An electron moves at 1.40 * 106 m/s through a region in which there is a magnetic field of unspecified direction and magnitude 7.40 * 10-2 T. (a) What are the largest and smallest possible magnitudes of the acceleration of the electron due to the magnetic field? (b) If the actual acceleration of the electron is one-fourth of the largest magnitude in part (a), what is the angle between the electron velocity and the magnetic field?
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An electron moves at 1.40 * 106 m/s through a region in which there is a magnetic field of unspecified direction and magnitude 7.40 * 10-2 T. (a) What are the largest and smallest possible magnitudes of the acceleration of the electron due to the magnetic field? (b) If the actual acceleration of the electron is one-fourth of the largest magnitude in part (a), what is the angle between the electron velocity and the magnetic field?
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- An alpha particle has a positive charge that is double the charge of a proton, and a mass of 6.64 ✕ 10−27 kg. At a particular instant, it is moving with a speed of 5.00 ✕ 106 m/s through a magnetic field. At this instant, its velocity makes an angle of 62° with the direction of the magnetic field at the particle's location. The magnitude of the field is 0.230 T. (a) What is the magnitude of the magnetic force (in N) on the particle? (b) What is the magnitude of the particle's acceleration (in m/s2) at this instant?What is the force on an electron moving in the negative z direction with a speed of 4.1 x 106 m/s in a magnetic field of (-3.6,5.2,-2.8) mT? Give the answer as a vectorAn electron beam is directed horizontally into a region where there is both an electric field and a magnetic field. The electric field points upward with a magnitude EE = 2.9 N/C, as shown in the figure. While moving through the region, the electron beam remains directed in a straight, horizontal line with a speed of 460 m/s. 1.) Express the magnitude of the electric force using the electric field E and the elementary charge e. 2.) Calculate the numerical value of the magnitude of the electric force in newtons. 3.) Express the magnitude of the magnetic force in terms of the elementary charge ee, electron speed vv, and BB, the magnitude of the magnetic field. 4.) If the electron is continuing in a horizontal straight line, express the magnitude of the magnetic field in terms of vv and EE, B = ? 5.) Calculate the magnitude of the magnetic field in tesla, if the electron continues in a horizontal straight line. 5.)
- A proton travels through uniform magnetic and electric fields. The magnetic field is in the negative x direction and has a magnitude of 2.12 mT. At one instant the velocity of the proton is in the positive y direction and has a magnitude of 2960 m/s. At that instant, what is the magnitude of the net force acting on the proton if the electric field is (a) in the positive z direction and has a magnitude of 3.55 V/m, (b) in the negative z direction and has a magnitude of 3.55 V/m, and (c) in the positive x direction and has a magnitude of 3.55 V/m? (a) Number i (b) Number i (c) Number i Units Units Units >A 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/mcan you please ans (a) (b) (c)?
- A proton experiences the greatest force as it travels 2.9 x 106 m/s in a magnetic field when it is moving northward. The force is upward and of magnitude 7.2 x10-13 N. What is the magnitude and direction of the magnetic field?A charged particle with a charge to mass ratio of 5.7E8 C/kg travels on a circular path that is perpendicular to a magnetic field whose magnitude is 0.84 T. How much time does it take for the particle to complete one revolution?Problem 2: = A particle with charge +2.0 C moves through a uniform magnetic field. At one instant the velocity vector is v (2î+4ĵ + 6k) m/sec and the magnetic force on the particle is F = (41-20ĵ+ 12k) N. The x and y components of the magnetic field are equal (Bx = By). Thus, the magnetic field is given by B = B×î + B×ĵ + B₂k. What are B× and B₂? Answer: Bx = -3 T and B₂ = −4 T.
- A proton enters a magnetic field in such a way that it is traveling perpendicularly to the magnetic field lines. If the proton moves in a circle of radius 125 mm, and feels a force of 5.88 x 10 -16 N, how fast is the proton traveling?A proton travels through uniform magnetic and electric fields. The magnetic field is in the negative x direction and has a magnitude of 3.15 mT. At one instant the velocity of the proton is in the positive y direction and has a magnitude of 2680 m/s. At that instant, what is the magnitude of the net force acting on the proton if the electric field is (a) in the positive z direction and has a magnitude of 4.98 V/m, (b) in the negative z direction and has a magnitude of 4.98 V/m, and (c) in the positive x direction and has a magnitude of 4.98 V/m? (a) Number 2.14752E-18 (b) Number i (c) Number i Units Units Units N N N