Will I get a shock holding a superconductive wire? Assume a current carrying superconductive wire is in front of me, with no voltage source attached.
If I hold it, will I get a shock? Forget for a minute that the superconducting wire will be too cold to touch.
My initial instinct was a maybe, for a second but ultimately no. As a normal shock occurs because of the potential difference around the wire (as the immediate electric field due to surface charges will cause a current to flow through my body). However without an emf to regain those surface charges, the current would stop as there is no emf?
 A: In order for there to be the potential for an "electric shock" at any level, there needs to be at minimum a sufficient potential difference (voltage) between the surfaces that you are touching. A superconductor carrying current has zero resistance, and therefore there is no potential difference between any two points on the superconductor itself.
On the other hand, if there is a sufficient potential difference between the superconductor and some other point, such as ground, then the potential for electric shock may exist between those two points. Whether or not the potential for electric shock actually exists between the two points depends on many other factors, too many to go into here.
Hope this helps.
A: Another way of thinking about Bob D's answer: when you grab a superconducting wire in two different places (e.g. with both hands), you can think of your body and the wire as being wired in parallel. There is zero resistance along the superconducting wire but positive resistance through your arms, so all of the current will travel down the zero-resistance branch - the superconducting wire - and no current will run through your arms. So you won't get a shock (although you might get frostbite).
A: Let's say the superconductor you grab is one turn of N turns of a 1 meter diameter by 1 meter long solenoidal magnet with an internal B=1 Tesla. The current in this closed loop is flowing with no resistance as long as the temperature stays at 4 degK.
Out of boundless curiosity, you grab the superconductor with your ~300 degK hand.  The conductor under your hand warms up and goes normal (ie: acquires a resistance).  The flowing current begins dissipating power in this resistance, heating it even more.  As the B field collapses, a big voltage $=N (1 meter)^2~dB/dt$ appears across your hand. The conductor and much of your hand turns to plasma as the energy stored in the B field $=(10^7/4\pi)(1~Tesla)^2(1~meter)^3 \approx 10^6~Joules \approx 1~Kg~TNT$ dissipates.  Fortunately you were wearing goggles and have one hand left!
