I Contradiction in formula for motional EMF

AI Thread Summary
The discussion centers on the application of the motional EMF formula to a rotating disk, revealing a sign contradiction in the calculations. The expressions derived for the motional EMF yield opposite signs when integrating the velocity cross magnetic field and when applying the double integral approach. The source of confusion is identified as the orientation of the area element, where the downward orientation of the area should be considered for a counterclockwise rotation. This oversight leads to the realization that the sign discrepancy arises from the incorrect assumption about the area element's orientation. Ultimately, correcting the orientation resolves the contradiction in the calculations.
masteralien
Messages
36
Reaction score
2
TL;DR Summary
There seems to be a contradiction in the sign of the motional EMF for a spinning disk depending in the formula used
The formula for motional EMF is
$$\oint({\bf{v}}\times{\bf{B}})d{\bf{l}}=-\frac{d}{dt}\int{{\bf{B}}\cdot{\bf{\hat{n}}}da}$$However applying this for a rotating disk of radius a there seems to be a sign contradiction
$${\bf{v}}\times{\bf{B}}=\omega s{\bf{\hat{\varphi}}}\times B{\bf{\hat{z}}}=B\omega s {\bf{\hat{s}}}$$

$$\int^{a}_0{B\omega s}ds=\frac{1}{2}B\omega a^2$$Now doing it with the Double Integral by moving the derivative inside
$$
-\frac{d}{dt}\int^{2\pi}_0\int^{a}_0{Bsdsd\varphi}$$

$$\\\frac{d\varphi}{dt}=\omega$$

$$\\-\int^{a}_0{B\omega sds}=-\frac{1}{2}B\omega a^2$$

These expressions are similar but have the opposite sign why is this.

My question is why is there this contradiction here did I do something wrong like these formulas should be the same.
 
Last edited:
Physics news on Phys.org
I take your OP illustrated in https://www.feynmanlectures.caltech.edu/II_17.html as 

1702169651553.png


Where is the area a or da of your RHS in this figure ?
 
Last edited:
anuttarasammyak said:
I take your OP illustrated in https://www.feynmanlectures.caltech.edu/II_17.html as 

View attachment 336999

Where is the area a or da of your RHS in this figure ?
Ur right if the disk rotates counterclockwise da should be negative as the curve which goes around the Area should have a downward orientation should have caught that.
 
Thread 'Gauss' law seems to imply instantaneous electric field propagation'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Thread 'A scenario of non-uniform circular motion'
(All the needed diagrams are posted below) My friend came up with the following scenario. Imagine a fixed point and a perfectly rigid rod of a certain length extending radially outwards from this fixed point(it is attached to the fixed point). To the free end of the fixed rod, an object is present and it is capable of changing it's speed(by thruster say or any convenient method. And ignore any resistance). It starts with a certain speed but say it's speed continuously increases as it goes...
Maxwell’s equations imply the following wave equation for the electric field $$\nabla^2\mathbf{E}-\frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2} = \frac{1}{\varepsilon_0}\nabla\rho+\mu_0\frac{\partial\mathbf J}{\partial t}.\tag{1}$$ I wonder if eqn.##(1)## can be split into the following transverse part $$\nabla^2\mathbf{E}_T-\frac{1}{c^2}\frac{\partial^2\mathbf{E}_T}{\partial t^2} = \mu_0\frac{\partial\mathbf{J}_T}{\partial t}\tag{2}$$ and longitudinal part...
Back
Top