A charged particle of mass m3.8 x 10-6 kg moves through a 0.8-m long velocity se
ID: 779003 • Letter: A
Question
A charged particle of mass m3.8 x 10-6 kg moves through a 0.8-m long velocity selector as shown in the figure below E, B A top view of the problem where q enters from the left. The magnetic field B points everywhere out of the plane and the electric field E points up only in the velocity selector After leaving the velocity selector, the particle moves along a semicircular path until it strikes a phosphorescent screen. If the entire motion (from start to end) takes T 2.0-seconds, determine the charge of the particle. Assume that B0.3 T and E5.0 V/m and assume that the entire motion takes place in deep space.Explanation / Answer
speed of charged particlce = v
from the figure the magentic force immediately after leaving the velcoity selector is directed upwards
In the velocity the magnetic force is upwards
by right hand the charge is negative
In the veleocity selector the charge moves straight
the net force on charge = 0
Fnet = 0
Fb + Fe = 0
magnetic force Fb = q*v*B ( upward)
electric force Fe = -E*q ( down ward)
q*v*B = E*q
speed v = E/B = 5/0.3 = 50/3 = 16.7 m/s
time taken for the charge to move in velocity selector t1 = d/v = 0.8/16.7 = 0.048 s
In the magnetic field
tiem spent t2 = T - t1 = 2 - 0.048 = 1.952 s
distance traveled = pi*r = v*t2
pi*r = 16.7*1.952
r = 10.4 m
In magnetic field
Fb = Fc
q*v*B = m*v^2/r
radius of circular path r = mv/qB
charge q = mv/(rB)
charge q = (3.8*10^-6*16.7)/(10.4*0.3) = 2.03*10^-5 C
charge q = -2.03*10^-5 C
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