A) [tex]3.25\cdot 10^7 m/s[/tex]
Assuming the electrons start from rest, their final kinetic energy is equal to the electric potential energy lost while moving through the potential difference [tex]\Delta V[/tex]:
[tex]K=\frac{1}{2}mv^2 = q\Delta V[/tex]
where
[tex]m=9.11\cdot 10^{-31}kg[/tex] is the mass of each electron
v is the final speed of the electrons
[tex]q=1.6\cdot 10^{-19}C[/tex] is the charge of the electrons
[tex]\Delta V=3.0 kV=3000 V[/tex] is the potential difference
Solving the equation for v, the speed, we find
[tex]v=\sqrt{\frac{2q\Delta V}{m}}=\sqrt{\frac{2(1.6\cdot 10^{-19}C)(3000 V)}{9.11\cdot 10^{-31} kg}}=3.25\cdot 10^7 m/s[/tex]
B) Centripetal acceleration, [tex]3.82\cdot 10^4 m/s^2[/tex], in units of g: 3898 g
When the electrons cross the region of the magnetic field, they experience a magnetic force which is perpendicular to their trajectory: therefore they start moving in a circular motion. The acceleration they experience is not tangential, but centripetal, and it is given by
[tex]a_c = \frac{v^2}{r}[/tex]
where v is the speed and r the radius of the trajectory.
We can equate the magnetic force exerted on the electrons to the centripetal force:
[tex]qvB=ma_c[/tex]
and isolate [tex]a_c[/tex] to find the centripetal acceleration:
[tex]a_c = \frac{qvB}{m}=\frac{(1.6\cdot 10^{-19} C)(3.25\cdot 10^7 m/s)(0.67 T)}{9.11\cdot 10^{-31} kg}=3.82\cdot 10^4 m/s^2[/tex]
And since [tex]g=9.81 m/s^2[/tex], the acceleration can be rewritten as
[tex]a_c = \frac{3.82\cdot 10^4 m/s^2}{9.81 m/s^2}=3898 g[/tex]
c) [tex]2.76\cdot 10^{10} m[/tex]
The radius of the circular trajectory can be found by using the formula for the centripetal acceleration:
[tex]a_c = \frac{v^2}{r}[/tex]
Solvign for r, we find
[tex]r=\frac{v^2}{a_c}=\frac{(3.25\cdot 10^7 m/s)^2}{3.82\cdot 10^4 m/s^2}=2.76\cdot 10^{10} m[/tex]
