001 10.0 points An electron in a vacuum is first accelerated by a voltage of 817
ID: 1323998 • Letter: 0
Question
001 10.0 points An electron in a vacuum is first accelerated by a voltage of 81700 V and then enters a region in which there is a uniform magnetic field of 0.334 T at right angles to the direction of the electron's motion. The mass of the electron is 9.11 X 10^-31 kg and its charge is 1.60218 x 10^-19 C. What is the magnitude of the force on the electron due to the magnetic field? Answer in units of N. 002 (part 1 of 4) 10.0 points A uniform magnetic field of magnitude 1.49 T acts in the positive z direction. Find the magnitude of the force exerted by the field on a proton if the proton's velocity is (2.4 Mm/s) i. Answer in units of pN. 003 (part 2 of 4) 10.0 points Find the magnitude of the force if the proton's velocity is (3 Mm/s)j. Answer in units of pN. 004 (part 3 of 4) 10.0 points Find the magnitude of the force for a velocity of (6.6 Mm/s) k. Answer in units of pN. 005 (part 4 of 4) 10.0 points Find the force if the velocity is (2 Mm/s)i + (4 Mm/s)j. Answer in units of pN.Explanation / Answer
energy of the electron on being accelerated = qE = 1.6 x 10^-19 x 81700 J = 1.3 x 10^-14 J
.5 x m x v^2 = 1.3 x 10^-14 J
v = 169.4 x 10^6 m/s
force in magnetic field = q V x B
= qVBsintheta
theta = 90
F = 1.6 x 10 ^-19 x 169.4 x 10^6 x 0.334
F = 9 x 10^-12 N
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