Concept explainers
An alpha particle (a He nucleus, containing two protons and two neutrons and having a mass of 6.64 × 10−27 kg) traveling horizontally at 35.6 km/s enters a uniform, vertical, 1.80-T magnetic field, (a) What is the diameter of the path followed by this alpha panicle? (b) What effect does the magnetic field have on the speed of the particle? (c) What are the magnitude and direction of the acceleration of the alpha particle while it is in the magnetic field? (d) Explain why the speed of the particle does not change even though an unbalanced external force acts on it.
Want to see the full answer?
Check out a sample textbook solutionChapter 27 Solutions
MASTERINGPHYSICS W/ETEXT ACCESS CODE 6
Additional Science Textbook Solutions
Essential University Physics: Volume 1 (3rd Edition)
College Physics: A Strategic Approach (4th Edition)
Conceptual Physics (12th Edition)
Cosmic Perspective Fundamentals
Physics: Principles with Applications
- A mass spectrometer (Fig. 30.40, page 956) operates with a uniform magnetic field of 20.0 mT and an electric field of 4.00 103 V/m in the velocity selector. What is the radius of the semicircular path of a doubly ionized alpha particle (ma = 6.64 1027 kg)?arrow_forwardA proton (charge +e, mass mp), a deuteron (charge +e, mass 2mp), and an alpha particle (charge +2e, mass 4mp) are accelerated from rest through a common potential difference V. Each of the particles enters a uniform magnetic field B, with its velocity in a direction perpendicular to B. The proton moves in a circular path of radius p. In terms of p, determine (a) the radius rd of the circular orbit for the deuteron and (b) the radius r for the alpha particle.arrow_forwardA long, straight, horizontal wire carries a left-to-right current of 20 A. If the wire is placed in a uniform magnetic field of magnitude 4.0105 T that is directed vertically downward, what is tire resultant magnitude of the magnetic field 20 cm above the wire? 20 cm below the wire?arrow_forward
- A magnetic field directed into the page changes with time according to B = 0.030 0t2 + 1.40, where B is in teslas and t is in seconds. The field has a circular cross section of radius R = 2.50 cm (see Fig. P23.28). When t = 3.00 s and r2 = 0.020 0 m, what are (a) the magnitude and (b) the direction of the electric field at point P2?arrow_forwardWhen the current through a circular loop is 6.0 A, the magnetic field at its center is 2.0104 T. What is the radius of the loop?arrow_forwardCalculate the magnitude of the magnetic field at a point 25.0 cm from a long, thin conductor carrying a current of 2.00 A.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_forwardOne long wire carries current 30.0 A to the left along the x axis. A second long wire carries current 50.0 A to the right along the line (y = 0.280 m, z = 0). (a) Where in the plane of the two wires is the total magnetic field equal to zero? (b) A particle with a charge of 2.00 C is moving with a velocity of 150iMm/s along the line (y = 0.100 m, z = 0). Calculate the vector magnetic force acting on the particle. (c) What If? A uniform electric field is applied to allow this particle to pass through this region undetected. Calculate the required vector electric field.arrow_forwardAcircularcoiofwireofradius5.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_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_forwardFigure CQ19.7 shows a coaxial cable carrying current I in its inner conductor and a return current of the same magnitude in the opposite direction in the outer conductor. The magnetic field strength at r = r0 is Find the ratio B/B0, at (a) r = 2r0 and (b) r = 4r0. Figure CQ19.7arrow_forwardA circular coil of radius 5.0 cm is wound with five turns and carries a current of 5.0 A. If the coil is placed in a uniform magnetic field of strength 5.0 T, what is the maximum torque on it?arrow_forward
- Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningPhysics for Scientists and Engineers with Modern ...PhysicsISBN:9781337553292Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning
- Physics for Scientists and EngineersPhysicsISBN:9781337553278Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningPhysics for Scientists and Engineers, Technology ...PhysicsISBN:9781305116399Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningCollege PhysicsPhysicsISBN:9781285737027Author:Raymond A. Serway, Chris VuillePublisher:Cengage Learning