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Practice HW 11 solution

# Practice HW 11 solution - song(shs546 Practice HW 11...

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song (shs546) – Practice HW 11 Solutions – weathers – (22202) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 0.0points A positively charged particle passes through a laboratory traveling in an easterly direction. There are both electric and magnetic field in the room and their effects on the charged particle cancel. If the electric field points upward, what must be the direction of the magnetic field? 1. east 2. west 3. south correct 4. north 5. upward 6. downward Explanation: The force on the object due to the magnetic field must point down in order to cancel the force due to the electric field. Using the right- hand rule, the magnetic field must point south in order for the magnetic force to be directed downward. 002 0.0points An electron in a vacuum is first accelerated by a voltage of 81700 V and then enters a region in which there is a uniform magnetic field of 0 . 444 T at right angles to the direction of the electron’s motion. The mass of the electron is 9 . 11 × 10 31 kg and its charge is 1 . 60218 × 10 19 C. What is the magnitude of the force on the electron due to the magnetic field? Correct answer: 1 . 20591 × 10 11 N. Explanation: Let : V = 81700 V , B = 0 . 444 T , m = 9 . 11 × 10 31 kg , q e = 1 . 60218 × 10 19 C . The kinetic energy K gained after acceler- ation is K = 1 2 m v 2 = q e V , so the velocity is v = radicalbigg 2 q e V m = radicalBigg 2 (1 . 60218 × 10 19 C)(81700 V) 9 . 11 × 10 31 kg = 1 . 6952 × 10 8 m / s . Then the force on it is f = q v B = (1 . 60218 × 10 19 C) × (1 . 6952 × 10 8 m / s) (0 . 444 T) = 1 . 20591 × 10 11 N . 003(part1of2)0.0points An electron is projected into a uniform mag- netic field given by vector B = B z ˆ k + B x ˆ ı , where B z = 4 . 9 T and B x = 1 . 7 T. The magnitude of the charge on an electron is 1 . 60218 × 10 19 C . z x y v = 3 . 9 × 10 5 m / s electron 4 . 9 T 1 . 7 T B Find the direction of the magnetic force when the velocity of the electron is v ˆ ı , where v = 3 . 9 × 10 5 m / s. 1. hatwide F = ˆ correct 2. hatwide F = ˆ

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song (shs546) – Practice HW 11 Solutions – weathers – (22202) 2 3. hatwide F = ˆ ı 4. hatwide F = ˆ ı 5. hatwide F = ˆ k 6. hatwide F = ˆ k 7. hatwide F = 1 2 parenleftBig ˆ k ı parenrightBig 8. hatwide F = 1 2 parenleftBig ˆ ı ˆ k parenrightBig 9. hatwide F = 1 2 parenleftBig ˆ k ˆ ı parenrightBig Explanation: Let : q = 1 . 60218 × 10 19 C , B z = 4 . 9 T , and B x = 1 . 7 T . Basic Concepts: Magnetic force on a mov- ing charge is given by vector F = qvectorv × vector B . Solution: vector B = (4 . 9 T) ˆ k + (1 . 7 T) ˆ ı v = (3 . 9 × 10 5 m / s) ˆ ı 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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Practice HW 11 solution - song(shs546 Practice HW 11...

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