An electron is accelerated from rest through a potential difference of 501V insi

Question

An electron is accelerated from rest through a potential difference of 501V inside an electron gun. This electron gun happens to be located inside a Helmholtz coil (R = 50cm, N = 101). At the moment the electron is launched, the Helmholtz coil has a current of 1.2A passing through it and is oriented so that the magnetic field it generates points in the positive z-direction (out of page).


If the electron is launched in positive x-direction (to the right), then what is the radius of the electron’s path?

I need detailed work shown please, for me to clearly understand it. Thank you.

Explanation / Answer

(1) First lets calculate the speed of the electron
Kinetic energy equals to the change of potential energy

mv^2/2=q*V
then v=sqrt(2qV/m)=sqrt (2*1.6*10-19*501/9.1*10-31) = 13.8*106 m/sec
(2)
magnetic field of the coil is given by
B=(µ*n*I/R)*(4/3)^1.5=0.000506 T

(3) Then given the electron with speed v in a magnetic field, lets calculate the radius
R=m*v/(q*B)=0.155 m

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