Question

The earth's magnetic field can affect the electron beam in an oscilloscope or a television tube. An electron that has been accelerated through a potential difference of 3.01 104 V has a horizontal initial velocity, directed north. Find the magnetic force on the electron when the earth's field has a horizontal component of 0.250 G and dips 60.0° below the horizontal (values typical in California).

Answer 1

The electron is accelerated through a potential difference of V = 3.01 x 104 V

The earth's field is B = 0.250 G = 0.250 x 1 x 10^-4 T = 0.250 x 10^-4 T

The earth's field is at an angle θ = 60 deg

The kinetic energy of the electron is equal to the work done on it when it is accelerated through a potential difference V.

Therefore,we get

(1/2)mv2 = q*V

v2 = (2q*V/m)

v = (2q* V/m)1/2

where q = 1.6 x 10-19 C and m = 9.1 x 10-31 kg

The magnetic force on the electron is

F = q*v*B * sinθ

F = q*(2q*V/m)1/2 * B*sinθ

F = [(1.6 x 10^-19)^2 x 3.01 x 10^4 x 0.250 x 10^-4 x sin(60)] / (9.1 x 10^-31)

F = 2.12 x 10^-6 N