Determine the magnitude and direction of the force on an electron traveling 8.75×10^5m/s horizontally to the east in a vertically upward magnetic field of strength 0.45 Tesla.
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A: Ve = 7.80 x 106 mB = 1.20 x 10-5 T
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A: Velocity of electron (v) = 3.90 ✕ 103 m/s Magnetic field (B) = 1.20 TForce(F) = 1.40 ✕ 10−16 N
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Q: 2. A proton moves perpendicular to a uniform magnetic field B at a speed of 1.00 x 107 m/s and…
A: Since we only answer up to 3 sub-parts, we’ll answer the first 3. Please resubmit the question and…
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A: Step 1:
Q: A proton travels through uniform magnetic and electric fields. The magnetic field is in the negative…
A: the force acting on a charged particle due to the magnetic field is given by FB=q(v×B) where q is…
Q: A proton travels through uniform magnetic and electric fields. The magnetic field is in the negative…
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Q: A proton travels through uniform magnetic and electric fields. The magnetic field is in the negative…
A:
Determine the magnitude and direction of the force on an electron traveling 8.75×10^5m/s horizontally to the east in a vertically upward magnetic field of strength 0.45 Tesla.
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- An electron moves in a circular path perpendicular to a magnetic field of magnitude 0.280 T. If the kinetic energy of the electron is 2.60 x 10-19 J, find the speed of the electron and the radius of the circular path. (a) the speed of the electron X Your response differs from the correct answer by more than 10%. Double check your calculations. m/s (b) the radius of the circular path X Your response differs from the correct answer by more than 10%. Double check your calculations. μmAS0.0-g metal ball having net charge Q = 5.50 µC is thrown out of a window horizontally north at a speed v = 22.0 m/s. The window is at a height h = 18.0 m above the ground. A uniform, horizontal magnetic field of magnitude B = 0.0100 T is perpendicular to the plane of the ball's trajectory and directed toward the west. (a) Assuming the ball follows the same trajectory as it would in the absence of the magnetic field, find the magnetic force acting on the ball just before it hits the ground. (Let the +x-direction be toward the north, the +y-direction be up and the +z-direction be east.) B. (b) Based on the result of part (a), is it justified for three-significant-digit precision to assume the trajectory is unaffected by the magnetic fleld? O Yes O No Explain.An ion carrying a single positive elementary charge (+e) has a mass of 4.22x10-27 kg. It is accelerated through an electric potential difference of 1.29 kV and then enters a uniform magnetic field of B = 0.2 T along a direction perpendicular to the field. What is the radius of the circular path of the ion in the magnetic field? (Units: cm)
- A uniform electric field of 7.42 V/m west and a uniform magnetic field of 7.78 T west exist in a region of space. Find the magnitude of the acceleration of an electron moving with velocity 1.52 m/s north in this region. Answer in units of m/s2.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 Na) At what speed (in m/s) will a proton move in a circular path of the same radius as an electron that travels at 7.45 ✕ 106 m/s perpendicular to the Earth's magnetic field at an altitude where the field strength is 1.25 ✕ 10−5 T? b) What would the radius (in m) of the path be if the proton had the same speed as the electron? c) What would the radius (in m) be if the proton had the same kinetic energy as the electron? d) What would the radius (in m) be if the proton had the same momentum as the electron?