I Motional EMF induced between the ends of the rotating rod

AI Thread Summary
The equation E(r) + v(r) × B(r) = 0 indicates that the Lorentz force is zero in a steady-state rod, allowing for the deduction of the electric field E(r) = -v(r) × B(r). The voltage between the rod's ends is represented by the line integral ∫ E(r) · dr. As the ends P and Q move in and out of the loop, the induced EMF changes direction, resulting in a voltage that alternates between V and -V based on the angular position. The induced EMF is proportional to the flux density, the rate of rotation, and half the length of the rod, leading to a quadratic relationship with voltage. The discussion highlights the dynamic nature of induced EMF in relation to the rod's motion and magnetic field interactions.
VVS2000
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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
16382054268036981608784320922012.jpg
 
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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.
 
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ergospherical said:
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?
 
VVS2000 said:
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!
 
ergospherical said:
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
 
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...
 
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ergospherical said:
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
 
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.
 
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Baluncore said:
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*}
 
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Thanks a lot everyone!
 
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