HW8-solutions

HW8-solutions - aliu (fa5484) HW8 mackie (20225) 1 This...

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Unformatted text preview: 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 (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 = 8 T and B y = 5 T. An electron moves into the field with a velocity vectorv = v x + v y + v z 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 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 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) { (4 m / s) (5 T) + (4 m / s) (8 T) + [(4 m / s) (5 T) (8 m / s) (8 T)] k } = (3 . 204 10 18 N) + ( 5 . 1264 10 18 N) + (7 . 0488 10 18 N) k 002 (part 2 of 3) 10.0 points What is the component of the force? Correct answer: 5 . 1264 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: 7 . 0488 10 18 N. Explanation: See the above part. 004 (part 1 of 4) 10.0 points 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 protons velocity is (2 . 4 Mm / s) . Correct answer: 0 . 542177 pN. Explanation: Let : q = 1 . 60218 10 19 C , vectorv = (2 . 4 Mm / s) = (2 . 4 10 6 m / s) , and vector B = (1 . 41 T) k . vector F = qvectorv vector B = (1 . 60218 10 19 C) bracketleftbig (2 . 4 10 6 m / s) bracketrightbig bracketleftBig (1 . 41 T) k bracketrightBig 10 12 pN 1 N = (0 . 542177 pN) bardbl vector F bardbl = . 542177 pN . 005 (part 2 of 4) 10.0 points Find the magnitude of the force if the protons velocity is (4 . 5 Mm / s) . Correct answer: 1 . 01658 pN. Explanation: Let : vectorv = (4 . 5 Mm / s) = (4 . 5 10 6 m / s) . aliu (fa5484) HW8 mackie (20225) 2 vector F = (1 . 60218 10 19 C) bracketleftbig (4 . 5 10 6 m / s) bracketrightbig bracketleftBig (1 . 41 T) k bracketrightBig 10 12 pN 1 N = (1 . 01658 pN) bardbl vector F bardbl = 1 . 01658 pN ....
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This note was uploaded on 04/27/2011 for the course PHYS 2202 taught by Professor Mackie during the Spring '11 term at Temple.

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HW8-solutions - aliu (fa5484) HW8 mackie (20225) 1 This...

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