A TV accelerated electrons for 1.5 cm through a potential difference of 5500V. T

Question

A TV accelerated electrons for 1.5 cm through a potential difference of 5500V. This electron is traveling towards the screen passing a magnetic field of 83 mT.

a) What is the velocity of our electrons in the TV? It is good exercise to calculate it again. Then look it up and compare.

b) Be an engineer! How could you produce a magnetic field of 83mT?

c) What is the minimum force the electron could feel in this field? This question is about the direction of the traveling electron with respect to the direction of the magnetic field.

Explanation / Answer

a)

Fe=Fm

q*E=q*v*B

====>

v=E/B

and

E=Vo/d

now,

v=Vo/B*d

=5500/(83*10^-3*1.5*10^-2)

=4.42*10^6 m/sec

b)

by using electomagnet we can produce magnetic field

c)

electric force on electron is

Fe=q*E

=1.6*10^-19*(V0/d)

=1.6*10^-19*(5500/1.5*10^-2)

=5.86*10^-14 N

and

magnetic force on electron is

FB=q*v*B

=1.6*10^-19*4.42*10^6*83*10^-3

=5.86*10^-14 N

Fnet=Fe+FB

=11.72*10^-14 N

direction tan(theta)=FB/FA

=====> tan(theta)=1

=====> theta=45 degrees

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