Question

A beam of electrons moving in the x-direction enters a region where a uniform 208-G magnetic field points in the y-direction. The boundary of the field region is perpendicular to the beam.If the beam penetrates 3.45 mm into the field region, what's the electrons' speed

Answer 1

Answer:

$1.26\cdot 10^7 m/s$

Explanation:

When a charged particle moves perpendicularly to a magnetic field, the force it experiences is:

$F=qvB$

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:

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

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

$v=\frac{qBr}{m}$

where in this problem we have:

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

$B=208 G=208\cdot 10^{-4}T$ 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,

$r=3.45 mm = 3.45\cdot 10^{-3} m$

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

So, the electron's speed is

$v=\frac{(1.6\cdot 10^{-19})(208\cdot 10^{-4})(3.45\cdot 10^{-3})}{9.11\cdot 10^{-31}}=1.26\cdot 10^7 m/s$