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An electron in a TV CRT moves with a speed of $6.00 \times 10^{7} \mathrm{m} / \mathrm{s},$ in a direction perpendicular to the Earth's field, which has a strength of $5.00 \times 10^{-5} \mathrm{T}$ . (a) What strength electric field must be applied perpendicular to the Earth's field to make the electron moves in a straight line? (b) If this is done between plates separated by $1.00 \mathrm{cm},$ what isthe voltage applied? (Note that TVs are usually surrounded by a ferromagnetic material to shield against external magnetic fields and avoid the need for such a correction.)
Part (a): The electric field needed to be applied is $3 \times 10^{3} \mathrm{V} / \mathrm{m}$Part $(\mathrm{b}) :$ The voltage applied between the plates is 30 $\mathrm{V}$
Physics 102 Electricity and Magnetism
Chapter 22
Magnetism
Magnetic Field and Magnetic Forces
Sources of Magnetic field
Electromagnetic Induction
Inductance
Rutgers, The State University of New Jersey
University of Michigan - Ann Arbor
Simon Fraser University
University of Sheffield
Lectures
08:42
In physics, a magnetic fie…
04:28
A magnetic field is a math…
02:28
An electron in a TV CRT mo…
05:46
A The picture tube in a te…
06:08
ssm The electrons in the b…
06:16
An electron in a cathode-r…
04:47
An electron moves at speed…
27:18
Deflection in a CRT. Catho…
06:37
Electrons in a television&…
05:04
(a) What value of magnetic…
15:30
An accelerating voltage of…
01:19
An electron travels with s…
03:16
A beam of electrons travel…
07:26
Crossed electric and magne…
04:04
ssm An electron is moving …
02:29
An electron of kinetic ene…
04:42
($a$) What value of magnet…
05:00
(II) An electron moving to…
06:07
An electron with a speed o…
06:31
04:32
At some instant the veloci…
05:44
for party. We know that the magnitude of the electric field is simply gonna be equal to the product between the magnitude of the electric field multiplied by the velocity. And so this would be equaling 5.0 times, 10 to the negative fifth. Tesla's multiplied by 6.0 times 10 to the seventh meters per second. And this is giving us 3.0 Well, say kilovolts per meter. Now, excuse me for part B. We know that the voltage a prat applied this would be essentially the voltage between the plates. We can then say that the voltage applied in this scenario would be equaling the electric field multiplied by the distance. And so this would be equaling 3.0 kilovolts per meter multiplied by the distance of one centimeter. Or we can say 10.1 meter. And we know that 3000 rather three kilovolts is simply 3000 volts per meter. And so we can most play this by 0.1 meter and we find that the voltage applied would be equaling 30.0 volts. This would be our final answer. That is the end of the solution. Thank you for watching
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