English Deutsch Français Italiano Español Português 繁體中文 Bahasa Indonesia Tiếng Việt ภาษาไทย
All categories

Electrons in a television's CRT are accelerated from rest by an electric field through a potential difference of 2.5kV. In contrast to an oscilloscope, where the electron beam is deflected by an electric field, the beam is deflected by a magnetic field.

What is the speed of the electrons?

The beam is deflected by a perpendicular magnetic field of magnitude 0.80T. What is the acceleration of the electrons while in the field?

What is the speed of the electrons after they travel 4.0mm through the magnetic field?

What strength electric field would give the electrons the same magnitude acceleration as in the acceleration of the electrons while in the field?

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?

2007-03-06 17:22:15 · 2 answers · asked by Sir Guitarist 2 in Science & Mathematics Physics

Sorry for not explaining. I tried doing this problem but I can't quite do it right. I have the formula sheet on front of me and I tried different aspects. I'm a non-physics major who is struggling to get through... it's weird how this is a core class for a medical major.

2007-03-06 17:30:39 · update #1

2 answers

You calculate the speed of the electrons by the equation e=½/mv²
You know m (mass of the electron) and e (energy is 2500 eV, which you convert to joules) so you calculate v.

Acceleration is force/mass.  The force is IxB.  If you consider a 1-meter length as containing 1 electron, I equals the electron speed in meters/second divided by coulombs per electron.  Use I to calculate the force; you already know m, so calculate A.

The electrons lose no speed in a constant magnetic field.  That's a trick question.

Since you've already got the force, calculate what it would take as an electric field (charge times field strength).

A moving magnetic field could accelerate electrons like an elecric field, but would be much more complex.

Sorry I can't put numbers on this, but I'm working without books at the moment.

2007-03-06 18:00:56 · answer #1 · answered by Engineer-Poet 7 · 0 0

I agree.........

2007-03-06 17:25:25 · answer #2 · answered by sassyred57 1 · 0 1

fedest.com, questions and answers