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Electrons start at rest pass through a +28-kV potential difference in an oscilloscope that is oriented...

Electrons start at rest pass through a +28-kV potential difference in an oscilloscope that is oriented so that the electron is moving perpendicular to Earth's 3.5 × 10-5-T magnetic field at the equator. The oscilloscope faces east.

Determine the speed of the electron.

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Answer #1

given are

V=28Kv

V=28,000V

B=3.5 × 10-5-T

so that when the magnetic field is perpendicular to the velocity, we simply have the magnitude of the force is, and the direction can be found using the right hand rule. It turns out this direction is perpendicular to the direction of velocity, which is the signature of a centripital forces that keep an object moving in a circle by changing the direction but not necessarily the magnitude of the velocity. This is what is happening in this problem, where we require that force provided by the magnetic field is the centripetal force we are looking for, i.e., where R is the radius of the circle the particle will end up traveling in should it have an unobstructed path in the magnetic field, and should the magnetic field be in a large enough region to contain the circle. Thus 9B For this problem, we write: mpUp where the velocity v is obtained by acceleration due to a common potential difference ? V

e charge of electon=1.6*10^-19C

mp=9.1*10^-31 Kg(mass of proton)

putting value in eqn we get

v=undrt(2* 1.6*10^-19*28000/9.1*10^-31)

v=undrt(9846.1*10^12)

v=99.2*10^6

v=9.92*10^5 m/sec--answer

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