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Unformatted text preview: homework 07 – KIM, JI – Due: Mar 2 2008, 4:00 am 1 Question 1, chap 30, sect 1. part 1 of 1 10 points Suppose a new particle is discovered, and it is found that a beam of these particles passes undeflected through “crossed” electric and magnetic fields, where E = 583 V / m and B = 0 . 0038 T. If the electric field is turned off, the particles move in the magnetic field in circular paths of radius r = 2 . 89 cm. Determine q m for the particles from these data. Correct answer: 1 . 39702 × 10 9 C / kg (toler ance ± 1 %). Explanation: Let : E = 583 V / m , q e = q p = 1 . 60218 × 10 − 19 C , m e = 9 . 10939 × 10 − 31 kg , m p = 1 . 67262 × 10 − 27 kg , B = 0 . 0038 T , and r = 0 . 0289 m . Since the particle passes undeflected, the electric force on it is equal to the magnetic force. When the electric field is turned off, only the magnetic force is exerted on it as the centripetal force. So we get two equations q E = B q v B q v = mv 2 r . So, the charge to mass ratio for this new particle is q m = E r B 2 = 583 V / m (0 . 0289 m) (0 . 0038 T) 2 = 1 . 39702 × 10 9 C / kg . For comparison, the electron chargetomass ratio is q e m e = 1 . 60218 × 10 − 19 C 9 . 10939 × 10 − 31 kg = 1 . 75882 × 10 11 C / kg , and the proton chargetomass ratio is q p m p = 1 . 60218 × 10 − 19 C 1 . 67262 × 10 − 27 kg = 9 . 57883 × 10 7 C / kg . Question 2, chap 29, sect 1. part 1 of 2 10 points An electron is projected into a uniform magnetic field given by vector B = B y ˆ + B z ˆ k , where B y = 4 . 2 T and B z = 2 T. The magnitude of the charge on an electron is 1 . 60218 × 10 − 19 C . y z x v = 390000 m / s electron 4 . 2 T 2 T B Find the direction of the magnetic force when the velocity of the electron is v ˆ k , where v = 390000 m / s. 1. hatwide F = − ˆ ı 2. hatwide F = 1 √ 2 parenleftBig ˆ − ˆ k parenrightBig 3. hatwide F = − ˆ 4. hatwide F = 1 √ 2 parenleftBig ˆ + ˆ k parenrightBig 5. hatwide F = ˆ 6. hatwide F = ˆ ı correct 7. hatwide F = ˆ k 8. hatwide F = − ˆ k 9. hatwide F = 1 √ 2 parenleftBig ˆ k − ˆ parenrightBig homework 07 – KIM, JI – Due: Mar 2 2008, 4:00 am 2 Explanation: Let : q = 1 . 60218 × 10 − 19 C , B y = 4 . 2 T , and B z = 2 T . Basic Concepts: Magnetic force on a mov ing charge is given by vector F = qvectorv × vector B . Solution: vector B = (4 . 2 T) ˆ + (2 T) ˆ k v = (390000 m / s) ˆ k for the electron. Find: The vector expression for the force on the electron. This solves both part 1 and part 2. We will go through two methods of doing the problem. The first is more mathematically oriented and the second uses more of a reasoning argu ment....
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This note was uploaded on 05/05/2008 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Physics, Charge, Mass, Work

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