Physics for Scientists and Engineers With Modern Physics
9th Edition
ISBN: 9781133953982
Author: SERWAY, Raymond A./
Publisher: Cengage Learning
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Question
Chapter 32, Problem 78CP
(a)
To determine
The reason for the configuration of conductors having an inductance.
(b)
To determine
The region which constitutes the flux loop.
(c)
To determine
The proof for the expression
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Physics for Scientists and Engineers With Modern Physics
Ch. 32.1 - A coil with zero resistance has its ends labeled a...Ch. 32.2 - Prob. 32.2QQCh. 32.3 - Prob. 32.3QQCh. 32.4 - Prob. 32.4QQCh. 32.5 - (i) At an instant of time during the oscillations...Ch. 32 - Prob. 1OQCh. 32 - Prob. 2OQCh. 32 - Prob. 3OQCh. 32 - Prob. 4OQCh. 32 - Prob. 5OQ
Ch. 32 - Prob. 6OQCh. 32 - Prob. 7OQCh. 32 - Prob. 1CQCh. 32 - Prob. 2CQCh. 32 - Prob. 3CQCh. 32 - Prob. 4CQCh. 32 - Prob. 5CQCh. 32 - Prob. 6CQCh. 32 - The open switch in Figure CQ32.7 is thrown closed...Ch. 32 - Prob. 8CQCh. 32 - Prob. 9CQCh. 32 - Prob. 10CQCh. 32 - Prob. 1PCh. 32 - Prob. 2PCh. 32 - Prob. 3PCh. 32 - Prob. 4PCh. 32 - Prob. 5PCh. 32 - Prob. 6PCh. 32 - Prob. 7PCh. 32 - Prob. 8PCh. 32 - Prob. 9PCh. 32 - Prob. 10PCh. 32 - Prob. 11PCh. 32 - Prob. 12PCh. 32 - Prob. 13PCh. 32 - Prob. 14PCh. 32 - Prob. 15PCh. 32 - Prob. 16PCh. 32 - Prob. 17PCh. 32 - Prob. 18PCh. 32 - Prob. 19PCh. 32 - Prob. 20PCh. 32 - Prob. 21PCh. 32 - Prob. 22PCh. 32 - Prob. 23PCh. 32 - Prob. 24PCh. 32 - Prob. 25PCh. 32 - Prob. 26PCh. 32 - Prob. 27PCh. 32 - Prob. 28PCh. 32 - Prob. 29PCh. 32 - Prob. 30PCh. 32 - Prob. 31PCh. 32 - Prob. 32PCh. 32 - Prob. 33PCh. 32 - Prob. 34PCh. 32 - Prob. 35PCh. 32 - Prob. 36PCh. 32 - Prob. 37PCh. 32 - Prob. 38PCh. 32 - Prob. 39PCh. 32 - Prob. 40PCh. 32 - Prob. 41PCh. 32 - Prob. 42PCh. 32 - Prob. 43PCh. 32 - Prob. 44PCh. 32 - Prob. 45PCh. 32 - Prob. 46PCh. 32 - Prob. 47PCh. 32 - Prob. 48PCh. 32 - Prob. 49PCh. 32 - Prob. 50PCh. 32 - Prob. 51PCh. 32 - Prob. 52PCh. 32 - Prob. 53PCh. 32 - Prob. 54PCh. 32 - Prob. 55PCh. 32 - Prob. 56PCh. 32 - Prob. 57PCh. 32 - Prob. 58PCh. 32 - Electrical oscillations are initiated in a series...Ch. 32 - Prob. 60APCh. 32 - Prob. 61APCh. 32 - Prob. 62APCh. 32 - A capacitor in a series LC circuit has an initial...Ch. 32 - Prob. 64APCh. 32 - Prob. 65APCh. 32 - At the moment t = 0, a 24.0-V battery is connected...Ch. 32 - Prob. 67APCh. 32 - Prob. 68APCh. 32 - Prob. 69APCh. 32 - Prob. 70APCh. 32 - Prob. 71APCh. 32 - Prob. 72APCh. 32 - Prob. 73APCh. 32 - Prob. 74APCh. 32 - Prob. 75APCh. 32 - Prob. 76APCh. 32 - Prob. 77APCh. 32 - Prob. 78CPCh. 32 - Prob. 79CPCh. 32 - Prob. 80CPCh. 32 - Prob. 81CPCh. 32 - Prob. 82CPCh. 32 - Prob. 83CP
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- Unreasonable results Frustrated by the small Hall voltage obtained in blood flow measurements, a medical physicist decides to increase the applied magnetic field strength to get a 0.500-V output for blood moving at 30.0 cm/s in a 1.50-cm-diameter vessel. (a) What magnetic field strength is needed? (b) What is unreasonable about this result? (C) Which premise is responsible?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_forwardA toroid has a major radius R and a minor radius r and is tightly wound with N turns of wire on a hollow cardboard torus. Figure P31.6 shows half of this toroid, allowing us to see its cross section. If R r, the magnetic field in the region enclosed by the wire is essentially the same as the magnetic field of a solenoid that has been bent into a large circle of radius R. Modeling the field as the uniform field of a long solenoid, show that the inductance of such a toroid is approximately L=120N2r2R Figure P31.6arrow_forward
- An electron in a TV CRT moves with a speed of 6.0107 m/s, in a direction perpendicular to Earth's field, which has a strength of 5.0105 T. (a) What strength electric field must be applied perpendicular to the Earth’s field to make the election moves in a straight line? (b) If this is done between plates separated by 1.00 cm, what is the voltage applied? (Note that TVs are usually surrounded by a ferromagnetic material to shield against external magnetic fields and avoid the need for such a collection,)arrow_forwardThe velocity vector of a singly charged helium ion (mHe = 6.64 1027 kg) is given by v=4.50105m/s. The acceleration of the ion in a region of space with a uniform magnetic field is 8.50 1012 m/s2 in the positive y direction. The velocity is perpendicular to the field direction. What are the magnitude and direction of the magnetic field in this region?arrow_forwardTwo 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_forward
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