Concept explainers
Each of the lettered points at the corners of the cube in Fig. Q27.12 represents a positive charge q moving with a velocity of magnitude ʋ in the direction indicated. The region in the figure is in a uniform magnetic field
Figure Q27.12
Want to see the full answer?
Check out a sample textbook solutionChapter 27 Solutions
University Physics with Modern Physics Plus Mastering Physics with eText -- Access Card Package (14th Edition)
Additional Science Textbook Solutions
Essential University Physics: Volume 2 (3rd Edition)
Conceptual Integrated Science
Cosmic Perspective Fundamentals
Introduction to Electrodynamics
Physics (5th Edition)
Physics: Principles with Applications
- A 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_forwardAt a particular instant an electron is traveling west to east with a kinetic energy of 10 keV. Earth's magnetic field has a horizontal component of 1.8105 T north and a vertical component of 5.0105 T down. (a) What is the path of the election? (b) What is the radius of curvature of the path?arrow_forwardA uniform magnetic field of magnitude is directed parallel to the z-axis. A proton enters the field with a velocity v=(4j+3k)106m/s and travels in a helical path with a radius of 5.0 cm. (a) What is the value of B? (b) What is the time required for one trip around the helix? (c) Where is the proton 5.0107s after entering the field?arrow_forward
- Acircularcoiofwireofradius5.Ocmhas2Otums and carries a current of 2.0 A. The coil lies in a magnetic field of magnitude 0.50 T that is directed parallel to the plane of the coil. (a) What is the magnetic dipole moment of the coil? (b) What is the torque on the coil?arrow_forwardTwo 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_forwardA proton moving in the plane of the page has a kinetic energy of 6.00 MeV. A magnetic field of magnitude H = 1.00 T is directed into the page. The proton enters the magnetic field with its velocity vector at an angle = 45.0 to the linear boundary of' the field as shown in Figure P29.80. (a) Find x, the distance from the point of entry to where the proton will leave the field. (b) Determine . the angle between the boundary and the protons velocity vector as it leaves the field.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_forwardAn 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 magnitude of the magnetic field 50 cm from a long, thin, straight wire is 8.0T . What is the current through the long wire?arrow_forward
- In 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_forwardAn infinitely long wire carrying a current I is bent at a right angle as shown in Figure P22.30. Determine the magnetic field at point P, located a distance x from the corner of the wire. Figure P22.30arrow_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_forward
- Physics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage LearningPrinciples of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningCollege PhysicsPhysicsISBN:9781305952300Author:Raymond A. Serway, Chris VuillePublisher:Cengage Learning
- College PhysicsPhysicsISBN:9781285737027Author:Raymond A. Serway, Chris VuillePublisher:Cengage LearningPhysics for Scientists and Engineers with Modern ...PhysicsISBN:9781337553292Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning