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Hi! I hope everything has been going well. I
need some help on the following:
I want to
calculate the amount of force on a NdfeB magnet.
The amount of force on each particle (or atom) in
the magnet is 0.000000000065024 newtons. to
calculate the atoms in the magnet I did:
grams/10.811 (10.811 is the atomic number of
boron) which equals 37.85958746. Next, I
multiplyed that by avagadros number: 37.85958746
times (6.02214 times 10 to the 23 power)and I got
2.27995736 times 10 to the 26 power. when I
multiply the number of atoms by the force on each
atom I get about: 1.482656617 times 10 to the 16
power newtons. That is A LOT of newtons! I dont
understand why I am getting so much for just a
magnet. I dont know what I may be doing wrong. I
hope you can help. Thanks. Hello, I have
some more information regarding my force problem:
The dimensions of the magnet that I am using are:
1.5" by 3" by 3/4" This is how I got 6.5024E-11 N
of force per atom: I used this formula: F=qv
times B f= force q= test particle of charge, q at
rest that point (proton, I believe) v= velocity
of particle sin= angle of particle or B, in other
words B = 12700 guass times (10to the power of -4
tesla divided by 1 guass) = 12.7 tesla, B= 12.7
tesla q= (proton) 1.60 times 10 to the power of
-19 coulombs v= 3.2 times 10 to the power of 7
m/s angle= 90 degrees, or 1.57 radians The
following are my calculations: F= [(1.60 time 10
to the - 19 power coulombs) times (3.2 times 10 to
the 7 power m/s)] times 12.7 tesla = 1.60 times
10 to the -19 power coulombs times 32000000m/s
times 12.7 tesla = 5.12 times 10 to the -12 power
times 12.7 tesla = 6.5024 times 10 to the -11
power N = 0.000000000065024 newtons (on one
particle).
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| Question Date: 2008-06-02 | | Answer 1:
You provided a lot of details for the
calculation, which actually are not necessary.
What is more useful to know is your motivation;
What is the problem you are trying to
solve? Why is your velocity a tenth of the
speed of light? Where did you get this number?
Why is the magnetic field of your magnet at 12.7
tesla? Thats extremely large and no natural
magnets can create that type of field (the largest
created by natural magnets are about 1 tesla).
And the highest magnetic field ever generated on
earth (for only a fraction of a second) was
probably around 25-30 tesla (I think 50 Tesla at
the very most, and it probably broke the
experimental setup right after). Also, just
to tell you if you're looking for the force on a
magnet sitting in a magnetic field, F=qvBsin
(theta) is not the way to go. The interaction is
more complicated than just that unfortunately, due
to the fact that applying that formula to complex
electron movements is quite difficult. However,
if the thing generating the B-field creates a
nearly uniform field in your region of interest,
you can use the dipole forcing in a uniform
magnetic field formula, which is relatively
simple. Good luck! | | Answer 2:
You don't give units in your calculation.Also,
you don't say how many grams you divided by the
molecular weight [= grams/mole]. It's important
to write the units for the numerator and the
denominator of each number you use and cross out
the ones that cancel each other to see if the
units of your answer are the units you wanted.
It's really easy to make mistakes in this type of
calculation, but they're important types of
calculation to make. Also, it's often hard to get
someone to really follow your line of thinking.
People find it a lot easier to tell you their line
of thinking. | | Answer 3:
Okay, I do not understand what the diamagnetic
properties of bismuth are or how they work
(evidently you know something that I don't
here!). However, your magnet is going to
consist of a large number of magnetic domains,
which won't all be in alignment with each-other,
so the force is going to get distributed in a
single direction. That will reduce the amount of
force by somewhat; it may in fact be enough. I am
also curious that the amount of force per atom is
that large... are you sure you've copied the
number down directly? Also, the strength of the
force is going to have to drop off with distance
from your lump of bismuth. Does your formula for
telling you how much force you will get include
distance in the equation? That said,
looking at the electric force, if you have one
Coulomb of charge separated from another Coulomb
of charge by a distance of a meter, the force
between them is about 1010 Newtons,
which is about twelve times the weight of a modern
aircraft carrier. So I wonder if you really have
only a tiny fraction of the atoms in your magnet
actually magnetized. Sorry I couldn't help you
more!
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