Answer:
[tex]1.26\cdot 10^7 m/s[/tex]
Explanation:
When a charged particle moves perpendicularly to a magnetic field, the force it experiences is:
[tex]F=qvB[/tex]
where
q is the charge
v is its velocity
B is the strength of the magnetic field
Moreover, the force acts in a direction perpendicular to the motion of the charge, so it acts as a centripetal force; therefore we can write:
[tex]qvB=m\frac{v^2}{r}[/tex]
where
m is the mass of the particle
r is the radius of the orbit of the particle
The equation can be re-arranges as
[tex]v=\frac{qBr}{m}[/tex]
where in this problem we have:
[tex]q=1.6\cdot 10^{-19}C[/tex] is the magnitude of the charge of the electron
[tex]B=208 G=208\cdot 10^{-4}T[/tex] is the strength of the magnetic field
The beam penetrates 3.45 mm into the field region: therefore, this is the radius of the orbit,
[tex]r=3.45 mm = 3.45\cdot 10^{-3} m[/tex]
[tex]m=9.11\cdot 10^{-31} kg[/tex] is the mass of the electron
So, the electron's speed is
[tex]v=\frac{(1.6\cdot 10^{-19})(208\cdot 10^{-4})(3.45\cdot 10^{-3})}{9.11\cdot 10^{-31}}=1.26\cdot 10^7 m/s[/tex]
