Problem 9 The electrons in the beam of a television tube have an energy of 7 keV

Question

Problem 9 The electrons in the beam of a television tube have an energy of 7 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 46.0 mu 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 die magnetic field b) What is die 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 59 deg. calculate the magnitude of the force on the electrons due to the horizontal component of the Earth's magnetic field.

Explanation / Answer

Note that

v = sqrt(2KE/m)

Here, KE = 7 keV = 1.1214E-15 J

As m = 9.11E-31 kg, then,

v = 4.962E7 m/s

Now, note that

F = q v B = m a

---> a = q v B /m

Thus, plugging in our values,

a = 4.01E14 m/s^2   [ANSWER, PART B]

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If the inclination of the magnetic field is 59 degrees, then this will diminish by a factor of cos59:

a = 2.07E14 m/s^2

Multiplying by the mass of the electron,

F = 1.88E-16 N   [ANSWER]

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