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# hw12b - aljabr(faa335 Hw12 Ross(89251 This print-out should...

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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 ˆ ı + B y ˆ , where B x = 9 T and B y = 3 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 = 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 ˆ ı 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 ˆ ı ˆ ˆ k v x v y v z B x B y 0 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle = q [ v z B y ˆ ı + v z B x ˆ + ( v x B y v y B x ) ˆ k ] = ( 1 . 602 × 10 19 C) {− (2 m / s) (3 T)ˆ ı + (2 m / s) (9 T)ˆ + [(8 m / s) (3 T) (2 m / s) (9 T)] ˆ k } = (9 . 612 × 10 19 N)ˆ ı + ( 2 . 8836 × 10 18 N)ˆ + ( 9 . 612 × 10 19 N) ˆ k 002 (part 2 of 3) 10.0 points What is the ˆ 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 ˆ 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: 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 ı 2. hatwide F = ˆ ı 3. hatwide F = + ˆ k correct 4. hatwide F = +ˆ 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 Right-hand rule for cross-products.

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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 ( ˆ ı ) , vectorv = v (+ˆ ) , and q > 0 , therefore , vector F = + | q | vectorv × vector B = + | q | v B [(+ˆ ) × ( ˆ ı )] = + | q | v B parenleftBig + ˆ k parenrightBig hatwide F = + ˆ k . This is the first of eight versions of the problem. 005 10.0 points A positively charged particle moving at 45 angles to both the z -axis and x -axis enters a magnetic field (pointing out of of the page), as shown in the figure below.
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hw12b - aljabr(faa335 Hw12 Ross(89251 This print-out should...

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