Dave wrote:
"Tracy Hall" wrote in message
oups.com...

I understand that coaxial cable does not radiate much energy because
symmetric opposing current sheets in the outer skin of the center
conductor and the inner skin of the shield essentially balance each
other.

Suppose I route coax through a strong DC magnetic field, such as in an
MRI, with the magnetic field perpendicular to the cable axis. Will the
current distribution in the center conductor then become biased "up"
with respect to the field (right hand rule), and the current
distribution in the shield then become biased "down," thereby
destroying symmetry?

If so, will the coax then radiate and become lossy?

Can anyone point me to an analysis of this problem?


analyze this way...
if the current in the coax is AC then each half cycle the electron flow
reverses so the forces would reverse, and since the actual electron velocity
is relatively low there would not be enough time for the electrons to really
move before the current reversed and they had to go the other direction. so
i would say that it would be unlikely to have any effect on coax carrying
AC. now, if you go to very low frequencies where the electrons have a
chance to move you might be able to measure something, but since the
wavelength increases with the lower frequency the relative size of any
imbalance becomes smaller. and essentially if it did completely separate
you still end up with balanced currents but in a twin lead arrangement and
you still have currents that cancel at any distance away from the line.
what would be more interesting is to calculate the torque on a piece of
twinlead carrying dc in a strong field.



To leading order, the superposition argument is right, but it assumes
that the current distribution in the coax is unchanged by the applied DC
field, which is not true in detail.

The Hall effect operates in metals, so an AC current in the shield will
indeed cause an AC voltage across the diameter of the coax shield,
outside as well as inside, and this will radiate as an electric dipole.

However, the Hall effect in metals is so small that it's difficult even
to measure it, and the electric dipole radiation from a source very
small compared to a wavelength is weak. The resulting leakage is
therefore (miniscule)**2, far smaller than that caused by the poor
shield coverage and poor shield continuity.

The shield conductors are just laid on top of each other, and since
their contacts are neither gas-tight nor self-wiping, they are somewhat
noisy and unreliable--we don't rely on that sort of contact elsewhere in
electronics.

Cheers,

Phil Hobbs