An electron beam moves toward an oscilloscope screen. The electron is accelerate

Question

An electron beam moves toward an oscilloscope screen. The electron is accelerated from rest through a potential difference of 3000 V toward a screen 0.50 m away. Assume Earth's magnetic field is aligned perpendicularly to the electron's trajectory and has the magnitude of4.0 × 105 T.An electron beam moves toward an oscilloscope screen. The electron is accelerated from rest through a potential difference of 3000 V toward a screen 0.50 m away. Assume Earth's magnetic field is aligned perpendicularly to the electron's trajectory and has the magnitude of4.0 × 105 T.

How much will the magnetic field deflect the electron's position at the screen.

Explanation / Answer

E=V/d=3000/0.5=6000 N/C

m=9.1*10^(-31) kg

total time taken to travel 0.50 m need to be calculated first

acceleration=a =eE/m=1.055*10^(15) m/s^2

by newton equation

S=ut+0.5at^2 (u=0,S=0.50,a=1.055*10^(15))

t=3.08*10^(-8) sec

velocity=a*t=eEt/m

magnetic force=qvB=e^2*Et*B/m

acceleration due to magnetic force=magnetic force/m=e^2*Et*B/m^2

a=dv/dt=e^2*EtB/m^2

dv=(e^2*EB/m^2)tdt

integrating both sides and putting v=0 at t=0

v=(e^2*EB/m^2)*t^2/2

v=dx/dt

dx=(e^2*EB/2m^2)t^2dt

again integrating both sides and puting x=0 at t=0sec

x=(e^2*EB/6m^2)t^3

putting t=3.08*10^(-8) sec and putting all other values

we got x=0.036 m

This is magnetic deflection

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