Suppose the electric field between the plates in Fig. 27.24 is 1.88 × 104 V/m and the magnetic field in both regions is 0.682 T. If the source contains the three isotopes of krypton. 82Kr, 84Kr, and 86Kr, and the ions are singly charged, find the distance between the lines formed by the three isotopes on the particle detector. Assume the atomic masses of the isotopes (in
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- A superconducting wire of diameter 0.25 cm carries a current of 1000 A. What is the magnetic field just outside the wire?arrow_forwardCosmic 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.arrow_forwardA 150-V battery is connected across two parallel metal plates of area 28.5 cm2 and separation 8.20 mm. A beam of alpha particles (charge +2e, mass 6.64 x 10-27 kg) is accelerated from rest through a potential difference of 75 kV and enters the region between the plates perpendicular to the electric field. What magnitude and direction of magnetic field are needed so that the alpha particles emerge undeflected from between the plates?arrow_forward
- An electron starts from rest near the negative vertical plate of a set of parallel plates and accelerates towards the positive plate through 150 V. After crossing this distance the electron passes through a hole. The electron then moves through a horizontal set of parallel plates where the top plate is negative. These horizontal plates are separated by 2.0 cm and have a potential difference of 240 V. The electron passes through these plates undeflected due to a magnetic field perpendicular to the page. The electron then passes out of the plates where it is acted upon by the same uniform magnetic field. Find the magnitude and direction of the uniform magnetic field and the radius of the circular path of the electron. Draw the complete path of the electron.arrow_forwardA 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.arrow_forwardA pellet which holds a charge of 10 coulombs is moving upwards (+Y) and driven by an electric field in the same direction with a magnitude of 50 V/m. There is a magnetic field with a magnitude of 25 Tesla’s pointing downwards (-Y). How fast does the pellet need to be going for the magnetic force to cancel the electric force? Group of answer choices 0.5 m/s 2 m/s 4 m/s The forces will cancel out at any speed the pellet may be travelling In this case, the electric force cannot be canceled out by the magnetic forcearrow_forward
- Trace the trajectory of the electron in the magnetic field. Show the direction of the electron in the field by using an arrow. Determine the diameter of the circle that the electrion makes in micrometers. v = 3*10^4 m/s B = 4.0*10^-2 Tarrow_forwardAn ion source is producing 6Li ions, which have charge +e and mass 9.99 × 10-27 kg. The ions are accelerated by a potential difference of 12 kV and pass horizontally into a region in which there is a uniform vertical magnetic field of magnitude B = 1.4 T. Calculate the strength of the smallest electric field, to be set up over the same region, that will allow the 6Li ions to pass through undeflected.arrow_forward8. A beam of electrons is accelerated through a potential difference of 9.0 kV before entering a velocity selector. If the B-field of the velocity selector is perpendicular to the velocity and has a value of 0.04 T, what value of the E-field is required (in the magnetic field region) if the particles are to be undeflected? V/m ceil6 ceilo il6 cci16 c 6 cci16 ceil cil6 ccil6 ccarrow_forward
- The beam of alpha-particles (m = 4.7 × 10−27 kg, q = 6.2 × 10−19 C) bends through a 180-degree region with a uniform magnetic field of 4.2 T. How much time does it take the alpha-particles to traverse the uniform magnetic field region?arrow_forwardUse the following constants if necessary. Coulomb constant, k = 8.987×10^9 N⋅m^2/C^2 . Vacuum permitivity, ϵ0= 8.854×10^−12 F/m. Magnetic Permeability of vacuum, μ0 = 12.566370614356×10^−7 H/m. Magnitude of the Charge of one electron, e = −1.60217662×10^−19 C. Mass of one electron, m_e = 9.10938356×10^−31 kg. Unless specified otherwise, each symbol carries their usual meaning. For example, μC means microcoulomb .arrow_forwardUse the following constants if necessary. Coulomb constant, k = 8.987×10^9 N⋅m^2/C^2 . Vacuum permitivity, ϵ0= 8.854×10^−12 F/m. Magnetic Permeability of vacuum, μ0 = 12.566370614356×10^−7 H/m. Magnitude of the Charge of one electron, e = −1.60217662×10^−19 C. Mass of one electron, m_e = 9.10938356×10^−31 kg. Unless specified otherwise, each symbol carries their usual meaning. For example, μC means microcoulomb .arrow_forward
- Physics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage Learning