# Percent correction for Magnetic Field vs. Electric Field!

1. Jun 10, 2014

### michaelle1991

Hi guys,
In the photo is the problem. We set Fb =Fe to show for E then out E into the Voltage equation to get V=vlB. However, if we won;t ignore the gravity force, what is the percent correction? (Note: q =Electron)

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2. Jun 10, 2014

### Simon Bridge

Welcome to PF;
What is the definition of "percent correction" - that will tell you how to work it out.
In the presence of gravity - draw a free-body diagram for the electron.

3. Jun 10, 2014

### michaelle1991

i did a little work, but idk if it is right?

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4. Jun 10, 2014

### Simon Bridge

You missed out defining what "percentage correction" means.
What are you correcting and why?

The fbd and calculation seems OK.

qvlB = qV + mgl

So V= vlB-gl(m/q) with gravity

Without gravity: V0=vlB

If you used B=V0/vl with V0=V then you have introduced a systematic error into your equation for determining B.

5. Jun 19, 2014