PhyIIHW6Solutions - hopkins(tlh982 – HW06 – criss...

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Unformatted text preview: hopkins (tlh982) – HW06 – criss – (4908) 1 This print-out should have 4 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points An electron in a vacuum is first accelerated by a voltage of 26300 V and then enters a region in which there is a uniform magnetic field of . 104 T at right angles to the direction of the electron’s motion. What is the force on the electron due to the magnetic field? Correct answer: 1 . 60268 × 10 − 12 N. Explanation: Let : V = 9 . 61842 × 10 7 m / s and B = 0 . 104 T . The kinetic energy gained after acceleration is KE = 1 2 m e v 2 = q e V , so the velocity is v = radicalbigg 2 q e V m = radicalBigg 2(1 . 60218 × 10 − 19 C)(26300 V) 9 . 10939 × 10 − 31 kg = 9 . 61842 × 10 7 m / s . Then the force on it is f = qvB = (1 . 60218 × 10 − 19 C) × (9 . 61842 × 10 7 m / s)(0 . 104 T) = 1 . 60268 × 10 − 12 N ....
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This note was uploaded on 10/29/2008 for the course PHY 2049 taught by Professor Criss during the Spring '08 term at University of South Florida.

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PhyIIHW6Solutions - hopkins(tlh982 – HW06 – criss...

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