Can someone resolve the (c) part, I got 3.74*10^-16 N but it is wrong Problem 3
ID: 779185 • Letter: C
Question
Can someone resolve the (c) part, I got 3.74*10^-16 N but it is wrong Problem 3 The electrons in the beam of a television tube have an energy of 18 keV. The tube is oriented so that the electrons move horizontally from south to north. The vertical component of the Earth's magnetic field at the location of the television has a magnitude of 53.0 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. Submit Answer Tries 0/6 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 Answer Tries 06 c) If the inclination of the earth's magnetic field near the TV is 29deg, calculate the magnitude of the force on the electrons due to the horizontal component of the Earth's magnetic field Submit Answer Tries 0/6 Due Monday March 26 11:59 am (EDTExplanation / Answer
a) East
b) let v is the speed of the elctron.
use, KE = (1/2)*m*v^2
==> v = sqrt(2*KE/m)
= sqrt(2*18*10^3*1.6*10^-19/(9.1*10^-31))
= 7.96*10^7 m/s
force acting on the elctron, F = q*v*B*sin(theta)
m*a = q*v*B*sin(90)
a = q*v*B/m
= 1.6*10^-19*9.56*10^7*53*10^-6/(9.1*10^-31)
= 8.91*10^14 m/s^2
c) F = q*v*Bh*sin(0)
= 0 (since v and B are parallel)