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Homework 11 Solutions

# Homework 11 Solutions - lim(kl9356 Homework 11...

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lim (kl9356) – 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 x -axis enters a magnetic field (point- ing out of of the page), as shown in the figure below. x y v z 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 3. hatwide F = ˆ k 4. hatwide F = +ˆ ı 5. hatwide F = ˆ correct 6. vector F = 0 ; no deflection 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 ( ˆ ı ) , and q < 0 , therefore , vector F = −| q | vectorv × vector B = −| q | v B bracketleftBig ( ˆ ı ) × parenleftBig + ˆ k parenrightBigbracketrightBig = −| q | v B ( ˆ ) hatwide F = ˆ . This is the eighth of eight versions of the problem. 002 10.0 points A proton moving at 3 . 4 × 10 6 m / s through a magnetic field of 6 . 6 T experiences a magnetic force of magnitude 2 . 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: 44 . 0555 . Explanation: Let : E = 2 . 5 × 10 12 N , B = 6 . 6 T , v = 3 . 4 × 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 (2 . 5 × 10 12 N) (1 . 60218 × 10 19 C) × 1 (3 . 4 × 10 6 m / s) (6 . 6 T) bracketrightbigg = 44 . 0555 .

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lim (kl9356) – 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 + B y ˆ , where B x = 2 T and B y = 2 T. An electron moves into the field with a velocity vectorv = v x ˆ ı + v y ˆ + v z ˆ k , where v x = 5 m / s, v y = 4 m / s and v
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Homework 11 Solutions - lim(kl9356 Homework 11...

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