PHY303L Solutions%208

# PHY303L - homework 08 – FAUSAK TAYLOR – Due 4:00 am 1 Question 1 chap 29 sect 2 part 1 of 1 10 points In an experiment designed to measure the

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Unformatted text preview: homework 08 – FAUSAK, TAYLOR – Due: Mar 17 2008, 4:00 am 1 Question 1, chap 29, sect 2. part 1 of 1 10 points In an experiment designed to measure the strength of a uniform magnetic field produced by a set of coils, electrons are accelerated from rest through a potential difference of 342 V. The resulting electron beam travels in a circle with a radius of 6 . 66 cm. The charge on an electron is 1 . 60218 × 10 − 19 C and its mass is 9 . 10939 × 10 − 31 kg. Assuming the magnetic field is perpendic- ular to the beam, find the magnitude of the magnetic field. Correct answer: 0 . 000936361 T (tolerance ± 1 %). Explanation: Let : e = 1 . 60218 × 10 − 19 C , r = 6 . 66 cm = 0 . 0666 m , V = 342 V , and m = m e = 9 . 10939 × 10 − 31 kg . Since K i = 0 and K f = 1 2 mv 2 , we have 1 2 mv 2 = | e | V v = radicalBigg 2 | e | V m e = radicalBigg 2 (1 . 60218 × 10 − 19 C) (342 V) 9 . 10939 × 10 − 31 kg = 1 . 09683 × 10 7 m / s . From conservation of energy, the increase in the electrons’ kinetic energy must equal the change in their potential energy | e | V : F = ev B = mv 2 r B = mv | e | r = (9 . 10939 × 10 − 31 kg) (1 . 60218 × 10 − 19 C) × (1 . 09683 × 10 7 m / s) (0 . 0666 m) = . 000936361 T . Question 2, chap 30, sect 1. part 1 of 6 10 points A particle of mass 5 . 478 × 10 − 26 kg and charge of magnitude 1 . 6 × 10 − 19 C is acceler- ated from rest in the plane of the page through a potential difference of 333 V between two parallel plates as shown. The particle is in- jected through a hole in the right-hand plate into a region of space containing a uniform magnetic field of magnitude 0 . 439 T oriented perpendicular to the plane of the page. The particle curves in a semicircular path and strikes a detector. q m Region of Magnetic Field B E hole What is the sign of the charge of the parti- cle? Neglect relativistic effects. 1. The charge q is negative, − . 2. The charge q is positive, + . correct 3. The charge q cannot be determined for information given. Explanation: The charge accelerates toward the nega- tive plate and away from the positive plate. Therefore the charge is positive, + . Question 3, chap 30, sect 1. part 2 of 6 10 points Which way does the magnetic field point? 1. to the left 2. out of the page correct homework 08 – FAUSAK, TAYLOR – Due: Mar 17 2008, 4:00 am 2 3. to the right 4. toward the bottom of page 5. toward the top of page 6. Cannot be determined 7. into the page Explanation: B + q m E hole + − − − − − + + + Because the particle curves down, the direc- tion of vectorv × vector B points down. By the right-hand rule, vector B must point out of the page . Question 4, chap 30, sect 1....
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## This note was uploaded on 03/30/2010 for the course CH 369 taught by Professor Kbrowning during the Spring '07 term at University of Texas at Austin.

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PHY303L - homework 08 – FAUSAK TAYLOR – Due 4:00 am 1 Question 1 chap 29 sect 2 part 1 of 1 10 points In an experiment designed to measure the

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