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I have taken a look at this question: Emf generated by a rotating rod

But my question is different:

A metallic rod is being rotated in a uniform magnetic field with a constant angular velocity $\omega$. The axis of rotation of the rod is vertical, and passes through its midpoint. If the specific charge of free electrons in the rod is $\frac{e}{m}$, then for what value of the vector $\omega$ will the potential difference between the tip of the rod and the midpoint be zero, considering the magnetic field to be upwards?

I'm having trouble understanding the physics behind it, much like the other question I linked.

Here's my attempt at this question:

Considering that the rod is moving perpendicular to the direction of the magnetic field, there should be a force applied on the electrons present in the conducting rod. This force will eventually get cancelled out when the charges get accumulated at the end of the rod and at its midpoint, which will generate some electric field that cancels out the magnetic force.

Thus: $q((\omega$ x$)\times$B)+q(E$_{in}$)=0 Here, x is the distance from the midpoint.

When we solve this equation, using $\Delta$V=$-\int$E$\cdot$dx, we get $\Delta$V=$\frac{B\omega l^{2}}{2}$. That can't be zero for any value of $\omega$ except zero. But, the answer is not zero...

Please guide me...

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The pd set up by the redistribution of electrons would indeed cancel the magnetic force on the free electrons in the rod, preventing further motion (apart from random thermal). The same applies inside an open-circuit battery when charges have built up on its terminals.

Your question (and it's one I've not seen presented this way before) is exploring a different effect. For the magnetic force to urge the electrons along the rod, even before the opposing electric field has built up, the magnetic force first has to provide the centripetal force on the electrons. I'll now leave it up to you...

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