Physics Asked by Jerry March on August 14, 2021
A metallic rod is rotating about a vertical axis through its end. For finding the EMF $E$ between the ends we simply equate $momega^2r=Eq$ then find $sqrt{-Edr}$. So for this calculation we consider centripetal force.
But when a uniform perpendicular magnetic field $B$ (parallel to axis) is switched on, due to the Lorentz force a redistribution of charge takes place and to calculate it we equate $Eq=qvB$ at static condition. So the EMF is $tfrac12 Bomega l^2$.
But why we didn’t consider centripetal concept here? I mean we should do something like $momega^2r=Eq-qvb$.
Please explain from basics. I am totally confused.
You are using $E$ to stand for both magnitude of electric field strength and emf. Using $mathscr E$ for emf, your 'rotational emf' in a rod of length $l$ is the 'work integral': $$mathscr E =int_0^l frac{momega^2r}{q} dr=frac{momega^2}{q}int_0^l r dr =frac{momega^2 l^2}{2q}$$ Putting in the mass and charge of an electron for $m$ and $q$, I find that for a 20 cm rod whirled at 50 revolutions per second about one end, $mathscr E approx 10$ nV. Perhaps that's why we usually ignore it!
Answered by Philip Wood on August 14, 2021
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