...

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