Problem 3 The electroms in the beam of a television tube have an energy of 20 ke
ID: 2030989 • Letter: P
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Problem 3 The electroms in the beam of a television tube have an energy of 20 keV. The tube is orientod so that the electrons move horizontally from south to morth. The vertical component of the Earth's magnetic field at the location of the television has a magnitude of 48.0 HT and is pointing down. a) In which direetion does the force on the electrons act (enter N for north, S foe Neglect a possible horlacetal composent of the magnetic field E for East, or W for West)? b) What is the magnitude of the acceleration due to the vertical component of the Earth's magnetic field of an electron in the beam? Submit er Tries 06 If the in lination of the earth's magnetic field near the TV is 47deg, calculane the magnitulc ofthe f ecs on do electrons dae to he horinnt I component of the Earth's magnetk field. ta ne Tries 06 Dus Monday March 26 11:59 am (DT) Problem 4Explanation / Answer
magnetic force Fb = q*( V X B ) x = cross product
kinetic energy K = (1/2)*m*v^2
given K = 20 KeV = 20*10^3*1.6*10^-19 J
20*10^3*1.6*10^-19 = (1/2)*9.1*10^-31*v^2
v = 83.8*10^6 m/s
velocity v is along north ( j)
magnetic field B is pointing dowm ( - k)
charge q = -1.6*10^-19 C
magnetic force Fb = -1.6*10^-19*( 83.8*10^6 j X 48*10^-6 -k )
Fb = 1.6*10^-19*83.8*10^6*48*10^-6 i
Fb = 6.43*10^-16 N i
(a)
direction East E
(b)
magnitude Fb = 6.43*10^-16 N
acceleration a = Fb/m = (6.43*10^-16)/(9.1*10^-31)
acceleration a = 7.06*10^14 <<<----------ANSWER
(c)
tantheta = Bv/Bh
Bh = Bv*tantheta
horizontal component of magentic field is along noth (j)
Bh = 48*10^-6*tan47 j
Fb = -1.6*10^-19*( 83.8*10^6 j X 48*10^-6*tan47 j )
Fb = 1.6*10^-19*83.8*10^6*48*10^-6 ( j X J ) since j X j = 0
Fb = 0
DONE please check the answer. any doubts post in comment box
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