a=5 b=4 m=75 n=30 Say that you have a charged particle moving through a velocity
ID: 2062285 • Letter: A
Question
a=5 b=4 m=75 n=30Say that you have a charged particle moving through a velocity selector that uses perpendicular electric and magnetic fields. If (a) is even, then the particle is a proton; if (a) is odd, then the particle is an electron. At the beginning, the particle is moving North at a speed of (m) km/s through a magnetic field that has a magnitude of (b/100) T. If (b) is even, then the magnetic field points UP; if (b) is odd, then the magnetic field points DOWN. Note: you may completely ignore gravity in this problem.
(a) What is the direction of the magnetic force on the charge?
(b) What electric field (magnitude and direction) would be required to cancel out the magnetic force on the charge?
(c) If the same charge were moving through the same electric field and magnetic field, but were moving at a speed of (n) km/s, what would the acceleration be? (Give both the magnitude of the acceleration and the direction of the acceleration.)
Explanation / Answer
a) is odd, then the particle is an electron if (b) is even, then the magnetic field points UP; and f = q(v cross b) so direction = westward ie along -x
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