Re: Stump The Physicist

From: Jim Black (ghytrfvbnmju7654_at_mail.com)
Date: 08/30/04 

Date: 29 Aug 2004 22:02:57 -0700

thoovler@excite.com (Igor) wrote in message news:<d434b6c6.0408291405.6a0beef1@posting.google.com>...
> keith__cochrane@hotmail.com (KC) wrote in message news:<8824f771.0408281008.20f3f803@posting.google.com>...
> > Knowing how much you people like to solve problems, I thought I'd post
> > a physics related trivia question, (based on my own observations and
> > resulting theory). No pulling out visual aides, or peeking at your
> > physics books or websites. The problem is a simple one that any
> > physicist should be able to answer off the top of thier respective
> > heads, so take a shot at it and good luck!
> >
> > Consider having two permanent magnets like the ones described below in
> > the ASCII drawing.
> > Magnet # 1 Measures 2" Long X 3/4" Wide at the poles and 1/4" Deep.
> > Magnet # 2 Measures 6" Long X 3/4" Wide at the poles and 1/4" Deep.
> > Both magnets configured as in the ASCII drawing below.
> >
> > Constants:
> > Magnet # 2 is held stationary, (unable to move or turn in any
> > direction), with its north magnetic pole facing upward, (as viewed in
> > the drawing), toward magnet # 1.
> > Magnet # 1 is able to move, (sliding upon a frictionless track), from
> > side to side, while remaining in a fixed upright position in relation
> > to magnet # 2, with magnet # 1's north magnetic pole always facing
> > left and south magnetic pole always facing right.
> >
> > Question: Based on the information, above and below, state which of
> > the following answers is correct, (A, B, C, or D), and explain why?
> >
> > A) Magnet # 1 will remain stationary.
> > B) Magnet # 1 will move to the right.
> > C) Magnet # 1 will move to the left.
> > D) State your own conclusion in its entirety.
> >
> > _
> > Magnet # 1 -------------> | |
> > Edge View --------------> | |
> > | |
> > North Magnetic Poles ---> | | <-- South Magnetic Pole
> > | | |
> > | Like ------> | |<------ Opposite
> > | Poles | | Poles
> > \|/ | | | |
> > \/ \/ |_| \/
> > ____________________________________________________
> > |____________________________________________________|
> > Magnet # 2 ^
> > /|\
> > |
> > South Magnetic Pole
> >
> > End of trivia question.
> >
> > I'll respond when I see someone get this right.
>
> The answer is (A) since it will experience a torque that will tend to
> align the two fields. If it were not restrained, it would rotate
> clockwise, but since it is only allowed to moved left or right, it
> can't move at all, since it will experience no net force along that
> line.

That's what I guessed when I first looked at it, too, but if you work
it out, you'll find that there is, in fact, a net force.

As a quick, cheap approximation, one can imagine that north and south
magnetic charges are spread out over the poles of the magnets. This
is not true, obviously, but it should produce about the right answer.

Ignoring the third dimension, the charges would be located at:

[where (u,v) is u to the right and v above the origin]

(a-s,y) (north #1)
(a, y) (south #1)

(x,-t) (south #2)
(x,0) (north #2)

Let x range from 0 to w, and y range from b to h.

The potential energy would then be some positive constant times:

integral dx integral dy

-1/sqrt[(a-s-x)^2+(y+t)^2] (north #1 & south #2)
+1/sqrt[(a-x)^2+(y+t)^2] (south #1 & south #2)
+1/sqrt[(a-s-x)^2+(y)^2] (north #1 & north #2)
-1/sqrt[(a-x)^2+(y)^2] (south #1 & north #2)

Making the assumption that s and t are small, the integrand is
approximately equal to:

s(d/dx)[t(d/dy)[-1/sqrt[(a-x)^2+(y)^2]]]

(Read d's as partial.)

Putting the st term into the positive constant, and integrating twice,
we get some positive constant times:

-1/sqrt[(a-w)^2+(h)^2]
+1/sqrt[(a-w)^2+(b)^2]
+1/sqrt[(a)^2+(h)^2]
-1/sqrt[(a)^2+(b)^2]

So the magnets should behave approximately as if there were like
charges located at (a,b) and (w,0), and a pair of like charges of the
opposite sign at (a,h) and (0,0).

Relevant Pages

  • Re: on the nature of gravity
    ... > I do maintain gravitation does result from ... > the absence of those charges but at the points ... > That's no more difficult than a kid pushing ... > a pair of repelling magnets together to store ...
    (sci.physics.relativity)
  • Re: OBarr 27 Nov 2004: Reasons for an ether.
    ... Not so with a pair of magnets. ... > in all directions? ... > They are your invention, ... All charges are entities but not all entities are charges. ...
    (sci.physics.relativity)
  • Re: Stump The Physicist
    ... > physics books or websites. ... > Both magnets configured as in the ASCII drawing below. ... > End of trivia question. ... Making the assumption that s and t are small, the integrand is ...
    (sci.physics)
  • Re: Does the strength of a magnet fall off at a greater rate than gravity?
    ... > non-dipole magnets so far, ... the behavior of charges, magnetic fields and electric ... > What do the magnetic field for these charges look like if they move? ... A moving particle creates a current. ...
    (sci.physics)
  • Re: New Cubic Atomic Model explains electron energy levels and bonding
    ... the references appear to support that there are energy levels ... Why do p shells hold 6 electrons? ... > Yes, you keep saying that, but this assumes that electric charges are ... > magnets are a stable structure. ...
    (sci.physics)