College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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a proton is moving through a magnetic field in a direction perpendicular to the field as shown below. The magnetic field has a magnitude of B= 0.13 T and the charge experiences a force of F= 9.98 x 10^-14 N. How fast and in what direction is it moving? The charge of a proton is qp= +1.6x 10^-19 C.
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- 2. A particle with charge 2.34 x 10^-6 C moves at 8.64 x 10^6 m/s through a magnetic field of strength 2.97 T. The angle between the particle s velocity and the magnetic field direction is 33.0 degrees and the particle undergoes an acceleration of 12.2 m/s^2. What is the particle s mass? Question 2 options: 15.9 kg 2.68 kg 2.4 kg 9.8 kgarrow_forwardA particle with positive charge q = 3.52 x 1018 C moves with a velocity v = (5î + 4ĵ – k) m/s through a region where both a uniform magnetic field and a uniform electric field exist. (a) Calculate the total force on the moving particle, taking B = (3î + 2ĵ + k) T and E = (3î - j - 2k) V/m. (Give your answers in N for each component.) Ey = Ey = F, = (b) What angle does the force vector make with the positive x-axis? (Give your answer in degrees counterclockwise from the +x-axis.) ° counterclockwise from the +x-axis (c) What If? For what vector electric field would the total force on the particle be zero? (Give your answers in V/m for each component.) E, = V/m E, = V/m E, = V/marrow_forwardA proton is moving under the combined influence of an electric field (E = 1747 V/m) and a magnetic field (B = 1.37 T), as shown in the figure. Assuming the proton is moving in the direction shown in the figure with speed 229 m/s at the instant it enters the crossed fields, what is the acceleration of the proton? (in m/s^2)arrow_forward
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