Answer:
a) 1.88*10^-18 N
b) 6.32*10^-19 N
c) 1.9*10^-18 N
Explanation:
The total force over the electron is given by:
[tex]\vec{F}=q\vec{E}+q\vec{v}\ X\ \vec{B}[/tex]
the first term is the electric force and the second one is the magnetic force.
You have that the velocity of the electron and the magnetic field are:
[tex]\vec{v}=2190\frac{m}{s}\ \hat{j}\\\\\vec{B}=-5.39*10^{-3}T\ \hat{i}[/tex]
by using the relation j X (-i) =- j X i = -(-k) = k, you obtain:
[tex]\vec{v} \ X\ \vec{B}=(2190m/s)(3.59*10^{-3}T)\hat{k}=7.862\ T m/s[/tex]
a) For an electric field of 3.91V/m in +z direction:
[tex]\vec{F}=q[\vec{E}+\vec{v}\ X\ \vec{B}]=(1.6*10^{-19})[3.91\hat{k}+7.862\ \hat{k}]N\\\\\vec{F}=1.88*10^{-18}N\hat {k}\\\\F=1.88*10^{-18}N[/tex]
b) E=3.91V/m in -z direction:
[tex]\vec{F}=(1.6*10^{-19})[-3.91\hat{k}+7.862\ \hat{k}]N\\\\\vec{F}=6.32*10^{-19}N\hat {k}\\\\F=6.32*10^{-19}N[/tex]
c) E=3.91 V/m in +x direction:
[tex]\vec{F}=(1.6*10^{-19})[3.91\hat{i}+7.862\ \hat{k}]N\\\\\vec{F}=[6.25*10^{-19}\ \hat{i}+1.25*10^{-18}\ \hat{k} ]N\\\\F=\sqrt{(6.25*10^{-19})^2+(1.25*10^{-18})^2}N=1.9*10^{-18}\ N[/tex]
