- #1
mcastillo356
Gold Member
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- TL;DR Summary
- Contradictory result
Faraday's law:
[tex]\epsilon=-N\dfrac{\Delta{\phi}}{\Delta{t}}=-N\dfrac{\Delta{(BA\cos{\theta})}}{\Delta{t}}=-N\dfrac{\Delta{(BA\cos{(\omega t)})}}{\Delta{t}}[/tex]
Applying calculus
[tex]\epsilon=NBA\omega\cos{(\omega t)}[/tex]
Shouldn't it be [tex]\epsilon=NBA\omega\sin{(\omega t)}[/tex], just if I apply limits?
Thanks
[tex]\epsilon=-N\dfrac{\Delta{\phi}}{\Delta{t}}=-N\dfrac{\Delta{(BA\cos{\theta})}}{\Delta{t}}=-N\dfrac{\Delta{(BA\cos{(\omega t)})}}{\Delta{t}}[/tex]
Applying calculus
[tex]\epsilon=NBA\omega\cos{(\omega t)}[/tex]
Shouldn't it be [tex]\epsilon=NBA\omega\sin{(\omega t)}[/tex], just if I apply limits?
Thanks