Physics for Scientists and Engineers With Modern Physics
9th Edition
ISBN: 9781133953982
Author: SERWAY, Raymond A./
Publisher: Cengage Learning
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Chapter 31, Problem 23P
To determine
The magnitude of the induced emf between the ends of the beam.
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Chapter 31 Solutions
Physics for Scientists and Engineers With Modern Physics
Ch. 31.1 - A circular loop of wire is held in a uniform...Ch. 31.2 - QUICK QUIZ 30.2 In Figure 30.8a, a given applied...Ch. 31.3 - Figure 30.12 (Quick Quiz 30.3) QUICK QUIZ 30.3...Ch. 31.5 - Prob. 31.4QQCh. 31.6 - Prob. 31.5QQCh. 31 - Prob. 1OQCh. 31 - Prob. 2OQCh. 31 - Prob. 3OQCh. 31 - Prob. 4OQCh. 31 - Prob. 5OQ
Ch. 31 - Prob. 6OQCh. 31 - Prob. 7OQCh. 31 - Prob. 8OQCh. 31 - Prob. 9OQCh. 31 - Prob. 10OQCh. 31 - Prob. 11OQCh. 31 - Prob. 1CQCh. 31 - Prob. 2CQCh. 31 - Prob. 3CQCh. 31 - Prob. 4CQCh. 31 - Prob. 5CQCh. 31 - Prob. 6CQCh. 31 - Prob. 7CQCh. 31 - Prob. 8CQCh. 31 - Prob. 9CQCh. 31 - Prob. 10CQCh. 31 - Prob. 1PCh. 31 - Prob. 2PCh. 31 - Prob. 3PCh. 31 - Prob. 4PCh. 31 - Prob. 5PCh. 31 - Prob. 6PCh. 31 - Prob. 7PCh. 31 - Prob. 8PCh. 31 - Prob. 9PCh. 31 - Scientific work is currently under way to...Ch. 31 - Prob. 11PCh. 31 - Prob. 12PCh. 31 - Prob. 13PCh. 31 - Prob. 14PCh. 31 - Prob. 15PCh. 31 - Prob. 16PCh. 31 - A coil formed by wrapping 50 turns of wire in the...Ch. 31 - Prob. 18PCh. 31 - Prob. 19PCh. 31 - Prob. 20PCh. 31 - Prob. 21PCh. 31 - Prob. 22PCh. 31 - Prob. 23PCh. 31 - A small airplane with a wingspan of 14.0 m is...Ch. 31 - A 2.00-m length of wire is held in an eastwest...Ch. 31 - Prob. 26PCh. 31 - Prob. 27PCh. 31 - Prob. 28PCh. 31 - Prob. 29PCh. 31 - Prob. 30PCh. 31 - Prob. 31PCh. 31 - Prob. 32PCh. 31 - Prob. 33PCh. 31 - Prob. 34PCh. 31 - Prob. 35PCh. 31 - Prob. 36PCh. 31 - Prob. 37PCh. 31 - Prob. 38PCh. 31 - Prob. 39PCh. 31 - Prob. 40PCh. 31 - Prob. 41PCh. 31 - Prob. 42PCh. 31 - Prob. 43PCh. 31 - Prob. 44PCh. 31 - Prob. 45PCh. 31 - Prob. 46PCh. 31 - Prob. 47PCh. 31 - Prob. 48PCh. 31 - The rotating loop in an AC generator is a square...Ch. 31 - Prob. 50PCh. 31 - Prob. 51APCh. 31 - Prob. 52APCh. 31 - Prob. 53APCh. 31 - Prob. 54APCh. 31 - Prob. 55APCh. 31 - Prob. 56APCh. 31 - Prob. 57APCh. 31 - Prob. 58APCh. 31 - Prob. 59APCh. 31 - Prob. 60APCh. 31 - Prob. 61APCh. 31 - Prob. 62APCh. 31 - Prob. 63APCh. 31 - Prob. 64APCh. 31 - Prob. 65APCh. 31 - Prob. 66APCh. 31 - Prob. 67APCh. 31 - A conducting rod moves with a constant velocity in...Ch. 31 - Prob. 69APCh. 31 - Prob. 70APCh. 31 - Prob. 71APCh. 31 - Prob. 72APCh. 31 - Prob. 73APCh. 31 - Prob. 74APCh. 31 - Prob. 75APCh. 31 - Prob. 76APCh. 31 - Prob. 77APCh. 31 - Prob. 78APCh. 31 - Prob. 79CPCh. 31 - Prob. 80CPCh. 31 - Prob. 81CPCh. 31 - Prob. 82CPCh. 31 - Prob. 83CP
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- A long, solid, cylindrical conductor of radius 3.0 cm carries a current of 50 A distributed uniformly over its cross-section. Plot the magnetic field as a function of the radial distance r from the center of the conductor.arrow_forwardA particle moving downward at a speed of 6.0106 m/s enters a uniform magnetic field that is horizontal and directed from east to west. (a) If the particle is deflected initially to the north in a circular arc, is its charge positive or negative? (b) If B = 0.25 T and the charge-to-mass ratio (q/m) of the particle is 40107 C/kg. what is ±e radius at the path? (c) What is the speed of the particle after c has moved in the field for 1.0105s ? for 2.0s?arrow_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_forward
- A magnetic field directed into the page changes with time according to B = 0.030 0t2 + 1.40, where B is in teslas and t is in seconds. The field has a circular cross section of radius R = 2.50 cm (see Fig. P23.28). When t = 3.00 s and r2 = 0.020 0 m, what are (a) the magnitude and (b) the direction of the electric field at point P2?arrow_forwardDetermine the initial direction of the deflection of charged particles as they enter the magnetic fields as shown in Figure P22.2. Figure P22.2.arrow_forwardWhat is the Earths magnetic flux through a. a basketball, b. a hula hoop standing up perpendicularly on its rim at the North Pole, and c. a hula hoop lying on the ground at the North Pole?arrow_forward
- Why is the following situation impossible? Figure P28.46 shows an experimental technique for altering the direction of travel for a charged particle. A particle of charge q = 1.00 C and mass m = 2.00 1015 kg enters the bottom of the region of uniform magnetic field at speed = 2.00 105 m/s, with a velocity vector perpendicular to the field lines. The magnetic force on the particle causes its direction of travel to change so that it leaves the region of the magnetic field at the top traveling at an angle from its original direction. The magnetic field has magnitude B = 0.400 T and is directed out of the page. The length h of the magnetic field region is 0.110 m. An experimenter performs the technique and measures the angle at which the particles exit the top of the field. She finds that the angles of deviation are exactly as predicted. Figure P28.46arrow_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_forwardCalculate 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_forward
- Two long coaxial copper tubes, each of length L, are connected to a battery of voltage V. The inner tube has inner radius o and outer radius b, and the outer tube has inner radius c and outer radius d. The tubes are then disconnected from the battery and rotated in the same direction at angular speed of radians per second about their common axis. Find the magnetic field (a) at a point inside the space enclosed by the inner tube r d. (Hint: Hunk of copper tubes as a capacitor and find the charge density based on the voltage applied, Q=VC, C=20LIn(c/b) .)arrow_forwardThe Hall effect finds important application in the electronics industry. It is used to find the sign and density of the carriers of electric current in semiconductor chips. The arrangement is shown in Figure P22.66. A semiconducting block of thickness t and width d carries a current I in the x direction. A uniform magnetic field B is applied in the y direction. If the charge carriers are positive, the magnetic force deflects them in the z direction. Positive charge accumulates on the top surface of the sample and negative charge on the bottom surface, creating a downward electric field. In equilibrium, the downward electric force on the charge carriers balances the upward magnetic force and the carriers move through the sample without deflection. The Hall voltage ΔVH = Vc − Va between the top and bottom surfaces is measured, and the density of the charge carriers can be calculated from it. (a) Demonstrate that if the charge carriers are negative the Hall voltage will be negative. Hence, the Hall effect reveals the sign of the charge carriers, so the sample can be classified as p-type (with positive majority charge carriers) or n-type (with negative). (b) Determine the number of charge carriers per unit volume n in terms of I, t, B, ΔVH, and the magnitude q of the carrier charge. Figure P22.66arrow_forwardWhen the current through a circular loop is 6.0 A, the magnetic field at its center is 2.0104 T. What is the radius of the loop?arrow_forward
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What is Electromagnetic Induction? | Faraday's Laws and Lenz Law | iKen | iKen Edu | iKen App; Author: Iken Edu;https://www.youtube.com/watch?v=3HyORmBip-w;License: Standard YouTube License, CC-BY