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College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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101. An electron of mass mmm and charge −e is moving through a uniform magnetic field B→=(Bx,0,0) in vacuum. At the origin, it has velocity v→=(vx,vy,0), where vx>0 and vy>0. A screen is mounted perpendicular to the x axis at a distance D from the origin.
*Throughout, you can assume that the effect of gravity is negligible.
What is the angular velocity ω of the electron associated with the circular component of its motion? Express the magnitude of the angular velocity in terms of Bx, the magnitude of the electric charge e, and other known quantities.
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B
screen"
Transcribed Image Text:y
B
screen
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- Q8.2arrow_forwardA particle of mass m=3.1×10−5kg and charge q=43μC moves with a velocity v⃗ =−2.1×104ms−1 i^ along the x axis. a) Find the magnitude of magnetic field created by this charged particle at a distance d=37cm from the x axis. Draw a schematic diagram. b) Now, if the charged particle moves in an uniform external magnetic field, B⃗ =−3.8T k^ , then calculate the magnetic force experienced by the charged particle in unit vector notation. x component of the magnetic force y component of the magnetic force z component of the magnetic force c) What will be the shape of the path of the charged particle moving in the uniform external magnetic field. Draw a schematic diagram and calculate a geometric quantity that can help you explain the path. d) Along with the external magnetic field, if we now apply an external electric field on the charged particle in a particular direction - the particle moves in a stright line. Explain why. And calculate the external electric field in unit vector notation.…arrow_forwardThe aurora is caused when electrons and protons, moving in the earth’s magnetic field of ≈5.0×10^−5 T, collide with molecules of the atmosphere and cause them to glow. a. What is the frequency of the circular orbit for an electron with speed 1.0×10^6 m/s? Assume that the electron moves in a plane perpendicular to the magnetic field. b. What is the frequency of the circular orbit for a proton with speed 5.0×10^4 m/s?arrow_forward
- A-4.80 uC charge is moving at a constant speed of 6.80 x 105 m/s in the +x-direction relative to a reference frame. At the instant when the point charge is at the origin, what is the magnetic field vector it produces at the following points a point (x 0.500 m, y = 0.500 m, z 0) Select one: B = -1.29x10-77 B = -4.62x10-7 R B = -7.15x10-7 R B = -2.12x10-7↑ B = 3.17x10 6arrow_forwardSuppose polarized light passes through a polarizing filter with its axis at an angle of 19° to the direction of polarization. To what percent of its original value is the intensity of the polarized light reduced?arrow_forwardQ1) Example 11.1: Page No. 498: (Practice) An alpha-particle of charge 3.2 x 10 19 C moves through a uniform magnetic field of 1.5 T. What is the maximum magnetic force on the alpha-particle when it is moving with a speed of 5.0 x 10 m/s? Q2) Problem 23: Page No. 525: (Practice) An electron moving at 4.00 x 10 m/s in a 1.25-T magnetic field experiences a magnetic force of 1.40 x 10 N. What angle does the velocity of the electron make with the magnetic field? Q3) Problem 33: Page No. 526: An alpha-particle (m-6.64x10"kg, q=3.2×101°C) travels in a circular path of radius 25 cm in a uniform magnetic field of magnitude 1.5 T. (a) What is the speed of the particle? (b) What is the kinetic energy in electron volts? (1 electron-volt = 1.602x10 9 joule) Q4) Problem 25: Page No. 525: (Practice) A cosmic-ray electron moves at 7.5x10 m/s perpendicular to Earth's magnetic field at an altitude where the ficld strength is 0.1 G. What is the radius of the circular path the electron follows? F00…arrow_forward
- The rod shown in the accompanying figure is moving through a uniform magnetic field of strength B = 0.85 T with a constant velocity of magnitude v = 4.5 m/s. What is the potential difference between the ends of the rod? Which end of the rod is at a higher potential? T. 5.0 cm Hint a. The magnitude of the potential difference between the ends of the rod is V. b. The Select an answer v of the rod is at a higher potential. Hint for (b)arrow_forwardNonearrow_forwardReview I Constants What is the largest possible magnitude of the acceleration of the electron due to the magnetic field? An electron moves at 2.20 x 10° m/s through a region in which there is a magnetic field of unspecified direction and magnitude 7.00 x 10-2 T. Express your answer with the appropriate units. HẢ ? Value Units a = Submit Request Answer Part B What is the smallest possible magnitude of the acceleration of the electron due to the magnetic field? Express your answer with the appropriate units. HẢ Value Units a = Submit Request Answer Part C 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? Express your answer in degrees to three significant figures.arrow_forward
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