# HW 20 - benavides (jjb2356) – homework 20 – Turner –...

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Unformatted text preview: benavides (jjb2356) – homework 20 – Turner – (59130) 1 This print-out should have 12 questions. Multiple-choice 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 . 8 × 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: 60 . 7143 μ T. Explanation: Let : E = 17000 N / C and v = 2 . 8 × 10 8 m / s . If the electrons move undeflected through the crossed fields v = E B then B = E v = 17000 N / C 2 . 8 × 10 8 m / s = 60 . 7143 T . 002 (part 1 of 2) 10.0 points A proton in a cyclotron is moving with a speed of 4 . 33 × 10 7 m / s in a circle of radius 0 . 62 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: 5 . 05011 × 10 − 12 N. Explanation: Let : v = 4 . 33 × 10 7 m / s , m p = 1 . 67 × 10 − 27 kg , r = 0 . 62 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)(4 . 33 × 10 7 m / s) 2 (0 . 62 m) = 5 . 05011 × 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 . 727951 T. Explanation: The force is due to the magnetic field. F B = B perp q e v , B = F q e v = . 727951 T . 004 (part 1 of 2) 10.0 points A proton travels with a speed of 1 . 97 × 10 6 m / s at an angle of 56 . 3 ◦ with a magnetic field of . 189 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: 4 . 96293 × 10 − 14 N. Explanation: Let : v = 1 . 97 × 10 6 m / s , θ = 56 . 3 ◦ , and B = 0 . 189 T . F = q v B sin θ = (1 . 60218 × 10 − 19 C)(1 . 97 × 10 6 m / s) × (0 . 189 T) sin56 . 3 ◦ = 4 . 96293 × 10 − 14 N. benavides (jjb2356) – 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 03/26/2012 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.

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HW 20 - benavides (jjb2356) – homework 20 – Turner –...

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