Question

The electrons in the beam of a television tube have an energy of 13 keV. The tube is oriented so that the electrons move horizontally from north to south. The vertical component of the Earth's magnetic field at the location of the television has a magnitude of 42.0 ?T and is pointing down.

a) In which direction does the force on the electrons act (enter N for north, S for South, E for East, or W for West)?  
Neglect a possible horizontal component of the magnetic field.

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?

c) If the inclination of the earth's magnetic field near the TV is 51deg, calculate the magnitude of the force on the electrons due to the horizontal component of the Earth's magnetic field.

Answer 1

magnetic force Fb = q*( V X B )    x = cross product

kinetic energy K = (1/2)*m*v^2

given K = 13 KeV = 13*10^3*1.6*10^-19 J

13*10^3*1.6*10^-19 = (1/2)*9.1*10^-31*v^2

v = 67.61*10^6 m/s

velocity v is along north ( -j)

magnetic field B is pointing down ( - k)

charge q = -1.6*10^-19 C

magnetic force Fb = -1.6*10^-19*( 67.61*10^6 -j X 42*10^-6 -k )

Fb = 1.6*10^-19*67.61*10^6*42*10^-6 (-i)

Fb = -4.54*10^-16 N i

(a)

direction West W

(b)

magnitude Fb =   4.54*10^-16 N

acceleration a = F/m = 4.54*10^-16/(9.1*10^-31)

acceleration a = 4.98*10^14 m/s^2    <<<<<----------ANSWER

(c)

tantheta = Bv/Bh

Bh = Bv*tantheta

horizontal component of magentic field is along north (j)

Bh = 48*10^-6*tan47 j

Fb = -1.6*10^-19*( 67.61*10^6 j X 42*10^-6*tan51 j )  

Fb = 1.6*10^-19*67.61*10^6*42*10^-6        since j X j = 0

Fb = 0    <<<<<--------ANSWER