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Unformatted text preview: terry (ect328) – homework 20 – Turner – (59130) 1 This printout should have 12 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The electrons in a beam are moving at 2 . 7 × 10 8 m / s in an electric field of 17000 N / C. What value must the magnetic field have if the electrons pass through the crossed fields undeflected? Correct answer: 62 . 963 μ T. Explanation: Let : E = 17000 N / C and v = 2 . 7 × 10 8 m / s . If the electrons move undeflected through the crossed fields v = E B then B = E v = 17000 N / C 2 . 7 × 10 8 m / s = 62 . 963 T . 002 (part 1 of 2) 10.0 points A proton in a cyclotron is moving with a speed of 5 . 34 × 10 7 m / s in a circle of radius 0 . 589 m. 1 . 67 × 10 − 27 kg is the mass of the pro ton, and 1 . 60218 × 10 − 19 C is its fundamental charge. What is the magnitude of the force exerted on the proton by the magnetic field of the cyclotron? Correct answer: 8 . 08507 × 10 − 12 N. Explanation: Let : v = 5 . 34 × 10 7 m / s , m p = 1 . 67 × 10 − 27 kg , r = 0 . 589 m , and q e = 1 . 60218 × 10 − 19 C . The magnetic force is the centripetal force which keeps the proton in circular motion. From the centripetal force equation, we have F = m p v 2 r = (1 . 67 × 10 − 27 kg)(5 . 34 × 10 7 m / s) 2 (0 . 589 m) = 8 . 08507 × 10 − 12 N . 003 (part 2 of 2) 10.0 points What is the magnitude of the magnetic field required to keep it moving in this circle? Correct answer: 0 . 945 T. Explanation: The force is due to the magnetic field. F B = B perp q e v , B = F q e v = . 945 T . 004 (part 1 of 2) 10.0 points A proton travels with a speed of 3 . 57 × 10 6 m / s at an angle of 41 . 6 ◦ with a magnetic field of . 366 T pointed in the y direction. The charge of proton is 1 . 60218 × 10 − 19 C. What is the magnitude of the magnetic force on the proton? Correct answer: 1 . 38989 × 10 − 13 N. Explanation: Let : v = 3 . 57 × 10 6 m / s , θ = 41 . 6 ◦ , and B = 0 . 366 T . F = q v B sin θ = (1 . 60218 × 10 − 19 C)(3 . 57 × 10 6 m / s) × (0 . 366 T) sin41 . 6 ◦ = 1 . 38989 × 10 − 13 N. terry (ect328) – homework 20 – Turner – (59130) 2 005 (part 2 of 2) 10.0 points The mass of proton is 1 . 67262 × 10 − 27 kg...
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This note was uploaded on 02/17/2010 for the course PHY 59130 taught by Professor Turner during the Fall '09 term at University of Texas.
 Fall '09
 Turner

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