Answer 2:
The trouble is, the specific question you're
asking can't be answered by a single
formula.Here's why. First, any magnet
experiences zero force if it's in a uniform
external field. It will only experience a torque:
it will try to twist to align itself with the
external magnetic field. This is why a compass
needle turns but doesn't go flying off to the
north pole.There's no net force on the compass,
only a torque which *turns* the needle. This is
true no matter how large or small the
"little"magnet is. So if the bismuth is in a
*uniform* magnetic field, like a small piece of Bi
at the center of a large magnet, the Bi will not
experience any force at all. So let's assume
the bismuth is not in a uniform field, but is on
the edge of a magnetic field, where the field
changes with position.Then we make an
approximation: we assume that the problem can be
calculated as if the magnet was only defined by
its north and south poles, and that they can be
represented as small points. (This is analogous
to positive and negative charges in an electric
field.)Here's the approximate formula for force
between two magnetic poles:
F=(mu_0)*qm1*qm2/(4*pi*r^{2}) where
mu_0 is magnetic permeability of free space
(4*pi*10^{7})Newton/Amp^{2}), r
is the distance between the two poles (like the
ends of two bar magnets), and qm1 & qm2 are the
pole strengths of each magnetic pole. Since every
magnet has 2 poles, this needs to be repeated 4
times: magnet 1 north to magnet 2
north magnet 1 north to magnet 2 south
magnet 1 south to magnet 2 north
magnet 1 south to magnet 2 south r and F will
be different for all 4 cases. These forces
are also indifferent directions, so you need to do
vector addition (in other words, add up the terms
in the "x" direction separately from those in the
"y" direction). Obviously, the distance
between the north pole & south pole on your
bismuth sample might be quite small, but it's
important. (If it were zero, then the magnetic
field would seem to be uniform at that scale,so
the total force would go to zero.) To say the
same thing a different way, if a single pole with
pole strength qm1 is in a uniform field B, the
force on that pole is F=(qm1)*B. Having two
opposite poles next to each other (north & south
on a small magnet)means F = (qm1)*B+(qm1)*B =
(qm1qm1)*B = 0, so zero total force. Now
how big is the pole strength of a magnet? Here's
the relation between magnetic field B created by a
single pole (either north or south) with pole
strength qm1: B (created by pole qm1) =
(mu_0) * (qm1) / (4*pi*r^{2})Obviously,
the field changes with distance r from the pole.
This gives a roughly approximation of B near a
magnet: B(north pole) minusB(south pole) at the
location where the bismuth will be. The distances
r_north & r_south may be different, depending on
where the bismuth is relative to the two poles.
And again, the two B's need to be added in vector
addition, since they pull in different
directions.For a real magnet, you would use this
backwards: measure B vs.distance near the pole,
calculate qm1, then use that to calculate B
everywhere else including both poles. It
sounds like you already know that diamagnetic
material is not magnetized when it's by itself but
becomes *temporarily* magnetized in the presence
of an external magnetic field. The magnetization
of diamagnetic materials is extremely weak.
Bismuth happens to be the strongest. In the
presence of a magnetic field H (in Tesla), them
agnetization of bismuth is: M =
.000166*H and by definition, B = mu_0
* (H + M) = mu_0 * (H  .000166*H)
= mu_0 * 0.999834 * H So you need to know
the actual magnetic field (H) in the location
where the bismuth will be. You can get a rough
idea from the calculations in the previous
paragraph, or from an experiment like this
one: magneticfields But
it's probably more accurate to find a Hall Effect
meter and measure the field directly.So
after all that, I'm still not sure if you've got
the answer to your question.
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