I Contradiction in formula for motional EMF

Click For 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
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 'Why higher speeds need more power if backward force is the same?'
Power = Force v Speed Power of my horse = 104kgx9.81m/s^2 x 0.732m/s = 1HP =746W Force/tension in rope stay the same if horse run at 0.73m/s or at 15m/s, so why then horse need to be more powerfull to pull at higher speed even if backward force at him(rope tension) stay the same? I understand that if I increase weight, it is hrader for horse to pull at higher speed because now is backward force increased, but don't understand why is harder to pull at higher speed if weight(backward force)...

Similar threads

Replies
1
Views
1K
Replies
6
Views
2K
Replies
11
Views
2K
Replies
4
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 3 ·
Replies
3
Views
406
  • · Replies 9 ·
Replies
9
Views
2K
Replies
1
Views
1K
  • · Replies 16 ·
Replies
16
Views
2K