Principles of Physics: A Calculus-Based Text
5th Edition
ISBN: 9781133104261
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Textbook Question
Chapter 23, Problem 29P
Within the green dashed circle shown in Figure P23.28, the magnetic field changes with time according to the expression B = 2.00t3 − 4.00t2 + 0.800, where B is in teslas, t is in seconds, and R = 2.50 cm. When t = 2.00 s, calculate (a) the magnitude and (b) the direction of the force exerted on an electron located at point P1, which is at a distance r1 = 5.00 cm from the center of the circular field region. (c) At what instant is this force equal to zero?
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Principles of Physics: A Calculus-Based Text
Ch. 23.1 - A circular loop of wire is held in a uniform...Ch. 23.1 - Prob. 23.2QQCh. 23.2 - You wish to move a rectangular loop of wire into a...Ch. 23.2 - Prob. 23.4QQCh. 23.3 - Prob. 23.5QQCh. 23.4 - In a region of space, a magnetic field is uniform...Ch. 23.6 - Prob. 23.7QQCh. 23.6 - Prob. 23.8QQCh. 23.7 - Prob. 23.9QQCh. 23 - Prob. 1OQ
Ch. 23 - Prob. 2OQCh. 23 - Prob. 3OQCh. 23 - A circular loop of wire with a radius of 4.0 cm is...Ch. 23 - A rectangular conducting loop is placed near a...Ch. 23 - Prob. 6OQCh. 23 - Prob. 7OQCh. 23 - Prob. 8OQCh. 23 - A square, flat loop of wire is pulled at constant...Ch. 23 - The bar in Figure OQ23.10 moves on rails to the...Ch. 23 - Prob. 11OQCh. 23 - Prob. 12OQCh. 23 - A bar magnet is held in a vertical orientation...Ch. 23 - Prob. 14OQCh. 23 - Two coils are placed near each other as shown in...Ch. 23 - A circuit consists of a conducting movable bar and...Ch. 23 - Prob. 17OQCh. 23 - Prob. 1CQCh. 23 - Prob. 2CQCh. 23 - Prob. 3CQCh. 23 - Prob. 4CQCh. 23 - Prob. 5CQCh. 23 - Prob. 6CQCh. 23 - Prob. 7CQCh. 23 - Prob. 8CQCh. 23 - Prob. 9CQCh. 23 - Prob. 10CQCh. 23 - Prob. 11CQCh. 23 - Prob. 12CQCh. 23 - Prob. 13CQCh. 23 - Prob. 14CQCh. 23 - Prob. 15CQCh. 23 - Prob. 16CQCh. 23 - Prob. 1PCh. 23 - An instrument based on induced emf has been used...Ch. 23 - A flat loop of wire consisting of a single turn of...Ch. 23 - Prob. 4PCh. 23 - Prob. 5PCh. 23 - Prob. 6PCh. 23 - A loop of wire in the shape of a rectangle of...Ch. 23 - When a wire carries an AC current with a known...Ch. 23 - Prob. 9PCh. 23 - Prob. 10PCh. 23 - Prob. 11PCh. 23 - A piece of insulated wire is shaped into a figure...Ch. 23 - A coil of 15 turns and radius 10.0 cm surrounds a...Ch. 23 - Prob. 14PCh. 23 - Figure P23.15 shows a top view of a bar that can...Ch. 23 - Prob. 16PCh. 23 - Prob. 17PCh. 23 - A metal rod of mass m slides without friction...Ch. 23 - Review. After removing one string while...Ch. 23 - Prob. 20PCh. 23 - The homopolar generator, also called the Faraday...Ch. 23 - Prob. 22PCh. 23 - A long solenoid, with its axis along the x axis,...Ch. 23 - Prob. 24PCh. 23 - Prob. 25PCh. 23 - Prob. 26PCh. 23 - A coil of area 0.100 m2 is rotating at 60.0 rev/s...Ch. 23 - A magnetic field directed into the page changes...Ch. 23 - Within the green dashed circle shown in Figure...Ch. 23 - Prob. 30PCh. 23 - Prob. 31PCh. 23 - Prob. 32PCh. 23 - Prob. 33PCh. 23 - Prob. 34PCh. 23 - Prob. 35PCh. 23 - Prob. 36PCh. 23 - Prob. 37PCh. 23 - Prob. 38PCh. 23 - Prob. 39PCh. 23 - Prob. 40PCh. 23 - Prob. 41PCh. 23 - Prob. 42PCh. 23 - Prob. 43PCh. 23 - Prob. 44PCh. 23 - Prob. 45PCh. 23 - Prob. 46PCh. 23 - Prob. 47PCh. 23 - Prob. 48PCh. 23 - Prob. 49PCh. 23 - Prob. 50PCh. 23 - Prob. 51PCh. 23 - Prob. 52PCh. 23 - Prob. 53PCh. 23 - Prob. 54PCh. 23 - Prob. 55PCh. 23 - Prob. 56PCh. 23 - Prob. 57PCh. 23 - Figure P23.58 is a graph of the induced emf versus...Ch. 23 - Prob. 59PCh. 23 - Prob. 60PCh. 23 - The magnetic flux through a metal ring varies with...Ch. 23 - Prob. 62PCh. 23 - Prob. 63PCh. 23 - Prob. 64PCh. 23 - Prob. 65PCh. 23 - Prob. 66PCh. 23 - Prob. 67PCh. 23 - Prob. 68PCh. 23 - Prob. 69PCh. 23 - Prob. 70PCh. 23 - Prob. 71PCh. 23 - Prob. 72PCh. 23 - Review. The use of superconductors has been...Ch. 23 - Prob. 74PCh. 23 - Prob. 75P
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- Two infinitely long current-carrying wires run parallel in the xy plane and are each a distance d = 11.0 cm from the y axis (Fig. P30.83). The current in both wires is I = 5.00 A in the negative y direction. a. Draw a sketch of the magnetic field pattern in the xz plane due to the two wires. What is the magnitude of the magnetic field due to the two wires b. at the origin and c. as a function of z along the z axis, at x = y = 0? FIGURE P30.83arrow_forwardIn Figure P22.43, the current in the long, straight wire is I1 = 5.00 A and the wire lies in the plane of the rectangular loop, which carries a current I2 = 10.0 A. The dimensions in the figure are c = 0.100 m, a = 0.150 m, and = 0.450 m. Find the magnitude and direction of the net force exerted on the loop by the magnetic field created by the wire. Figure P22.43 Problems 43 and 44.arrow_forwardWithin the green dashed circle show in Figure P30.21, the magnetic field changes with time according to the expression B = 2.00t3 4.00t2 + 0.800, where B is in teslas, t is in seconds, and R = 2.50 cm. When t = 2.00 s, calculate (a) the magnitude and (b) the direction of the force exerted on an electron located at point P, which is at a distance r = 5.00 cm from the center of the circular field region. (c) At what instant is this force equal to zero? Figure P30.21arrow_forward
- Calculate the magnitude of the magnetic field at a point 25.0 cm from a long, thin conductor carrying a current of 2.00 A.arrow_forwardOne long wire carries current 30.0 A to the left along the x axis. A second long wire carries current 50.0 A to the right along the line (y = 0.280 m, z = 0). (a) Where in the plane of the two wires is the total magnetic field equal to zero? (b) A particle with a charge of 2.00 C is moving with a velocity of 150iMm/s along the line (y = 0.100 m, z = 0). Calculate the vector magnetic force acting on the particle. (c) What If? A uniform electric field is applied to allow this particle to pass through this region undetected. Calculate the required vector electric field.arrow_forwardFigure CQ19.7 shows a coaxial cable carrying current I in its inner conductor and a return current of the same magnitude in the opposite direction in the outer conductor. The magnetic field strength at r = r0 is Find the ratio B/B0, at (a) r = 2r0 and (b) r = 4r0. Figure CQ19.7arrow_forward
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Magnets and Magnetic Fields; Author: Professor Dave explains;https://www.youtube.com/watch?v=IgtIdttfGVw;License: Standard YouTube License, CC-BY