Purcell Electromagnetism p.301-303: "the longitudinal current I flows, in effect, on the cylinder itself"?

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The term "longitudinal current" refers to the current flowing along the z-axis of the cylinder. This aligns with the understanding that the current is in the axial direction. The discussion confirms that the longitudinal current is indeed in the axial direction rather than the azimuthal direction. Clarification is provided that the current's flow is consistent with the cylinder's geometry. Understanding this concept is essential for grasping the principles of electromagnetism as presented in Purcell's text.
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Homework Statement
I quote "As for the field inside the solenoid, the longitudinal current
I flows, in effect, on the cylinder itself. Such a current distribution, a
uniform hollow tube of current, produces zero field inside the cylinder and the fact that a circular path inside the tube encloses
no current), leaving unmodified the interior field we calculated before. If
you follow a looping field line from inside to outside to inside again,
you will discover that it does not close on itself. Field lines generally
don’t. You might find it interesting to figure out how this picture would be changed if the wire that leads the current I away from the coil were
brought down along the axis of the coil to emerge at the bottom."
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I don't understand what the book meant as 'longitudinal current'. Is this in the axial direction (z direction) or the azimuthal direction (\phi direction)? It would only make sense if the current is in the axial direction. A confirmation of my guess would be appreciated. A picture from the book is attached.

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Longitudinal is the direction along the z-axis of the cylinder as you surmised.
 
Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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