A 10.9 μC particle with a mass of 2.80x10-5 kg movesperpendicular to a 0.910 T magnetic field in a circular path ofradius 26.9 mFor the steps and strategies involved in solving a similarproblem, you may view the following Quick Example 22-8 video:REASONING AND SOLUTION. Calculate the time 7' required for a particle of mass and charge qto complete a circular orbit in a magnetic field of strengthFor any object in uniform circular motion7=²7For a charged object moving in a uniform magnetic field:Part AHow fast is the particle moving?Express your answer in meters per second.IVE ΑΣΦ9.84V =?m/s

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A 10.9 μC particle with a mass of 2.80x10-5 kg moves perpendicular to a 0.910 T magnetic field in a circular path of radius 26.9 m. For the steps and strategies involved in solving a similar problem, you may view the following Quick Example 22-8 video: Part A How fast is the particle moving?

A 10.9 μC particle with a mass of 2.80x10-5 kg moves perpendicular to a 0.910 T magnetic field in a circular path of radius 26.9 m. For the steps and strategies involved in solving a similar problem, you may view the following Quick Example 22-8 video: Part A How fast is the particle moving?

Modern Physics

3rd Edition

ISBN:9781111794378

Author:Raymond A. Serway, Clement J. Moses, Curt A. Moyer

Publisher:Raymond A. Serway, Clement J. Moses, Curt A. Moyer

Chapter14: Nuclear Physics Applications

Section: Chapter Questions

Problem 32P

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The magnetic field in a cyclotron is 1.25 T, and the maximum orbital radius of the circulating protons is 0.40 m. (a) What is the kinetic energy of the protons when they are ejected from the cyclotron? (b) What Is this energy in MeV? (c) Through what potential difference would a proton have to be accelerated to acquire this kinetic energy? (d) What is the period of tire voltage source used to accelerate the piotons? (e) Repeat tire calculations for alpha-particles.
A mystery particle enters the region between the plates of a Thomson apparatus as shown in Figure 4.6. The deflection angle θ is measured to be 0.20 radians (downwards) for this particle when V = 2000 V, ℓ = 10.0 cm, and d = 2.00 cm. If a perpendicular magnetic field of magnitude 4.57 × 10−2 T is applied simultaneously with the electric field, the particle passes through the plates without deflection. (a) Find q/m for this particle. (b) Identify the particle. (c) Find the horizontal speed with which the particle entered the plates. (d) Must we use relativistic mechanics for this particle?
Cosmic rays are high-energy charged particles produced by astronomical objects. Many of the cosmic rays that make their way to the Earth are trapped by the Earths magnetic field and never reach the surface. These trapped cosmic rays are found in the Van Allen beltsdonut-shaped zones over the Earths equator (Fig. 30.34). These cosmic rays are mostly protons with energies of about 30 MeV. The inset in the figure shows a cosmic ray proton as it is about to enter the Earths magnetic field. The cosmic rays velocity is initially perpendicular to the field. Three students discuss what happens to the incoming cosmic ray. Decide which student or students are correct. Figure 30.34 The Van Allen belts are donut-shaped zones of trapped cosmic rays above the Earths surface. Inset: What happens to this cosmic ray as it enters the Earths magnetic field? Shannon: The velocity is perpendicular to the magnetic field, so the cosmic ray just passes through the field and hits the Earths atmosphere. Avi: What you are saying is that the magnetic field exerts no force on the cosmic ray. Actually, it exerts a huge force because the velocity is perpendicular to the magnetic field. The force will be into the page. Cameron: Avi is right. The cosmic ray proton is going to feel a huge magnetic force. Because it is positively charged, it will be pushed upward along the magnetic field lines. Shannon: I never said the force was zero. There is a force, but the force is perpendicular to the magnetic field lines. In this case, thats to the lefttoward the Earth. Avi: The force is perpendicular to the magnetic field, but it also has to be perpendicular to the velocity. Because B and v are both in the plane of the page, the force must be perpendicular to the page.
Viewers of Star Trek hear of an antimatter drive on the Starship Enterprise. One possibility for such a futuristic energy source is to store antimatter charged particles in a vacuum chamber, circulating in a magnetic field, and then extract them as needed. Antimatter annihilates with normal matter, producing pure energy. What strength magnetic field in T is needed to hold antiprotons, moving at 1.2 × 107 m/s, in a circular path 1.85 m in radius? Antiprotons have the same mass as protons but the opposite (negative) charge.
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Explain path of charged particle in magnetic field at,theta=0°,90°,180°
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Alpha particles (mass = 6.7 × 10^-27 kg, q = +2e) are accelerated from rest through a potential difference of 3.0 kV, then enter a uniform magnetic field B = 0.40 T perpendicular to their direction of motion. Calculate the radius of the path.
A 64-pC hypothetical charge of mass 44 fg travels at 81 km/s. It is ejected southward thus entering a uniform magnetic field with unknown magnitude and direction. If tha path is traced clockwise, a. determine the magnitude and direction of the magnetic field which causes the charge to undergo a semicircular path (given that the diameter is 51 m) b. calculate the required time for the charge to complete a semi-circualar patth from K to L. c. calculate the amgnitude and direction of the magnetic force at L.