Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
expand_more
expand_more
format_list_bulleted
Concept explainers
Question
Chapter 27, Problem 77P
To determine
The magnetic field at point
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solutionStudents have asked these similar questions
(B)
An electron beam passes through a magnetic field 2 x 10³Wb/m² and an electric field of 3.4 x
104 V/m, both fields are normal to each other and acting simultaneously in the same region.
The path of electrons remains unchanged. Calculate the electron speed. If the electric field is
switched off, what will be radius of the circular path?
An electric field is restricted to a circular area of diameter d = 10.3 cm as shown in the figure.
Éout
•P
At the instant shown, the field direction is out of the page, its magnitude is 300 V/m, and its magnitude is increasing at a
rate of 21.4 V/(m. s).
(a) What is the direction of the magnetic field at the point P, r = 15.1 cm from the center of the circle?
upwards
downwards
(b) What is the magnitude of the magnetic field (in T) at the point P, r = 15.1 cm from the center of the circle?
1.51*10**-16
X
Apply the extended form of Ampere's law. Note that I = 0 in this case. What is the electric flux? What is the rate of
change of the electric flux? T
(c) What If? As before, at the moment shown in the figure, the electric field within the circle has a magnitude of
300 V/m and is increasing at a rate of 21.4 V/(ms). In addition, suppose that the radius of the circular area of the
electric field increases at a rate of 1.00 cm/s. What would the magnitude of the magnetic field be at point P…
In region M directly above the large upper wire, the magnetic field would point where?
Chapter 27 Solutions
Physics for Scientists and Engineers
Ch. 27 - Prob. 1PCh. 27 - Prob. 2PCh. 27 - Prob. 3PCh. 27 - Prob. 4PCh. 27 - Prob. 5PCh. 27 - Prob. 6PCh. 27 - Prob. 7PCh. 27 - Prob. 8PCh. 27 - Prob. 9PCh. 27 - Prob. 10P
Ch. 27 - Prob. 11PCh. 27 - Prob. 12PCh. 27 - Prob. 13PCh. 27 - Prob. 14PCh. 27 - Prob. 15PCh. 27 - Prob. 16PCh. 27 - Prob. 17PCh. 27 - Prob. 18PCh. 27 - Prob. 19PCh. 27 - Prob. 20PCh. 27 - Prob. 21PCh. 27 - Prob. 22PCh. 27 - Prob. 23PCh. 27 - Prob. 24PCh. 27 - Prob. 25PCh. 27 - Prob. 26PCh. 27 - Prob. 27PCh. 27 - Prob. 28PCh. 27 - Prob. 29PCh. 27 - Prob. 30PCh. 27 - Prob. 31PCh. 27 - Prob. 32PCh. 27 - Prob. 33PCh. 27 - Prob. 34PCh. 27 - Prob. 35PCh. 27 - Prob. 36PCh. 27 - Prob. 37PCh. 27 - Prob. 38PCh. 27 - Prob. 39PCh. 27 - Prob. 40PCh. 27 - Prob. 41PCh. 27 - Prob. 42PCh. 27 - Prob. 43PCh. 27 - Prob. 44PCh. 27 - Prob. 45PCh. 27 - Prob. 46PCh. 27 - Prob. 47PCh. 27 - Prob. 48PCh. 27 - Prob. 49PCh. 27 - Prob. 50PCh. 27 - Prob. 51PCh. 27 - Prob. 52PCh. 27 - Prob. 53PCh. 27 - Prob. 54PCh. 27 - Prob. 55PCh. 27 - Prob. 56PCh. 27 - Prob. 57PCh. 27 - Prob. 58PCh. 27 - Prob. 59PCh. 27 - Prob. 60PCh. 27 - Prob. 61PCh. 27 - Prob. 62PCh. 27 - Prob. 63PCh. 27 - Prob. 64PCh. 27 - Prob. 65PCh. 27 - Prob. 66PCh. 27 - Prob. 67PCh. 27 - Prob. 68PCh. 27 - Prob. 69PCh. 27 - Prob. 70PCh. 27 - Prob. 71PCh. 27 - Prob. 72PCh. 27 - Prob. 73PCh. 27 - Prob. 74PCh. 27 - Prob. 75PCh. 27 - Prob. 76PCh. 27 - Prob. 77PCh. 27 - Prob. 78PCh. 27 - Prob. 79PCh. 27 - Prob. 80PCh. 27 - Prob. 81PCh. 27 - Prob. 82PCh. 27 - Prob. 83PCh. 27 - Prob. 84PCh. 27 - Prob. 85PCh. 27 - Prob. 86PCh. 27 - Prob. 87PCh. 27 - Prob. 88PCh. 27 - Prob. 89PCh. 27 - Prob. 90PCh. 27 - Prob. 91PCh. 27 - Prob. 92PCh. 27 - Prob. 93PCh. 27 - Prob. 94PCh. 27 - Prob. 95PCh. 27 - Prob. 96PCh. 27 - Prob. 97PCh. 27 - Prob. 98P
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.Similar questions
- Repeat the previous problem, but with the loop lying flat on the ground with its current circulating counterclockwise (when viewed from above) in a location where Earth’s field is north, but at an angle 45.0° below the horizontal and with a strength of 6.0105T.arrow_forwardIs B constant in magnitude for points that lie on a magnetic field line?arrow_forwardThe magnetic field between the poles of a horseshoe electromagnet is uniform and has a cylindrical symmetry about an axis from the middle of the South Pole to the middle of the North Pole. The magnitude of the magnetic field changes as a rate of dB/dt due to the changing current through the electromagnet, Determine the electric field at a distance r from the center.arrow_forward
- Check Your Understanding The wire loop forms a full circle of radius R and current I. What is the magnitude of the magnetic field at the center?arrow_forwardHow is the percentage change in the strength of the magnetic field across the face of the toroid related to the percentage change in the radial distance from the axis of the toroid?arrow_forwardA particle of charge q and mass m is accelerated from rest through a potential difference V, after silica it encounters a uniform magnetic field B. If the particle moves in a plane perpendicular to B, shaft is the radius of its circular orbit?arrow_forward
- An electron moving with a velocity v=(4.0i+3.0j+2.0k)106m/s enters a region where there is a uniform electric field and a uniform magnetic field. The magnetic field is given by v=(1.0i2.0j+4.0k)102T. If the electron travels through a region without being deflected, what is the electric field?arrow_forwardA long, straight, cylindrical conductor contains a cylindrical cavity whose axis is displaced by n from the axis of the conductor, as shown in the accompanying figure. The current density in the conductor is given by J=J0k, where J0 is a constant and k is along the axis of the conductor. Calculate the magnetic field at an arbitrary point P in the cavity by superimposing the field of a solid cylindrical conductor with radius R1and current density Jonto the field of a solid cylindrical conductor with radius R2and current density J . Then use the fact that the appropriate azimuthal unit vectors can be expressed as 1=kr1and 2=kr2 to show that everywhere inside the cavity the magnetic field is given by the constant B=120J0ka , where a=r1r2 and r1=r1r1 is the position of P relative to the center of the conductor and r2=r2r2 is the position of P relative to the center of the cavity.arrow_forwardCheck Your Understanding In what orientation would a magnetic dipole have to be to produce (a) a maximum torque in a magnetic field? (b) A maximum energy of the dipole?arrow_forward
- A solenoid of radius r = 1.25 cm and length { = 31.0 cm has 330 turns and carries 12.0 A. R (a) Calculate the flux through the surface of a disk-shaped area of radius R = 5.00 cm that is positioned perpendicular to and centered on the axis of the solenoid as in the figure (a) above. 7.87 v µWb (b) Figure (b) above shows an enlarged end view of the same solenoid. Calculate the flux through the tan area, which is an annulus with an inner radius of a = 0.400 cm and outer radius of b = 0.800 cm. 2.14 What is the area of the shaded annulus? µWbarrow_forwardA solenoid 2.91E-2m in diameter and 0.287m long has 304 turns and carries 12.0A. R (b) Calculate the flux through the surface of a disk of radius 5.00E-2m that is positioned perpendicular to and centred on the axis of the solenoid. Figure b) shows an enlarged end view of the same solenoid as in the last question. Calculate the flux through the blue area, which is defined by an annulus that has an inner radius of 0.356cm and an outer radius of 0.712cm.arrow_forwardAn electric field is restricted to a circular area of diameter d = 10.3 cm as shown in the figure. "out •P At the instant shown, the field direction is out of the page, its magnitude is 300 V/m, and its magnitude is increasing at a rate of 21.4 V/(m. s). (a) What is the direction of the magnetic field at the point P, r = 15.1 cm from the center of the circle? upwards downwards (b) What is the magnitude of the magnetic field (in T) at the point P, r = 15.1 cm from the center of the circle? 2.09*10**-17 X Apply the extended form of Ampere's law. Note that I = 0 in this case. What is the electric flux? What is the rate of change of the electric flux? T (c) What If? As before, at the moment shown in the figure, the electric field within the circle has a magnitude of 300 V/m and is increasing at a rate of 21.4 V/(ms). In addition, suppose that the radius of the circular area of the electric field increases at a rate of 1.00 cm/s. What would the magnitude of the magnetic field be at point P…arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
- Glencoe Physics: Principles and Problems, Student...PhysicsISBN:9780078807213Author:Paul W. ZitzewitzPublisher:Glencoe/McGraw-Hill
Glencoe Physics: Principles and Problems, Student...
Physics
ISBN:9780078807213
Author:Paul W. Zitzewitz
Publisher:Glencoe/McGraw-Hill
Magnets and Magnetic Fields; Author: Professor Dave explains;https://www.youtube.com/watch?v=IgtIdttfGVw;License: Standard YouTube License, CC-BY