Electrons in a television's CRT are accelerated from rest by an electric field through a potenti

Step 1: To find the speed of the electrons accelerated through a potential difference of $2.5 ma

Electrons in a television's CRT are accelerated from rest by...

Electrons in a television's CRT are accelerated from rest by an electric field through a potential difference of $2.5 \mathrm{kV}$. In contrast to an oscilloscope, where the electron beam is deflected by an electric field, the beam is deflected by a magnetic field. (a) What is the speed of the electrons? (b) The beam is deflected by a perpendicular magnetic field of magnitude $0.80 \mathrm{~T}$. What is the magnitude of the acceleration of the electrons while in the field? (c) What is the speed of the electrons after they travel $4.0 \mathrm{~mm}$ through the magnetic field? (d) What strength electric field would give the electrons the same magnitude acceleration as in (b)? (e) Why do we have to use an electric field in the first place to get the electrons up to speed? Why not use the large acceleration due to a magnetic field for that purpose?

Transcript

00:01 Hi, here in this given problem, potential difference applied in cathode ray tube, potential difference which is used to accelerate electrons, that is given as 2 .5 kilo volt or we can say this is 2 ,500 volt.

00:23 In the first part of the problem, we have to find speed gained by the electrons under the influence of this much potential difference for which we use energy conservation kinetic energy gained by the electrons that will be equal to electrostatic potential energy gained by them under the influence of this much potential difference so we get an expression for the speed and that is the square root of twice of e v divided by m so plugging in all the known values here this is two times of charge carried by an electron 1 .6 into 10 to the power minus 19 into potential difference 2 ,500 divided by mass of the electron 9 .1 into 10 dash to the power minus 31.

01:20 So finally it comes out to be equal to 2 .96 into 10 raised to the power 7 meter per second answer for the first part of this given problem here then in the second part magnetic field given is 0 .80 tesla so in order to find acceleration gained by the electron here using an expression for the magnetic lawrence force experienced by this electron f equals to e v b and then using homes law this is equal to m into a so an expression for the acceleration will be a is equal to e v b by m plugging in all the known values here this is again electronic charge 1 .6 into 10 power minus 19 speed we have found in the first part of the problem 2 .96 into 10 dash per 7 magnetic field 0 .8 0 and and divided by m, mass of the electron 9 .1 into 10 dash to the power minus 31.

02:42 So it comes out to be equal to this acceleration gained by the electron, this is 4 .16 into 10 dash to power 18 meter per second square.

02:54 Answer for the second part of this given problem here.

02:59 Then in the third part of the problem, distance covered, displacement covered by the second.

03:07 Electron is 4 .00 millimeter or we can say this is 4 .00 into 10 dash per minus 3 meter under the influence of this acceleration.

03:19 So to find final velocity achieved using third equation of motion vf squared equals to v i square plus 2 a s and for s this is delta x.

03:34 So, here it will come out to be equal to 0 plus twice of acceleration 4 .16 into 10 dash per 18 into 4 into 10 dash per minus 3...

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