A charged particle (m = 5.0 g, q = -70 mu C) moves horizontally at a constant sp
ID: 1453488 • Letter: A
Question
A charged particle (m = 5.0 g, q = -70 mu C) moves horizontally at a constant speed of 30 km/s in a region where the free fall gravitational acceleration is 9.8 m/s^2 downward, the electric field is 700 N/C upward, and the magnetic field is perpendicular to the velocity of the particle. What is the magnitude of the magnetic field in this region? An electron follows a circular path (radius = 15 cm) in a uniform magnetic field (magnitude = 3.0 G). What is the period of this motion? A proton with a kinetic energy of 0.20 keV follows a circular path in a region where the magnetic field is uniform and has a magnitude of 60 mT. What is the radius of this path?Explanation / Answer
15) Net force along veritcal direction is
weight of particle + electric force = magnetic force
mg +qE = qvB
(-700*70*10^-6) - (5*10^-3*9.8) = (-70*10^-6*30*10^3*B)
B = 47 mT
16) r =15 m , B = 3G = 3*10^-4 T , electron mass m = 9.1*10^-31 kg , q = 1.6*10^-19 C
Centripetal force = magnetic force
mv^2/r = qvB
mv/r = qB
v/r = w =2pi/T
m*2pi/T = qB
(9.1*10^-31*2*3.14)/T = (1.6*10^-19*3*10^-4)
T =0.12 us
17) K =0.2 keV , B = 60 mT , m =1.67*10^-27 kg ,
K =(1/2)mv^2
0.2*10^3*1.6*10^-19 = 0.5*1.67*10^-27*v^2
v = 195763.5 m/s
Centripetal force = magnetic force
mv^2/r = qvB
mv/r = qB
(1.67*10^-27*195763.5)/r = (1.6*10^-19*0.06)
r = 3.4 cm
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