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At a given instant, an electron and a proton are moving with the same velocity in a constant magnetic field. Compare the magnetic forces on these particles. Compare their accelerations.
The magnitude of the proton and electron magnetic forces are the same since they have the same amount of charge. The direction of these forces however are opposite of each other. The accelerations are opposite in direction and the electron has a larger acceleration than the proton due to its smaller mass.
Does increasing the magnitude of a uniform magnetic field through which a charge is traveling necessarily mean increasing the magnetic force on the charge? Does changing the direction of the field necessarily mean a change in the force on the charge?
An electron passes through a magnetic field without being deflected. What do you conclude about the magnetic field?
The magnetic field must point parallel or anti-parallel to the velocity.
If a charged particle moves in a straight line, can you conclude that there is no magnetic field present?
How could you determine which pole of an electromagnet is north and which pole is south?
A compass points toward the north pole of an electromagnet.
A cosmic-ray electron moves at $7.5\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{6}\text{m/s}$ perpendicular to Earth’s magnetic field at an altitude where the field strength is $1.0\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-5}}\text{T}.$ What is the radius of the circular path the electron follows?
4.27 m
(a) Viewers of Star Trek have heard of an antimatter drive on the Starship Enterprise . One possibility for such a futuristic energy source is to store antimatter charged particles in a vacuum chamber, circulating in a magnetic field, and then extract them as needed. Antimatter annihilates normal matter, producing pure energy. What strength magnetic field is needed to hold antiprotons, moving at $5.0\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{7}\text{m/s}$ in a circular path 2.00 m in radius? Antiprotons have the same mass as protons but the opposite (negative) charge. (b) Is this field strength obtainable with today’s technology or is it a futuristic possibility?
(a) An oxygen-16 ion with a mass of $2.66\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-26}}\text{kg}$ travels at $5.0\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{6}\text{m/s}$ perpendicular to a 1.20-T magnetic field, which makes it move in a circular arc with a 0.231-m radius. What positive charge is on the ion? (b) What is the ratio of this charge to the charge of an electron? (c) Discuss why the ratio found in (b) should be an integer.
a. $4.80\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-19}}\phantom{\rule{0.2em}{0ex}}\text{C};$ b. 3; c. This ratio must be an integer because charges must be integer numbers of the basic charge of an electron. There are no free charges with values less than this basic charge, and all charges are integer multiples of this basic charge.
An electron in a TV CRT moves with a speed of $6.0\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{7}\text{m/s},$ in a direction perpendicular to Earth’s field, which has a strength of $5.0\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-5}}\text{T}.$ (a) What strength electric field must be applied perpendicular to the Earth’s field to make the electron moves in a straight line? (b) If this is done between plates separated by 1.00 cm, what is the voltage applied? (Note that TVs are usually surrounded by a ferromagnetic material to shield against external magnetic fields and avoid the need for such a correction.)
(a) At what speed will a proton move in a circular path of the same radius as the electron in the previous exercise? (b) What would the radius of the path be if the proton had the same speed as the electron? (c) What would the radius be if the proton had the same kinetic energy as the electron? (d) The same momentum?
a. $4.09\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{m/s};$ b. $7.83\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{m};$ c. $1.75\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}\text{m/s},$ then, $1.83\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{2}\phantom{\rule{0.2em}{0ex}}\text{m};$ d. $4.27\phantom{\rule{0.2em}{0ex}}\text{m}$
(a) What voltage will accelerate electrons to a speed of $6.00\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-7}}\phantom{\rule{0.2em}{0ex}}\text{m/s}?$ (b) Find the radius of curvature of the path of a proton accelerated through this potential in a 0.500-T field and compare this with the radius of curvature of an electron accelerated through the same potential.
An alpha-particle $(m=6.64\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-27}}\phantom{\rule{0.2em}{0ex}}\text{kg,}$ $q=3.2\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-19}}\phantom{\rule{0.2em}{0ex}}\text{C})$ 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? (c) Through what potential difference must the particle be accelerated in order to give it this kinetic energy?
a. $1.8\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{7}\text{m/s};$ b. $6.8\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{6}\text{eV};$ c. $6.8\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{6}\text{V}$
A particle of charge q and mass m is accelerated from rest through a potential difference V , after which it encounters a uniform magnetic field B . If the particle moves in a plane perpendicular to B , what is the radius of its circular orbit?
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