Question: the solar wind is a thin, hot gas given off by the sun. charged particles in this gas enter the magnetic field of the earth and can experience a magnetic force. suppose a charged particle traveling with a speed of 9.0e6 m/s encounters the earth's magnitude of 1.2e-7 T. assuming that the particle's velocity is perpendicular to the magnetic field at an altitude where the magnitude of 1.2e-7 T. assuming that the particle's velocity is perpendicular to magnetic field find the radius of...Find (a) an electron and (b) a proton
My Approach:
(a) electron
r = (9.11e-31)(9.0e6)/ (1.6e-19)(1.2e-7) = 427 m
(b) proton
r = (1.67e-27)(9.0e6)/ (1.6e-19)(1.2e-7) = 7.83e5 m
Book Answer: I don't have the answer to (a); however, for part (b) it is supposed to be 7.83e8 m. it's the reason why im posthing this question! appreciate the help
2007-02-09 05:47:03 · 2 answers · asked by Jimmy 3 in Science & Mathematics ➔ Physics
Radius of what?
2007-02-10 02:24:32 · answer #1 · answered by Edward 7 · 0⤊ 0⤋
sure, this information is stable sufficient in physics. In around action, centripetal stress is continuous and constantly perpendicular to hurry. So in case you have a difficulty of a relentless speed and a relentless stress perpendicular to the path of action, the opposite end applies - the direction is around. this would possibly not be this form of information that a mathematician want, besides the indisputable fact that that is elementary and sufficient.
2016-09-28 21:22:35 · answer #2 · answered by ? 4 · 0⤊ 0⤋