an electron accelerated from rest through a voltage of 790 v enters a region of constant magnetic field. part a if the electron follows a circular path with a radius of 21 cm , what is the magnitude of the magnetic field? express your answer using two significant figures. b



Answer :

The electron is accelerated through a potential difference of , so the kinetic energy gained by the electron is equal to its variation of electrical potential energy:

1/2mv²=eΔV

where

m is the electron mass

v is the final speed of the electron

e is the electron charge

is the potential difference

Re-arranging this equation, we can find the speed of the electron before entering the magnetic field:

v=√2eΔv/m=√2(1.6-10-19c)(790v)/9.1-10-31=1.66.10 m/s

Now the electron enters the magnetic field. The Lorentz force provides the centripetal force that keeps the electron in circular orbit:

evb=mv²/r

where B is the intensity of the magnetic field and r is the orbital radius. Since the radius is r=25 cm=0.25 m, we can re-arrange this equation to find B:

b=m/er =3.8.10 t

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