Understanding a Homopolar Generator

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jisbon
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Homework Statement
Give an expression for the induced current in a homopolar generator as a function of angular velocity ##\omega##
Relevant Equations
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So I was searching up on the homopolar generator and found this explanation for the generator, as well as the proposed solution.

Schematic of a homopolar generator: a conducting cylinder spinning about its vertical axis with angular velocity omega_M, a rotating disk contact on top with omega_d, radial lead R, a voltmeter V, and magnetic field B lines passing vertically through the cylinder.


Equation chain: V = integral from 0 to r of E dot dl = integral from 0 to r of (v cross B) dot dl = integral from 0 to r of ((r cross omega_d) cross B) dot dl = B r^2 omega_d / 2.


However, I don't really understand what the solution is trying to do here.

Ok, I understand to find the current, you will need the potential, which is the integration of emf. However, I'm stuck at the last part. How did they exactly derive ##dl ##and what is ##dl## exactly?

Any explanation will be appreciated. Thanks
 
Hi.
##d\mathbf{l}## is along the line between shaft and contact point. ##d\mathbf{l}## and emf working on the line have same direction, with plus or minus signature.
 
I think the integral "overloads" the symbol ##r## as it uses it both at the integral boundary (from 0 to r) and to denote the position vector along the path of integration which is along a radius R. I think it would be more proper to denote the ##\vec{r}## as ##\vec{l}## and then everything falls into place cause then ##d\vec{l}## gains its meaning inside the integral.