Answer:
[tex]8.6\cdot 10^{-18} m[/tex]
Explanation:
Initially, the electron is travelling undeflected at constant speed- this means that the electric force and the magnetic force acting on the electron are balanced. So we can write
q E = q v B
where
q is the electron's charge
[tex]E=8.3\cdot 10 V/m[/tex] is the electric field magnitude
v is the electron's speed
[tex]B=7.3\cdot 10^3 T[/tex] is the magnitude of the magnetic field
Solving for v,
[tex]v=\frac{E}{B}=\frac{8.3 \cdot 10 V/m}{7.3\cdot 10^3 T}=0.011 m/s[/tex]
Then the electric field is turned off, so the electron (under the influence of the magnetic field only) will start moving in a circle of radius r. Therefore, the magnetic force will be equal to the centripetal force:
[tex]qvB= m \frac{v^2}{r}[/tex]
where
[tex]q=1.6\cdot 10^{-19} C[/tex] is the electron's charge
[tex]m=9.11\cdot 10^{-31} kg[/tex] is the electron's mass
Solving for r, we find the radius of the electron's orbit:
[tex]r=\frac{mv}{qB}=\frac{(9.11\cdot 10^{-31} kg)(0.011 m/s)}{(1.6\cdot 10^{-19} C)(7.3\cdot 10^3 T)}=8.6\cdot 10^{-18} m[/tex]
