A velocity selector has a magnetic field of magnitude 0.28T in the positive x di
ID: 1971559 • Letter: A
Question
A velocity selector has a magnetic field of magnitude 0.28T in the positive x direction and an electric field of magnitude 0.46 MV/m in the negative z. direction, (a) Draw a vector diagram that includes the vectors for the magnetic field, the electric field, the velocity of the particle, the force of the magnetic field and the force of the electric field. Label all vectors for full credit, (b) Calculate the speed a particle must have to pass through the region containing the magnetic and electric fields undeflected. (c) Calculate the energy a proton must have to pass undeflected. (d) Calculate the energy an electron must have to pass undeflected.Explanation / Answer
Magnetic field B = 0.28 TElectric field E = 0.46 MV/m. = 0.46 x 10 6 V / m (a) The speed of a particle be for it to pass through undeflected v = E / B = 1.642 x 10 6 m / s (b) Kinetic energy must protons have to pass through undeflected K = ( 1/ 2) mv 2 Where m = mass of proton = 1.67 x 10 -27 kg Substitute values we get K = 1.2 x 10 -15 J = ( 2.25 x 10 -15 ) / ( 1.6 x 10 -19 ) eV = 14.08 x 10 3 eV = 14.08keV (c). The kinetic energy must electrons have to pass through undeflected K ' = ( 1/ 2) m' v 2 Where m ' = mass of electron = 9.11 x 10 -31 kg Substitute values we get K' = 1.23 x 10 -18 J = (1.23 x 10 -18 ) / ( 1.6 x 10 -19 ) eV = 7.7 eV Where m ' = mass of electron = 9.11 x 10 -31 kg Substitute values we get K' = 1.23 x 10 -18 J = (1.23 x 10 -18 ) / ( 1.6 x 10 -19 ) eV = 7.7 eV