(a) What is the speed of the ? -particle? m/s (b) What is the magnitude of the m

Question

(a) What is the speed of the ?-particle?
m/s

(b) What is the magnitude of the magnetic force on it?
N

(c) What is the radius of its circular path?
m

An ?-particle has a charge of +2e and a mass of 6.64 x 10^6 V and then enters a uniform magnetic field whose magnitude is 1.60 T. The ?-particle moves perpendicular to the magnetic field at all times. (a) What is the speed of the ?-particle? m/s (b) What is the magnitude of the magnetic force on it? N (c) What is the radius of its circular path? m x 10^-27 kg. It is accelerated from rest through a potential difference that has a value of 1.20

Explanation / Answer

W = K.E...

q*V = (0.5*m*v^2)

(2*1.6*10^-19*1.2*10^6) = 0.5*6.64*10^-27*v^2.....
3.84*10^-13 = 3.32e-27*v^2

v = sqrt((3.84*10^-13)/(3.32e-27))

v = 10.75*10^6 m/sec...

B) F = q*v**sin(90)=

F = q*v*B = (2*1.6*10^-19*10.75*10^6*1.6)= 5.504e-12 N

C) r = (m*v)/(q*B)= (6.64*10^-27*10.75*10^6)/(2*1.6*10^-19*1.6) = 0.1394 m = 13.94 cm

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