In a line made up of two parallel coax cables with connected shields and
with the shields floating at both ends, let's suppose that one center
conductor (cable A) is carrying 1A and the other (cable B) 3A, exactly
out of phase with each other -- that is, the line is not balanced. The
inside of the cable A shield will have 1A and the inside of the cable
shield B will have 3A, again exactly out of phase, and each one out of
phase with its corresponding center conductor current.

Let's consider electrically short cables for simplicity -- short enough
that the currents don't change much along the length.

As Tom said, the fields are contained within the coax cables -- on the
inside of each one is a canceling pair of currents. But at the ends of
the cables, the shield currents can escape to the outside, 1A from cable
A, and an out of phase 3A from cable B. The result is a net 2A flowing
on the outside of the shield. This current will cause the combined cable
to radiate exactly the same amount as if the shield wasn't there at all!
All we've done is to physically separate the differential and common
mode currents, but we haven't changed their values.

So what happens when the conductor currents are balanced, i.e., equal
and opposite? Well, then there's no net current on the outside of the
shield. Now there's no line radiation, again exactly the same as if the
shield wasn't there. The shield likewise has no effect on reception of
unwanted signals -- it provides no shielding at all. The only way to
make it provide shielding is to block the current path from the inside
to the outside of the cables at both ends. In the case of an antenna
feedline, however, this would require putting the antenna into a
shielded box.

The shield has an impact on the differential mode impedance, but it
doesn't help line balance or provide shielding at all.

Roy Lewallen, W7EL