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Solved: A square, conducting, wire loop of side L, total

University Physics with Modern Physics (1) | 14th Edition | ISBN: 9780321973610 | Authors: Hugh D. Young Roger A. Freedman ISBN: 9780321973610 228

Solution for problem 29.7 Chapter 29

University Physics with Modern Physics (1) | 14th Edition

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University Physics with Modern Physics (1) | 14th Edition | ISBN: 9780321973610 | Authors: Hugh D. Young Roger A. Freedman

University Physics with Modern Physics (1) | 14th Edition

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Problem 29.7

A square, conducting, wire loop of side L, total mass m, and total resistance R initially lies in the horizontal xy-plane, with corners at 1x, y, z2 = 10, 0, 02, 10, L, 02, 1L, 0, 02, and 1L, L, 02. There is a uniform, upward magnetic field BS= Bk n in the space within and around the loop. The side of the loop that extends from 10, 0, 02 to 1L, 0, 02 is held in place on the x-axis; the rest of the loop is free to pivot around this axis. When the loop is released, it begins to rotate due to the gravitational torque. (a) Find the net torque (magnitude and direction) that acts on the loop when it has rotated through an angle f from its original orientation and is rotating downward at an angular speed v. (b) Find the angular acceleration of the loop at the instant described in part (a). (c) Compared to the case with zero magnetic field, does it take the loop a longer or shorter time to rotate through 90? Explain. (d) Is mechanical energy conserved as the loop rotates downward? Explain.

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Day 1 ­ 1/4/2016 Mondays 5­6:30 Sections with the Professor Adriane Steinacker First Homework posted this Friday Lectures are Webcast Login: phys5b Password: Y0ung2sl1t Class Materials: rulers, compass, scientific calculator Yellow is her favorite color 1) Density (mass density) lett rho ρ ρ = mass over volume [ρ] = kg v m3 1 L (Liter) = 100cm​ 1m​3 = 1000L 1 g kg converting 1 cm3= m 3 Assume air is made of Nitrogen molecules N​ atomic mass z = 7 7 protons and 7 electrons 2 A = 14 so 7p & 7e Mass of proton = 1.67 x 10​ ­27 Mass of neutron = M​ p M​N2​ x 14 x 1.67 x 10​

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Chapter 29, Problem 29.7 is Solved
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Textbook: University Physics with Modern Physics (1)
Edition: 14
Author: Hugh D. Young Roger A. Freedman
ISBN: 9780321973610

University Physics with Modern Physics (1) was written by and is associated to the ISBN: 9780321973610. This textbook survival guide was created for the textbook: University Physics with Modern Physics (1), edition: 14. The answer to “A square, conducting, wire loop of side L, total mass m, and total resistance R initially lies in the horizontal xy-plane, with corners at 1x, y, z2 = 10, 0, 02, 10, L, 02, 1L, 0, 02, and 1L, L, 02. There is a uniform, upward magnetic field BS= Bk n in the space within and around the loop. The side of the loop that extends from 10, 0, 02 to 1L, 0, 02 is held in place on the x-axis; the rest of the loop is free to pivot around this axis. When the loop is released, it begins to rotate due to the gravitational torque. (a) Find the net torque (magnitude and direction) that acts on the loop when it has rotated through an angle f from its original orientation and is rotating downward at an angular speed v. (b) Find the angular acceleration of the loop at the instant described in part (a). (c) Compared to the case with zero magnetic field, does it take the loop a longer or shorter time to rotate through 90? Explain. (d) Is mechanical energy conserved as the loop rotates downward? Explain.” is broken down into a number of easy to follow steps, and 191 words. Since the solution to 29.7 from 29 chapter was answered, more than 271 students have viewed the full step-by-step answer. The full step-by-step solution to problem: 29.7 from chapter: 29 was answered by , our top Physics solution expert on 01/09/18, 07:46PM. This full solution covers the following key subjects: . This expansive textbook survival guide covers 44 chapters, and 4574 solutions.

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Solved: A square, conducting, wire loop of side L, total