A proton moves through a magnetic field at 25.7% of the speed of light. At one location, the field has a magnitude of 0.00691 Tand the proton's velocity makes an angle of 123° with the field.Use c = 2.998 × 108 m/s for the speed of light and e = 1.602 × 10-19 C as the elementary charge.What is the magnitude F of the magnetic force acting on theproton at that location?FB =3.918x10-14IncorrectN

The magnitude of the magnetic force on the proton is 7.15 x 10-14 NExplanation:Given:Speed of proton…

Answered: A proton moves through a magnetic field…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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A negative charge of q = -2.2 * 10-17 C and m = 2.3 * 10-26 kg enters a magnetic field B = 1.9 T with initial velocity v = 270 m/s as shown in the attached image. The magnetic field points into the screen. Express the radius R, of the circular motion in terms of the centripetal acceleration a and the speed v. Calculate the numerical value of the radius R in meters.
A particle with a charge of -1.2 x 10-8 C is moving with instantaneous velocityv = (4.2 x 104 m/s) i + (-3.8 x 104 m/s) j. What are the components of the force exerted on this particle by a magnetic field B = (1.4 T) i + (2.4 T) k ?
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/m
An electron encounters an E and B fields.B field is given by B = 0.1 j T.Electron experiences a force F = (9.6 x 10^-14 i – 9.6 x 10^-14 k)Find the electric field encountered by the electron. The velocity of the electron isv = 5 x 10^6 i m/s.
Some charged particle travels at a constant speed of 12e5 m/s to the right into a magnetic field of strength 1.7 T pointed out of the page. This causes the particle to start traveling in a circular arc of radius 0.315 m. What is the charge to mass ratio (q/m) of this particle 5.23e5 C/kg None of the above 2.24e6 C/kg 2.99e6 C/kg 7.47e5 C/kg
A particle with charge 9.49 x 10^-6 C moves at 1.86 x 10^6 m/s through a magnetic field of strength 2.95 T. The angle between the particle s velocity and the magnetic field direction is 48.3 degrees and the particle undergoes an acceleration of 17.4 m/s^2. What is the particle s mass? 2.23 kg 8.7 kg 15.7 kg 22.6 kg
(d) Would the electron undergo the same acceleration? O Yes O No
In a mass spectrometer, a doubly charged ion having a particular velocity is selected by using a magnetic filed of 130 mT perpendicular to an electric field of 6.9 kV/m. The same magnetic field is used to deflect the ion in a circular path with a radius of 93 mm. What is the mass of the ion? The mass, m = Incorrect Units Correct .
A positive charge traveling with a speed of 9,678 m/s enters the deflection chamber of a mass spectrometer. In the deflection, chamber exists a uniform magnetic field of 0.55 T. The charge undergoes a deflection with a radius of curvature that is 0.024 m. If the charge is 1.602 x 10-19 C, what is its mass in atomic mass units*?
Q. 3: A proton moves through a uniform magnetic field given by B = (30 î – 205) mT. At a time t1, the proton has a velocity given by = the proton is FB (Vz î + (2000m/s)ĵ) and the magnetic force of (4 * 10-17N) k. At that instant, what is the velocity vx?
A carbon-14 ion with a charge of +6.408x10^-19 C and a mass of 2.34x10^-26 kg is sent through a mass spectrometer and hits a detector at a point 10.0 cm to the left of where the beam leaves the velocity selector. The velocity selector and the detector are both in a region of magentic field of strength 0.500 T. What is the speed of the particle after it leaves the velocity selecctor? Then, what is the accelerating potential needed to reach the speed found in that part if the ions start from rest? Please also explain and show the steps you used to get there/the physics behind why/how you got to the answer to help me better understand. Thank you soo much. Also, the work and the explanation or most important because I already have the correct answer - I'm just unsure of how to get there.
Q. 2: The velocity of an electron at an instant moving in a magnetic field B = (3.0 î + 4.0ĵ)mT is given by v = (300 î – 400ĵ)km/s. Calculate the magnetic force acting on the electron. What is the radius of the electron orbit?

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