COLLEGE PHYSICS
2nd Edition
ISBN: 9781464196393
Author: Freedman
Publisher: MAC HIGHER
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Question
Chapter 19, Problem 56QAP
To determine
The magnitude of the magnetic field in the interior of the coils
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COLLEGE PHYSICS
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- A square loop whose sides are 6.0-cm long is made with copper wire of radius 1.0 mm. If a magnetic field perpendicular to the loop is changing at a rate of 5.0 mT/s, what is the current in the loop?arrow_forwardSolenoid A has length L and N turns, solenoid B has length 2L and N turns, and solenoid C has length L/2 and 2N turns. If each solenoid carries the same current, rank the magnitudes of the magnetic fields in the centers of the solenoids from largest to smallest.arrow_forwardA toroid with a square cross section 3.0cm3.0cm has an inner radius of 25.0 cm. It is wound with 500 turns of wire, and it carries a current of 2.0 A. What is the strength of the magnetic field at the center of the square cross section?arrow_forward
- A toroid with an inner radius of 20 cm and an outer radius of 22 cm is tightly wound with one layer of wire that has a diameter of 0.25 mm. (a) How many turns are there on the toroid? (b) If the current through the toroid windings is 2.0 A, what is the strength of the magnetic field at the center of the toroid?arrow_forwardA cosmic-ray electron moves at 7.5 × 106 m/sinches perpendicular to Earth’s magnetic field at an altitude queer the field strength is 1.0 × 105T. What is the radius of the circular path the electron follows?arrow_forwardA toroid has 250 trims of wire and carries a current of 20 A. Its inner and outer radii are 8.0 and 9.0 cm. What are the values of its magnetic field at r = 8.1, 8.5, and 8.9 cm?arrow_forward
- An 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_forwardElectrons in Earths upper atmosphere have typical speeds near 6.00 105 m/s. (a) Calculate the magnitude of Earths magnetic field if an electrons velocity is perpendicular to the magnetic field and its circular path has a radius of 7.00 102 m. (b) Calculate the number of times per second that an electron circles around a magnetic field line.arrow_forwardRank the magnitudes of the following magnetic fields from largest to smallest, noting any cases of equality. (a) the field 2 cm away from a long, straight wire carrying a current of 3 A (b) the Held at the center of a flat, compact, circular coil, 2 cm in radius, with 10 turns, carrying a current of 0.3 A (c) the field at the center of a solenoid 2 cm in radius and 200 cm long, with 1 000 turns, carrying a current of 0.3 A (d) the field at the center of a long, straight, metal bar, 2 cm in radius, carrying a current of 300 (e) a field of 1 mTarrow_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_forwardDetermine the initial direction of the deflection of charged particles as they enter the magnetic fields shown in Figure P29.2.arrow_forwardThe accompanying figure shows a cross-section of a long, hollow, cylindrical conductor of inner radius r1= 3.0 cm and outer radius r2= 5.0 cm. A 50-A current distributed uniformly over the cross-section flows into the page. Calculate the magnetic field at r = 2.0 cm. r = 4.0 cm. and r = 6.0 cm.arrow_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