Question

An electron travels undeflected in a path that is perpendicular to an electric feld of 8.3 x 10 v/m. It is also moving perpendicular to a magnetic field with a magnitude of 7.3 x 103 T. If the electric field is turned off, at what radius would the electron orbit? O 124 x 10*m 889 x 104 m O 9.85 x 104m O 1.06 x 10o m

Answer 1

Answer:

$8.6\cdot 10^{-18} m$

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

$E=8.3\cdot 10 V/m$ is the electric field magnitude

v is the electron's speed

$B=7.3\cdot 10^3 T$ is the magnitude of the magnetic field

Solving for v,

$v=\frac{E}{B}=\frac{8.3 \cdot 10 V/m}{7.3\cdot 10^3 T}=0.011 m/s$

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:

$qvB= m \frac{v^2}{r}$

where

$q=1.6\cdot 10^{-19} C$ is the electron's charge

$m=9.11\cdot 10^{-31} kg$ is the electron's mass

Solving for r, we find the radius of the electron's orbit:

$r=\frac{mv}{qB}=\frac{(9.11\cdot 10^{-31} kg)(0.011 m/s)}{(1.6\cdot 10^{-19} C)(7.3\cdot 10^3 T)}=8.6\cdot 10^{-18} m$