From which direction (left or right) would a proton need to enter in order to pass through the velocity selector below? If the magnetic field produced is 2.67 × 10-5 T and it is possible to shoot protons travelling at 2.0% of the speed of light through the filter, what is the required strength of the electric field? d= 2.5 cm x x x x x x
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- A velocity selector uses a fixed electric field of magnitude E and the magnetic field is varied to select particles of various speeds. If the electric field strength is 8.4 x 10oª N/C, what should be the value of the magnetic field (in tesla) to select protons of velocity 4.2 x 105 m/s? Mass of proton = 1.67 x 10-27 kg. Charge of proton = 1.6 x 10-19 C.Determine the velocity of a beam of electrons that goes un-deflected when moving perpendicular to an electric and a magnetic field. E and B are also perpendicular to each other and have magnitudes 7.7 x103 V/m and 7.5x10-3T, respectively.-25 1.58 x 10 A phosphate molecule PO of mass m = difference |AV| = 4.35 V and enters the mass spectrometer with magnetic field of magnitude |B| = 0.090 T. The particle then follows a semi-circular path shown in the picture before kg is accelerated across the potential reaching the detector located to the left of its entry point. В d Δν a) Sketch the described situation (you can start with the figure shown in the problem) CLEARLY indicating the direction of the magnetic field inside the mass spectrometer b) Determine the speed of the particle as it enters the region of magnetic field. c) Determine the distance between the entry point of the particle and the detector.
- 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 forceAn electric field with initial magnitude 210 V/m directed into the page and increasing at a rate of 13.0 V/(ms) is confined to the area of radius R = 15.0 cm in the figure below. If r = 45.0 cm, what are the magnitude and direction of the magnetic field at point A? magnitude direction ---Select--- Ø Ø A O RCrossed electric and magnetic fields have magnitudes of 4.65 V/m and 74.5 mT. Find the speed at which electrons moving perpendicular to both fields can travel through them without being deflected?
- A beam of electrons with velocity vx = 5 x105 m/s is introduced into a uniform magnetic field Bz = 30 mTesla. An electric field Ey is applied perpendicular to both the initial velocity and the magnetic field. What value of electric field must be applied to leave the electron beam undeflected?ASAPwrong