An ionized deuteron (a hydrogen atom with a neutron in the nucleus) passes to the east into a velocity selector built with an electric field of E = 11.5 kV/m and a magnetic field of B = 55 mT. How fast is the ion moving if it goes straight through the velocity selector?

Problem 5: An electron (q= -e = -1.60 × 10-19 C and me = 9.11 × 10-31 kg) in a cathode-ray tube is accelerated through a potential difference of 10 kV. Thus, the electron leaves the cathode-ray tube with a speed v = 5.93 × 107 m/sec. The electron then passes through the 2.0 cm wide region of uniform magnetic field (shown in the Figure above on the right). What field strength will deflect the electron by 10°? Answer: B = 0.0029 T. OV 10 kV 2.0 cm 10°

A proton, that is accelerated from rest through a potential of 55 kV enters the velocity filter, consisting of a parallel-plate capacitor and a magnetic field, shown below. What is the proton's speed? Take mp 1.67 × 10-27 kg and e = = 1.6 × 10-¹⁹ C. P = E The magnetic field, B = (x) B (x) The speed, v = The E-field between the plates is 2.2 × 105 N/C. What B-field is required so that the protons are not deflected? Units Select an answer ✓ X Units Select an answer ✓

A 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 force

A velocity selector consists of electric and magnetic fields described by the expressions  E= Ek  and  B= B ĵ, with B = 17.0 mT. Find the value of E such that a 800-eV electron moving in the negative x direction is undeflected. kV/m

A proton is moving under the combined influence of an electric field (E = 2569 V/m) and a magnetic field (B = 2.45 T), as shown in the figure. Assuming the proton is moving in the direction shown in the figure with speed 255 m/s at the instant it enters the crossed fields, what is the acceleration of the proton? (units are in m/s^2)

A particle passes through a mass spectrometer as illustrated in the figure below. The electric field between the plates of the velocity selector has a magnitude of 8060 V/m, and the magnetic fields in both the velocity selector and the deflection chamber have magnitudes of 0.0920 T. In the deflection chamber the particle strikes a photographic plate 39.6 cm removed from its exit point after traveling in a semicircle. The diagram depicts a velocity selector. The plane of the page is filled with symmetrically arranged crosses labeled vector B0, in above a horizontal plate, and vector Bin below the horizontal plate. Below the horizontal plate, two vertical plates are placed parallel to each other in the right position of the field. The left plate is positively charged and the right plate is negatively charged. The electric field vector E points from the positive plate to the right toward the negative plate. A positively charged particle of charge q and velocity vector v pointing upward…

A uniform electric field of 7.42 V/m west and a uniform magnetic field of 7.78 T west exist in a region of space. Find the magnitude of the acceleration of an electron moving with velocity 1.52 m/s north in this region. Answer in units of m/s2.

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.

The Bainbridge spectrometer works by ionizing particles, sending them through a velocity selector, and then detecting a given charge-to-mass ratio based on its trajectory in a uniform magnetic field (See Ch 11.7 in your book). Suppose your spectrometer works by altering the electric field of the velocity selector to scan for ions. If a singly-charged ion of mass 140.9 amu required an electric field of 1.9 N/C, what electric field would be necessary to detect a doubly-charged ion of mass 161.9? Give your answer in units of N/C.