Re: [Phys-l] Magnetic force and work

["Snyder, Tom" <Tom.Snyder@llcc.edu>]

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.

 

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Stop reading here if you want to discover the solution for

the spinning ring yourself.

 

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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).