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HW8-solutions

# HW8-solutions - aliu(fa5484 HW8 mackie(20225 This print-out...

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aliu (fa5484) – HW8 – mackie – (20225) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001(part1of3)10.0points The magnetic field over a certain range is given by vector B = B x ˆ ı + B y ˆ , where B x = 8 T and B y = 5 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 = 4 m / s, v y = 8 m / s and v z = 4 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: 3 . 204 × 10 18 N. Explanation: Let : v x = 4 m / s , v y = 8 m / s , v z = 4 m / s , q = 1 . 602 × 10 19 C , B x = 8 T , and B y = 5 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) {− (4 m / s) (5 T)ˆ ı + (4 m / s) (8 T)ˆ + [(4 m / s) (5 T) (8 m / s) (8 T)] ˆ k } = (3 . 204 × 10 18 N)ˆ ı + ( 5 . 1264 × 10 18 N)ˆ + (7 . 0488 × 10 18 N) ˆ k 002(part2of3)10.0points What is the ˆ component of the force? Correct answer: 5 . 1264 × 10 18 N. Explanation: See the above part. 003(part3of3)10.0points What is the ˆ k component of the force? Correct answer: 7 . 0488 × 10 18 N. Explanation: See the above part. 004(part1of4)10.0points A uniform magnetic field of magnitude 1 . 41 T acts in the positive z direction. Find the magnitude of the force exerted by the field on a proton if the proton’s velocity is (2 . 4 Mm / s) ˆ ı . Correct answer: 0 . 542177 pN. Explanation: Let : q = 1 . 60218 × 10 19 C , vectorv = (2 . 4 Mm / s) ˆ ı = (2 . 4 × 10 6 m / s) ˆ ı , and vector B = (1 . 41 T) ˆ k . vector F = qvectorv × vector B = (1 . 60218 × 10 19 C) × bracketleftbig (2 . 4 × 10 6 m / s) ˆ ı bracketrightbig × bracketleftBig (1 . 41 T) ˆ k bracketrightBig × 10 12 pN 1 N = (0 . 542177 pN) ˆ bardbl vector F bardbl = 0 . 542177 pN . 005(part2of4)10.0points Find the magnitude of the force if the proton’s velocity is (4 . 5 Mm / s) ˆ . Correct answer: 1 . 01658 pN. Explanation: Let : vectorv = (4 . 5 Mm / s) ˆ = (4 . 5 × 10 6 m / s) ˆ  .

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aliu (fa5484) – HW8 – mackie – (20225) 2 vector F = (1 . 60218 × 10 19 C) × bracketleftbig (4 . 5 × 10 6 m / s) ˆ bracketrightbig × bracketleftBig (1 . 41 T) ˆ k bracketrightBig × 10 12 pN 1 N = (1 . 01658 pN) ˆ ı bardbl vector F bardbl = 1 . 01658 pN . 006(part3of4)10.0points Find the magnitude of the force for a velocity of (7 . 5 Mm / s) ˆ k . Correct answer: 0 pN. Explanation: Let : vectorv = (7 . 5 Mm / s) ˆ k = (7 . 5 × 10 6 m / s) ˆ k .
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