hw12b - aljabr (faa335) Hw12 Ross (89251) 1 This print-out...

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Unformatted text preview: aljabr (faa335) Hw12 Ross (89251) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 3) 10.0 points The magnetic field over a certain range is given by vector B = B x + B y , where B x = 9 T and B y = 3 T. An electron moves into the field with a velocity vectorv = v x + v y + v z k , where v x = 8 m / s, v y = 2 m / s and v z = 2 m / s. The charge on the electron is 1 . 602 10 19 C. What is the component of the force ex- erted on the electron by the magnetic field? Correct answer: 9 . 612 10 19 N. Explanation: Let : v x = 8 m / s , v y = 2 m / s , v z = 2 m / s , q = 1 . 602 10 19 C , B x = 9 T , and B y = 3 T . The force exerted on the electron by the mag- netic field is given by vector F = qvectorv vector B = q vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle k v x v y v z B x B y vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle = q [ v z B y + v z B x + ( v x B y v y B x ) k ] = ( 1 . 602 10 19 C) { (2 m / s) (3 T) + (2 m / s) (9 T) + [(8 m / s) (3 T) (2 m / s) (9 T)] k } = (9 . 612 10 19 N) + ( 2 . 8836 10 18 N) + ( 9 . 612 10 19 N) k 002 (part 2 of 3) 10.0 points What is the component of the force? Correct answer: 2 . 8836 10 18 N. Explanation: See the above part. 003 (part 3 of 3) 10.0 points What is the k component of the force? Correct answer: 9 . 612 10 19 N. Explanation: See the above part. 004 10.0 points A positively charged particle moving paral- lel to the y-axis enters a magnetic field (point- ing into of the page), as shown in the figure below. y z v x vector B vector B + q Figure: is in the x-direction, is in the y-direction, and k is in the z-direction. What is the initial direction of deflection? 1. hatwide F = + 2. hatwide F = 3. hatwide F = + k correct 4. hatwide F = + 5. vector F = 0 ; no deflection 6. hatwide F = k 7. hatwide F = Explanation: Basic Concepts: Magnetic Force on a Charged Particle: vector F = qvectorv vector B Right-hand rule for cross-products. aljabr (faa335) Hw12 Ross (89251) 2 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 ( ) , vectorv = v (+ ) , and q > , therefore , vector F = + | q | vectorv vector B = + | q | v B [(+ ) ( )] = + | q | v B parenleftBig + k parenrightBig hatwide F = + k ....
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This note was uploaded on 01/27/2011 for the course PHYS 251 taught by Professor Gavrine during the Spring '10 term at Purdue University-West Lafayette.

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hw12b - aljabr (faa335) Hw12 Ross (89251) 1 This print-out...

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