Answer:
a) 1.62*10^{-18}N
b) 2.84*10^{-19}N
c) 1.16*10^{-18}N
Explanation:
The net force is given by the expression:
[tex]\vec{F}=q(\vec{E}+\vec{v}\ X\ \vec{B})[/tex]
By the right hand rule, we know that the direction of the magnetic force is
j X -i = k
[tex]\vec{v}X\vec{B}=vB\hat{k}[/tex]
Hence, the direction of the magnetic force is the +z direction.
(a) E=4.19V/m k
[tex]F=(1.6*10^{-19}C)[4.19\frac{V}{m}+(2540\frac{m}{s})(2.35*10^{-3}T)]=1.62*10^{-18}N[/tex]
where we have used q=9.1*10^{-31}kg.
(b) E=-4.19V/m k
[tex]F=(1.6*10^{-19}C)[-4.19\frac{V}{m}+(2540\frac{m}{s})(2.35*10^{-3}T)]=2.84*10^{-19}N[/tex]
(c) E=4.19V/m i
In this case the net force will have two components:
[tex]F=(1.6*10^{-19}C)[4.19\frac{V}{m}\hat{i}+(2540\frac{m}{s})(2.35*10^{-3}T)\hat{k}]\\\\F=[6.704*10^{-19}\hat{i}+9.55*10^{-19}\hat{k}]N[/tex]
and its magnitude will be:
[tex]F=\sqrt{(6.704*10^{-19})^2+(9.55*10^{-19})^2}=1.16*10^{-18}N[/tex]
hope this helps!!
