A negatively charged particle moves in a uniform magnetic field that has a positive x- direction. The magnetic force on the particle is in the negative z-direction. What can you conclude about the z-component of the particle's velocity?
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- A laboratory electromagnet produces a magnetic field of magnitude 1.51 T. A proton moves through this field with a speed of 5.72 x 106 m/s. (a) Find the magnitude of the maximum magnetic force that could be exerted on the proton. N (b) What is the magnitude of the maximum acceleration of the proton? m/s² (c) Would the field exert the same magnetic force on an electron moving through the field with the same speed? Yes No Explain. (d) Would the electron experience the same acceleration? Yes No Explain.A charged particle moves in a region of uniform magnetic field along a helical path (radius= 5.0 cm, pitch = 12 cm, period = 27.6 ms). What is the speed of this particle? V p Pitch -2r- -BTwo particles enter a region in space where there is a constant magnetic field B. Particle 1 has a mass of 2m, speed 3v, and unknown charge of q1. Particle 2 has a mass of m, speed v, and a charge with magnitude q2 = +2e. See figure. While in the field, the radius of motion for particle 1 is 3R and the radius of motion for particle 2 is R. See figure. Find B and q1. The knowns are m, v, R, and e. radius = 3R 3v ● B =? radius= R
- An iron (density ρ) rod with length L, cross sectional area A, spans across two parallel, metal train tracks. The tracks are connected to a power supply and have a potential ∆V across them. Between the tracks are placed magnets such that the B-field points directly upwards with strength B. What is the acceleration of the iron rod be the moment it starts from rest? What will acceleration be as a function of speed as it continues? Assume the contact is frictionless between the tracks and the rod so that no force of friction needs to be overcome. What will the top speed of the rod be under these conditions?(d) Would the electron undergo the same acceleration? O Yes O NoA charged particle moves into a region of uniform magnetic field, goes through half a circle, and then exits that region. The particle is either a proton or an electron (you must decide which). It spends 130 ns in the region. What is the magnitude of B? Give your answer in T. (mproton=1.672x1027 kg, melectron-9.1lx10-3! kg, e-1.602x10-19 C) OB
- In vacuum, magnetic force on a moving charged particle provides an acceleration to the particle. Thus, it also changes the kinetic energy of the particle. O True O False The direction of magnetic force is perpendicular both to the velocity of the particle and to the magnetic field. True O False The net magnetic force acting on any closed current-carrying loop in a uniform magnetic field is zero. True O False All electric field lines are continuous and always form closed loops. OTrue FalseAn electron in a magnetic field moves along a circle with a radius of 25.4 m with a speed that follows: v(t)=V0e−bt where b = 0.56 s-1and V0 = 170 m/s. I need to find the angular acceleration at t=3.7s, but I don't know how I'm supposed to get that becuase the formula that we have says that it is the second derivative of angular position, but when I calculate angular position all I get is 6.57516 radians. I don't understand how I'm supposed to take a derivative of that since it doesn't have any variables. I don't know if it is needed for this, but I already calculated the centripetal acceleration as 18.0446m/s^2