Motional EMF induced between the ends of the rotating rod

[VVS2000]

TL;DR Summary

Suppose I have a rod which is rotating about a point on one of the edges of a rectangular region. A uniform magnetic field exists in the region and it is coming out of the plane of the region
How to find the induced emf between the ends of the rotating rod?
Picture given below

 

[ergospherical]

The equation $\mathbf{E}(\mathbf{r}) + \mathbf{v}(\mathbf{r}) \times \mathbf{B}(\mathbf{r})= \mathbf{0}$ holds everywhere in the rod; this is a statement that the Lorentz force is zero in the steady-state. This let's you deduce the electric field $\mathbf{E}(\mathbf{r}) = - \mathbf{v}(\mathbf{r}) \times \mathbf{B}(\mathbf{r})$ everywhere in the rod. The voltage between the ends is nothing but the line integral $\int \mathbf{E}(\mathbf{r}) \cdot d\mathbf{r}$ along the rod.

 

[VVS2000]

The equation $\mathbf{E}(\mathbf{r}) + \mathbf{v}(\mathbf{r}) \times \mathbf{B}(\mathbf{r})= \mathbf{0}$ holds everywhere in the rod; this is a statement that the Lorentz force is zero in the steady-state. This let's you deduce the electric field $\mathbf{E}(\mathbf{r}) = - \mathbf{v}(\mathbf{r}) \times \mathbf{B}(\mathbf{r})$ everywhere in the rod. The voltage between the ends is nothing but the line integral $\int \mathbf{E}(\mathbf{r}) \cdot d\mathbf{r}$ along the rod.

But your answer doesn't seem to depend on time because clearly, as P and Q(The ends) go in and out of the loop, is'nt there a change in the direction of induced emf?

 

[ergospherical]

as P and Q(The ends) go in and out of the loop, is'nt there a change in the direction of induced emf?

Yes!

 

[VVS2000]

Yes!

So the answer won't be an integral because of the discontinuity that exists everytime P and Q go in and out of the plane

 

[ergospherical]

That equation is true at any given time. The voltage will be $V$ until $\pi / \omega$, and then $-V$ until $2\pi / \omega$, and then...

 

[VVS2000]

That equation is true at any given time. The voltage will be $V$ until $\pi / \omega$, and then $-V$ until $2\pi / \omega$, and then...

Is there a way How you arrived at that answer? Intuitively, I can understand but can't do it mathematically

 

[Baluncore]

The EMF induced is proportional to the rate flux is cut by the conductive rod.

As the angular area is swept, the flux is cut at a constant rate, so the rod EMF V, will be a square wave, alternating between ±V . The polarity will flip every π radians.

The induced EMF will be proportional to flux density, multiplied by the rate of rotation, multiplied by half the length of the rod.

 

[ergospherical]

The induced EMF will be proportional to flux density, multiplied by the rate of rotation, multiplied by half the length of the rod.

The last dependence is quadratic, viz
\begin{align*}
V = -\int \mathbf{E}(\mathbf{r}) \cdot d\mathbf{r} = \int \mathbf{v}(\mathbf{r}) \times \mathbf{B}(\mathbf{r}) \cdot d\mathbf{r} = \omega B \int_0^{l/2} r dr = \dfrac{1}{8} \omega B l^2
\end{align*}

 

[VVS2000]

Thanks a lot everyone!