What is the z-component of the net force acting on a particle (so you can ignore gravity) of charge 467.4 uC moving at (437.9, -520.7, -526.6) m/s in an electric field = (4,778.8, -4,132.8, -4,712.2) N/C and a magnetic field = (-0.8, -0.7, 0.2) T? Assume that nothing is in contact with the charge.

What is the z-component of the net force acting on a particle (so you can ignore gravity) of charge 467.4 uC moving at (437.9, -520.7, -526.6) m/s in an electric field = (4,778.8, -4,132.8, -4,712.2) N/C and a magnetic field = (-0.8, -0.7, 0.2) T? Assume that nothing is in contact with the charge.

Inquiry into Physics

8th Edition

ISBN:9781337515863

Author:Ostdiek

Publisher:Ostdiek

Chapter12: Relativity, Particle Physics, And Cosmology

Section: Chapter Questions

Problem 2C

See similar textbooks

Similar questions

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.
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.
Nobel laureate Richard Feynman (19181088) once said that if two persons stood at arms length from each other and each person had 1% more electrons than protons, the force of repulsion between them would be enough to lift a weight equal to that of the entire Earth. Carry out an order-of-magnitude calculation to substantiate this assertion.
What is the repulsive electrical force between two protons 5.0x10^-15 m apart from each other in an atomic nucleus ?
A point charge for which Q = 5x10-16 C and m = 2 × 10-26 kg is moving in the combined fields E = 100ax – 300ay + 200az V/m and B = -ax + 2ay – 3az mT. If the charge velocity at t = 0 is v(0) = (4ax – 3ay - 2az)10³ m/s (a) give the unit vector showing the direction in which the charge is accelerating at t = 0; (b) using the electric field intensity what is the force at t=0, (c) using magnetic flux density what force is produce at t = 0. %3D
-12 F/m. Magnetic 12.566370614356 × 10¬7 H/m. Magnitude of the Charge of one electron, e = -1.60217662 × 10¬19 C. Mass of one electron, me = 9.10938356 × 10¬31 kg. Unless specified otherwise, each symbol carries their usual meaning. For example, µC means micro coulomb 8.987 x 10° N · m² /C². Vacuum permitivity, €o = 8.854 × 10 Use the following constants if necessary. Coulomb constant, k Permeability of vacuum, µo The flux oE = 17 N · m² /C through the open surface A = 7i + 4j+ 2k is given. Find the value of a if the electric field is E = -ai +7k Value of "a" Give your answer up to at least three significance digits.
A carbon-14 ion with a charge of +6.408x10^-19 C and a mass of 2.34x10^-26 kg is sent through a mass spectrometer and hits a detector at a point 10.0 cm to the left of where the beam leaves the velocity selector. The velocity selector and the detector are both in a region of magentic field of strength 0.500 T. What is the speed of the particle after it leaves the velocity selecctor? Then, what is the accelerating potential needed to reach the speed found in that part if the ions start from rest? Please also explain and show the steps you used to get there/the physics behind why/how you got to the answer to help me better understand. Thank you soo much. Also, the work and the explanation or most important because I already have the correct answer - I'm just unsure of how to get there.
A carbon-14 ion with a charge of +6.408x10^-19 C and a mass of 2.34x10^-26 kg is sent through a mass spectrometer and hits a detector at a point 10.0 cm to the left of where the beam leaves the velocity selector. The velocity selector and the detector are both in a region of magentic field of strength 0.500 T. If a carbon-12 ion (q=+6.408x10^-19 C, m= 1.99x10^-26 kg) is accelerated to the same velocity as the carbon-14 ion, where would you need to place the detector to detect it? Please also explain and show the steps you used to get there/the physics behind why/how you got to the answer to help me better understand. Thank you soo much. Also, the work and the explanation or most important because I already have the correct answer - I'm just unsure of how to get there.
What is the repulsive force between two protons that are located .120 nm apart ? ( Ans: + 1.6 X 10 -8 N)
In the Rutherford model of the hydrogen atom, a proton (mass M, charge Q) is the nucleus and an electron (mass m, charge q) moves around the proton in a circle of radius r. Let k denote the Coulomb force constant (1/4TTE0) and G the universal gravitational constant. The ratio of the electrostatic force to the gravitational force between electron and proton is: O kMm/GQq O kQq/GMm O GQq/kMm O GMm/kQq O kQq/GMmr²
Use the following constants if necessary. Coulomb constant, k = 8.987 × 10º N · m²/C² . Vacuum permitivity, €o = 8.854 × 10-12 F/m. Magnitude of the Charge of one electron, e = –1.60217662 × 10¬19 C. Mass of one electron, me = 9.10938356 x 10¬31 kg. Unless specified otherwise, each symbol carries their usual meaning. For example, µC means micro coulomb . Battery Air Fuel A fuel gauge uses a capacitor to determine the height of the fuel in a tank. The effective dielectric constant kchanges from a value of 1 when the tank is empty, to a value of 7, when the tank is full. The appropriate electronic circuitry can determine the effective dielectric constant of the combined air and fuel between the capacitor plates. The dimension of the fuel tank is width w = 19 cm, length L = 12 cm and breadth d = 18 cm. The height of the fuel between the plates is h = 7 cm. A battery with e.m.f of 12 v is connected across the plates. For gasoline, dielectric constant K, You can ignore any fringing effects.…
What is the force acting on a 1.6(10)–19 C charge when it is moving at 3(10)7m/s perpendicular to a magnetic field whose strength is 0.02 T? What is the radius of the particle’s trajectory if the mass of the particle is 3.3(10)-27 kg?