2. -4 points My Notes Ask Your Teacher An electron is accelerated from rest by a
ID: 1401271 • Letter: 2
Question
2. -4 points My Notes Ask Your Teacher An electron is accelerated from rest by a potential difference of 350 V. It then enters a uniform magnetic field of magnitude 200 mT with its velocity perpendicular to the field. (a) Calculate the speed of the electron as it enters the magnetic field. (b) Calculate the radius of its path in the magnetic field. (c) Suppose another elementary partical, a muon (mass 206.8 x mass of electron and same charge) is accelerated from rest in the same apparatus. Calculate the speed of the muon as it enters the magnetic field. (d) Calculate the radius of the path of the muon in the magnetic field.Explanation / Answer
here,
potential difference , V = 350 V
(a)
let the speed of electron be v
using work energy theorm
work done = change in kinetic energy
V * e = 0.5 * me * v^2
350 * 1.6*10^-19 = 0.5 * 9.1*10^-31 * v^2
v = 1.11 * 10^7 m/s
the speed of electron as it enters the magnetic feild is 1.11 * 10^7 m/s
(b)
magnetic feild , B = 200 mT
B = 0.2 T
radius , r = me*v/(B * e)
r = 9.1*10^-31 * 1.11 * 10^7/(0.2 * 1.6*10^-19)
r = 3.16*10^-4 m
radius of the path in magnetic feild is 3.16*10^-4 m
(c)
mass of muon , M = 206.8 * me
let the speed of electron be u
using work energy theorm
work done = change in kinetic energy
V * e = 0.5 *206.8 * me * u^2
350 * 1.6*10^-19 = 0.5 * 206 .8 * 9.1*10^-31 * u^2
u = 7.72 * 10^5 m/s
the speed of muon as it enters the magnetic feild is 7.72 * 10^5 m/s
(d)
magnetic feild , B = 200 mT
B = 0.2 T
radius , R = 206.8 * me*u/(B * e)
R = 206.8 * 9.1*10^-31 * 7.72 * 10^5/(0.2 * 1.6*10^-19)
R = 4.54*10^-3 m
radius of the path of muon in magnetic feild is 4.54*10^-3 m
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