Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN: 9781133939146
Author: Katz, Debora M.
Publisher: Cengage Learning
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Chapter 31.4, Problem 31.4CE
Magnetic Field Due to a Long, Straight Wire
In a laboratory, you measure the magnitude of the magnetic field generated by a long, straight wire, and you plot your results—B as a function of position r. Which of the graphs in Figure 31.15 best represents the magnetic field due to a long, straight wire?
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Chapter 31 Solutions
Physics for Scientists and Engineers: Foundations and Connections
Ch. 31.1 - CASE STUDY Measuring the Magnetic Field Near a Bar...Ch. 31.2 - Prob. 31.2CECh. 31.3 - Prob. 31.3CECh. 31.4 - Magnetic Field Due to a Long, Straight Wire In a...Ch. 31.5 - Prob. 31.5CECh. 31 - Review Suppose you want to use a small, positively...Ch. 31 - Prob. 3PQCh. 31 - Prob. 5PQCh. 31 - Plot the deflection angle of the compass needle in...Ch. 31 - Prob. 7PQ
Ch. 31 - Prob. 8PQCh. 31 - Prob. 9PQCh. 31 - What is the Earths magnetic flux through a. a...Ch. 31 - Prob. 11PQCh. 31 - Prob. 12PQCh. 31 - Figure P31.13 shows a uniform magnetic field. a....Ch. 31 - Prob. 14PQCh. 31 - Figure P31.13 shows a uniform magnetic field. a....Ch. 31 - Prob. 16PQCh. 31 - Prob. 17PQCh. 31 - Prob. 18PQCh. 31 - Prob. 19PQCh. 31 - Prob. 20PQCh. 31 - Prob. 21PQCh. 31 - Prob. 22PQCh. 31 - A steady current I flows through a wire of radius...Ch. 31 - Prob. 24PQCh. 31 - A magnetic field of 4.00 T is measured at a...Ch. 31 - Prob. 27PQCh. 31 - Sketch a plot of the magnitude of the magnetic...Ch. 31 - Prob. 29PQCh. 31 - Prob. 31PQCh. 31 - Prob. 32PQCh. 31 - Prob. 33PQCh. 31 - Prob. 34PQCh. 31 - Prob. 35PQCh. 31 - Prob. 36PQCh. 31 - Prob. 37PQCh. 31 - Prob. 38PQCh. 31 - Prob. 39PQCh. 31 - Prob. 40PQCh. 31 - Prob. 41PQCh. 31 - Prob. 42PQCh. 31 - Prob. 43PQCh. 31 - Prob. 44PQCh. 31 - Prob. 45PQCh. 31 - Prob. 46PQCh. 31 - Prob. 47PQCh. 31 - Prob. 48PQCh. 31 - Prob. 49PQCh. 31 - Prob. 50PQCh. 31 - Prob. 51PQCh. 31 - Prob. 52PQCh. 31 - Prob. 53PQCh. 31 - Prob. 54PQCh. 31 - Prob. 55PQCh. 31 - Prob. 58PQCh. 31 - A uniform magnetic field B=5.44104iT passes...Ch. 31 - Prob. 60PQCh. 31 - A solenoid 1.25 m long with a current of 5.00 A in...Ch. 31 - Prob. 63PQCh. 31 - Prob. 64PQCh. 31 - Prob. 65PQCh. 31 - Prob. 66PQCh. 31 - Prob. 67PQCh. 31 - Prob. 68PQCh. 31 - Prob. 69PQCh. 31 - Prob. 70PQCh. 31 - Prob. 71PQCh. 31 - Prob. 72PQCh. 31 - Prob. 74PQCh. 31 - Prob. 75PQ
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- Figure P31.13 shows a uniform magnetic field. a. Can you find an Amprian loop that gives a circulation integral of zero? If so, draw the loop and the field. If not, explain why not. b. Can you find an Amprian loop that gives a nonzero circulation integral? If so, draw the loop and the field. If not, explain why not. Figure P31.13arrow_forwardTwo long, straight, parallel wires carry current as shown in Figure P30.18. If the currents are equal, find an expression for the magnetic field at point C. Use the indicated coordinate system to write your answer in component form. FIGURE P30.18arrow_forwardUnreasonable results A charged particle having mass 6.641027kg (that of a helium atom) moving at 8.70105m/s perpendicular to a 1.50-T magnetic field travels in a circular path of radius 16.0 mm. (a) What is the charge of the particle? (b) What is unreasonable about this result? (c) Which assumptions are responsible?arrow_forward
- In Figure P28.28, the cube is 40.0 cm on each edge. Four straight segments of wireab, bc, cd, and daform a closed loop that carries a current I = 5.00 A in the direction shown. A uniform magnetic field of magnitude B = 0.020 0 T is in the positive y direction. Determine the magnetic force vector on (a) ab, (b) bc, (c) cd, and (d) da. (c) Explain how you could find the force exerted on the fourth of these segments from the forces on the other three, without further calculation involving the magnetic field. Figure P28.28arrow_forwardTwo long, straight wires carry the same current as shown in Figure P30.22. One wire is parallel to the z axis and the other wire is parallel to the x axis as shown. Find an expression for the magnetic field at the origin.arrow_forwardA very large parallel-plate capacitor has uniform charge per unit area + on the upper plate and on the lower plate. The plates are horizontal, and both move horizontally with speed v to the right. (a) What is the magnetic field between the plates? (b) What is the magnetic field just above or just below the plates? (c) What are the magnitude and direction of the magnetic force per unit area on the upper plate? (d) At what extrapolated speed v will the magnetic force on a plate balance the electric force on the plate? Suggestion: Use Amperes law and choose a path that closes between the plates of the capacitor.arrow_forward
- Determine 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_forwardDetermine the initial direction of the deflection of charged particles as they enter the magnetic fields shown in Figure P29.2.arrow_forwardAn alpha-particle ( m=6.641027kg , q=3.21019C ) travels in a circular path of radius 25 cm in a uniform magnetic field of magnitude 1.5 T. (a) What is the speed of the particle? (b) What is the kinetic energy in electron-volts? (c) Through what potential difference must the particle be accelerated in order to give it this kinetic energy?arrow_forward
- A 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_forwardWhy 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_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
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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