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

`8.6\cdot 10^(-18) m`

Step-by-step 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=(E)/(B)=(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 (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=(mv)/(qB)=((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`