Electromotive force and trigonometry

[mcastillo356]

TL;DR

Contradictory result

Faraday's law:
\epsilon=-N\dfrac{\Delta{\phi}}{\Delta{t}}=-N\dfrac{\Delta{(BA\cos{\theta})}}{\Delta{t}}=-N\dfrac{\Delta{(BA\cos{(\omega t)})}}{\Delta{t}}
Applying calculus
\epsilon=NBA\omega\cos{(\omega t)}
Shouldn't it be \epsilon=NBA\omega\sin{(\omega t)}, just if I apply limits?
Thanks

 

[member 587159]

Assuming A,B don't depend on t, I think you are right!

 

[mcastillo356]

There is only one way: if t_0=NBA\omega. But this is not calculus. Or yes?

 

[mfb]

The point of t=0 is generally arbitrary, it doesn't matter if your angle follows a sine, a cosine, or anything else with a constant phase shift.

 

[mcastillo356]

Thanks, mfb!

 

[mcastillo356]

Thank's, Math_QED