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Re: [Phys-l] Magnetic force and work
From
: "Snyder, Tom" <
Tom.Snyder@llcc.edu
>
Date
: Fri, 24 Mar 2006 12:13:14 -0600
In pondering Joel's question I started thinking about a simple model
that
Might be somewhat relevant.
Take a circular ring with a net positive charge uniformly spread around
the
ring. Let the ring be rotating about an axis through the center of the
ring
and perpendicular to the plane of the ring. Call this the x-axis. Now
add a time-independent magnetic field that has a radially outward
component at each element of the ring (as well as perhaps a component
parallel to the x-axis). By radial I mean perpendicular to the x-axis.
Such
a field could be produced, for example, by a bar magnet placed along the
x-axis at some distance from the center of the ring with the north pole
closer to ring. However, to make the model simple, I imagine that the
radial component of the magnetic field is independent of x.
The magnetic interaction will cause the spinning charged ring to acquire
translational motion parallel to the x-axis even though no work is done
by the magnetic force. It is interesting to work out the complete
motion
(neglecting radiation losses!) for a ring of radius R, mass M, charge Q,
and angular velocity W starting with no translational motion. I give
the
solution below (at least what I think is the solution).
This simple model shows how a magnetic interaction can cause an object
to gain translational KE without any work being done by the magnetic
force.
However, this model still doesn't describe what's going on when a magnet
is attracted to another magnet where you have to worry about what's
going
on at the atomic level with the electron spins. I don't have any
insight into
that.
---------------------------------------------------------------------
Stop reading here if you want to discover the solution for
the spinning ring yourself.
---------------------------------------------------------------------
I find that the ring executes a translational simple harmonic motion
parallel to
the x-axis. If B represents the radial component of the external
magnetic
field, the amplitude of the SHM is WRM/(QB). The angular frequency of
the
SHM is QB/M. In addition, a "rotational SHM" motion occurs such that
the
ring's angular velocity slows down as the translational velocity
increases and
vice versa. Any gain in translational kinetic energy is balanced by a
corresponding loss of rotational kinetic energy and vice versa. At the
instant
of maximum translational KE, there is no rotational KE.
So, the magnetic interaction allows an interchange of translational and
rotational KE without the magnetic field doing any work.
For reasonable values of the parameters for a macroscopic ring, the
build-up
of translational motion occurs very slowly. For example, let the
initial rotation
speed be 100 rev per sec, Q = 1 micro-Coulomb, B = 0.1 T, M = 10 grams,
and R = 10 cm. There is then about 20 J of mechanical KE in the system.
I find the period of the SHM to be about a week and the amplitude of the
translational SHM to be about 4 thousand miles! The maximum
translational
speed is about 60 m/s (which is also the maximum rotational speed of a
point
of the ring).
Follow-Ups
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Re: [Phys-l] Magnetic force and work
From:
John Denker <jsd@av8n.com>
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