If the changing voltage is now removed and the dielectric is pulled out capacito

Question

If the changing voltage is now removed and the dielectric is pulled out capacitor is the work done in pulling out the dielectric very slowly at constant speed? An alpha particle trapped in the earth's van Allen belt is gyrating in a helical path around magnetic lines of force having a strength B = 4 times 10^8 t if the alpha particles have's KE of 150 mev and come in at an angle of 53 degree to the direction of the magnetic field B, find a) the total speed of the protons their transverse and parallel components of velocity to the field B. b) the radius of the helix c) the angular frequency and period of the transverse circular d) the pitch of the helical motion e) If we treat the distance between Earth pales as 1.9 times 10^4 km and assume a constant parallel velocity to the field what is the time for alpha particles to go between poles?

Explanation / Answer

given., B = 4*10^-5 k T
v = vo(cos(53)k + sin(53) i) [ initial velocity ]

now, hence, only the component of velocity perpendicular to the magnetic field is responsible for gyration
and 150 MeV = 0.5mvo^2
m = alpha particle = 4u = 4*1.602*10^-27 kg
150*10^6*1.6*10^-19 = 0.5*4*1.602*10^-27*Vo^2
Vo = 8.6548*10^7 m/s

so, velocity compoennt perpendicular to magentic field = Vosin(53) = 6.912*10^7 m/s

a) Total speed = 8.6548*10^7 m/s
b) radius = r
mv^2/r = qvB
r = mv/qB = 4*1.602*10^-27*6.912*10^7 / 4*1.6*10^-19 * 4*10^-5 = 17301.6 m
c) w = 2*pi/T = 2*pi*v/2*pi*r = v/r = 3995 rad/s
d) pitch = vocos(53)*2*pi*r/vosin(53) = 2*pi*r/tan(53) = 81876.697 m
e) d = 1.9*10^4 km
T = 2*pi*r/vocos(53)
v' = Vocos(53)
T' = 1.9*10^4*10^3/Vocos(53) = 0.364 s

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