- #1
Percent correction for Magnetic Field vs. Electric Field
- Thread starter michaelle1991
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Homework Help Overview
The discussion revolves around calculating the percent correction for the magnetic field and electric field in a scenario involving an electron, particularly when considering the effects of gravity on the system.
Discussion Character
- Exploratory, Assumption checking, Problem interpretation
Approaches and Questions Raised
- Participants explore the relationship between the forces acting on the electron, questioning the definition and implications of "percent correction." There is an attempt to derive a voltage equation that incorporates gravitational effects alongside magnetic and electric forces.
Discussion Status
Some participants have provided guidance on drawing free-body diagrams and clarifying the meaning of percent correction. There appears to be an ongoing exploration of the equations involved, with multiple interpretations of how to incorporate gravity into the calculations.
Contextual Notes
There is mention of a systematic error introduced when adjusting the equations for the magnetic field, indicating that the discussion is considering the implications of various forces and their corrections.
- #2
Simon Bridge
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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.
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
- #4
Simon Bridge
Science Advisor
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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.
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
berkeman
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michaelle1991 said:
Please check your PMs...
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